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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Calculate the kinetic energy of a tractor crawler belt of mass `m` if the tractor moves with a velocity `v`. There is no slipping. Neglect the size of the wheels. |
| Answer» Since the lower part of the belt is in contact with the rigid floor, velocity of this part becomes zero. The crawler moves with velocity v, hence the velocity of upper part of the belt becomes `2v` by the rolling condition and kinetic energy of upper part `=1/2(m/2)(2v)^2=mv^2`, which is also the sought kinetic energy, assuming that the length of the belt is much larger than the radius of the wheels. | |
| 2. |
A uniform cylinder of radius `r` and mass `m` can rotate freely about a fixed horizontal axis. A thin cord of length l and mass `m_(0)` is would on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length `x` of the hanging part of the end. the wound part of the cord is supposed to have its centre of gravity on the cylinder axis is shown in figure. |
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Answer» Let us use the equation `(dM_z)/(dt)=N_z` relative to the axis through O (1) For this purpose, let us find the angular momentum of the system `M_z` about the given rotation axis and the corresponding torque `N_x`. The angular momentum is `M_z=Iomega+mvR=(m_0/2+m)R^2omega` [where `I=(m_0)/(2)R^2` and `v=omegaR` (no cord slipping)] So, `(dM_z)/(dt)=((MR^2)/(2)+mR^2)beta_z` (2) The downward pull of gravity on the overhanging part is the only external force, which exertes a torque about the z-axis, passing through O and is given by, `N_z=(m/l)xgR` Hence from the equation `(dM_z)/(dt)=N_z` `((MR^2)/(2)+mR^2)beta_z=m/lxgR` Thus, `beta_z=(2mgx)/(lR(M+2m))gt0` Note: We may solve this problem using conservation of mechanical energy of the system (cylinder + thread) in the uniform field of gravity. |
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| 3. |
A smooth light horizontal rod AB can rotate about a vertical axis passing through its end A. The rod is fitted with a small slieeve of mass m attached to the end A by a weightless spring of length `l_0` and stiffness `x`. What work must be performed to slowly get this system going and reaching the angular velocity `omega`? |
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Answer» Let the deformation in the spring be `Deltal`, when the rod AB has attained the angular velocity `omega`. From the second law of motion in projection form `F_n=mw_n`. `kDeltal=momega^2(l_0+Deltal)` or, `Deltal=(momega^2l_0)/(k-momega^2)` From the energy equation, `A_(ext)=1/2mv^2+1/2kDeltal^2` `=1/2momega^2(l_0+Deltal)^2+1/2kDeltal^2` `=1/2momega^2(l_0+(momega^2l_0)/(k-momega^2))^2+1/2k((momega^2l_0^2)/(k-momega^2))^2` On solving `A_(ext)=k/2(l_0^2eta(1+eta))/((1-eta)^2)`, where `eta=(momega^2)/(k)` |
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| 4. |
Two relativistic particles move at right angles to each other in a laboratory frame of reference, one with the velocity `v_1` and the other with the velocity `v_2`. Find their relative velocity. |
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Answer» The approach velocity is defined by `vecV_(approach)=(dvecr_1)/(dt)-(dvecr_2)/(dt)=V_1-vecV_2` in the laboratory frame. So `V_(approach)=sqrt(v_1^2+v_2^2)` On the other hand, the relative velocity can be obtained by using the velocity addition formula and has the componets `[-v_1, v_2sqrt(1-(v_1^2/c^2))]` so `V_r=sqrt(v_1^2+v_2^2-(v_1v_2^2)/(c^2))` |
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| 5. |
A point traversed half the distance with a velocity `v_0`. The remaining part of the distance was covered with velocity `v_1` for half the time, and with velocity `v_2` for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion. |
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Answer» Correct Answer - `< < v > > =2v_0(v_1+v_2)//(2v_0+v_1+v_2)` Let s be the total distance traversed by the point and `t_1` the time taken to cover half the distance. Further let `2t` be the time to cover the rest half of the distance. Therefore `s/2=v_0t_1` or `t_1=(s)/(2v_0)` (1) and `s/2=(v_1+v_2)t` or `2t=(s)/(v_1+v_2)` (2) Hence the sought average velocity `lt v gt =(s)/(t_1+2t)=(s)/([s//2v_0]+[s//(v_1+v_2)])=(2v_0(v_1+v_2))/(v_1+v_2+2v_0)` |
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| 6. |
The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as `T=as^2`, where a is ticle as a function of s. |
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Answer» We have `T=1/2mv^2=as^2` or, `v^2=(2as^2)/(m)` (1) Differentiating Eq. (1) with respect to time `2vw_t=(4as)/(m)v` or, `w_t=(2as)/(m)` (2) Hence net acceleration of the particle `w=sqrt(w_t^2+w_n^2)=sqrt(((2as)/(m))^2+((2as)/(mR))^2)=(2as)/(m)sqrt(1+(s//R)^2)` Hence the sought force, `F=mw=2assqrt(1+(s//R)^2)` |
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| 7. |
A locomotive of mass m starts moving so that its velocity varies according to the law `V=alphasqrts,` where `alpha` is a constant and s is the distance covered. Find the total work done by all the forces acting on the locomotive during the first t seconds after the beginning of motion. |
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Answer» Differentiating `v(s)` with respect to time `(dv)/(dt)=(a)/(2sqrts)(ds)/(dt)=(a)/(2sqrts)asqrts=a^2/2=w` (As locomative is in unidirectional motion) Hence force acting on the locomotive `F=mw=(ma^2)/(2)` Let, at `v=0` at `t=0` then the distance covered during the first t seconds `s=1/2wt^2=1/2a^2/2t^2=a^2/4t^2` Hence the sought work, `A=Fs=(ma^2)/(2)((a^2t^2))/(4)=(ma^4t^2)/(8)` |
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| 8. |
A solid body starts rotating about a stationary axis with an angular acceleration `alpha=(2.0xx10^(-2))trad//s^(2)` here `t` is in seconds. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle `theta=60^(@)` with its velocity vector? |
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Answer» Angle `alpha` is related with `|w_t|` and `w_n` by means of the formula: `tan alpha=(w_n)/(|w_t|)`, where `w_n=omega^2R` and `|w_t|=betaR` (1) where R is the radius of the circle which an arbitrary point of the body circumscribes. From the given equation `beta=(domega)/(dt)=at` (here `beta=(domega)/(dt)`, as `beta` is positive for all values of t) Integrating within the limit `underset(0)overset(omega)int domega=a underset(0)overset(t)int dt`, or, `omega=1/2at^2` So, `w_n=omega^2R=((at^2)/(2))^2 R=(a^2t^4)/(4)R` and `|w_t|=betaR=atR` Putting the values of `|w_t|` and `w_n` in Eq. (1), we get, `tan alpha=(a^2t^4R//4)/(atR)=(at^3)/(4)` or, `t=[(4/a)tanalpha]^(1//3)` |
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| 9. |
At the moment `t=0` a stationary particle of mass m experiences a time-dependent force `F=at(tau-t)`, whera a is a constant vector, `tau` is the time during which the given force acts. Find: (a) the momentum of the particle when the action of the force discontinued: (b) the distance covered by the particle while the force acted. |
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Answer» From the equation of the given time dependence force `vecF=vecat(pi-t)` at `t=pi`, the force vanishes, (a) Thus `Deltavecp=vecp=underset(0)overset(tau)int vecFdt` or, `vecp=underset(0)overset(tau)int vecat(tau-t)dt(vecatau^3)/(6)` but `vecp=mvecv` so `vecv=(vecatau^3)/(6m)` (b) Again from the equation `vecF=mvecw` `vecat(tau-t)=m(dvecv)/(dt)` or, `veca(ttau-t^2)dt=mdvecv` Integrating within the limits for `vecv(t)`, `underset(0)overset(t)int veca(t tau-t^2)dt=m underset(0)overset(vecv)intdvecv` or, `vecv=(veca)/(m)((taut^2)/(2)-(t^3)/(3))=(vecat^2)/(m)(tau/2-t/3)` Thus `v=(at^2)/(m)(tau/2-t/3)` for `tletau` Hence distance covered during the time interval `t=tau`, `s=underset(0)overset(tau)int v dt` `=underset(0)overset(tau)(at^2)/(m)(tau/2-t/3)dt=a/m(tau^4)/(12)` |
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| 10. |
A cannon and a target are `5.10 km` apart and located at the same level. How soon will the shell launched with the initial velocity `240 m//s` reach the target in the absence of air drag? |
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Answer» Correct Answer - `0.41` or `0.71min` later, depending on the initial angle. Total time of motion `tau=(2v_0sin alpha)/(g)` or `sin alpha=(taug)/(2v_0)=(9*8tau)/(2xx240)` (1) and horizontal range `R=v_0cos alpha tau` or `cos alpha=(R)/(v_0tau)=(5100)/(240tau)=(85)/(4tau)` (2) From Eqs. (1) and (2) `((9*8)^2tau^2)/((480)^2)+((85)^2)/((4tau^2)^2)=1` On simplifying `tau^4-2400tau^2+1083750=0` Solving for `tau^2` we get: `tau^2=(2400+-sqrt(1425000))/(2)=(2400+-1194)/(2)` Thus `tau=42.39s=0.71min` and `tau=24.55s=0.41min` depending on the angle `alpha`. |
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| 11. |
At the moment `t=0` a particle leaves the origin and moves in the positive direction of the x-axis. Its velocity varies with time as `v=v_0(1-t//tau)`, where `v_0` is the initial velocity vector whose modulus equals `v_0=10.0cm//s`, `tau=5.0s`. Find: (a) the x coordinate of the particle at the moments of time `6.0`, `10`, and `20s`, (b) the moments of time when the particles is at the distance `10.0cm` from the origion, (c) the distance s covered by the particle during the first `4.0` and `8.0s`, draw the approximate plot `s(t)`. |
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Answer» Correct Answer - (a) `x=v_0t(1-t//2pi)`, `x=0.24`, 0 and `-4.0m`; (b) `1.1, 9` and `11s`; (c) `{((1-t//2tau)v_0t, for, t, <=, tau),([1+(1-t//pi)^2]v_0t//2, for, t, >=,tau):}` 24 and 34cm respectively. (a) As the particle leaves the origin at `t=0` So, `Deltax=x-int v_xdt` (1) As `vecv=vec(v_0)(1-t/tau)`, where `vec(v_0)` is directed towards the `+ve` x-axis So, `v_x=v_0(1-t/tau)` (2) From (1) and (2), `x=underset0oversettintv_0(1-t/tau)dt=v_0t(1-(t)/(2pi))` (3) Hence x coordinate of the particle at `t=6s`. `x=10xx6(1-(6)/(2xx5))=24cm=0*24m` Similary at `x=10xx10(1-(10)/(2xx5))=0` and at `x=10xx20(1-(20)/(2xx5))=-200cm=-2m` (b) At the moments the particle is at a distance of `10cm` from the origin, `x=+-10cm`. Putting `x=+10` in Eq. (3) `10=10t(1-(t)/(10))` or, `t^2-10t+10=0`, So, `t=t=(10+-sqrt(100-40))/(2)=5+-sqrt(15)s` Now putting `x=-10` in Eqn (3) `-10=10(1-t/10)`, On solving, `t=5+-sqrt(35)s` As t cannot be negative, so, `t=(5+sqrt(35))s` Hence the particle is at a distance of `10cm` from the origin at three moments of time: `t=5+-sqrt(15)s, 5+sqrt(35)s` (c) we have `vecv=vec(v_0)(1-t/tau)` So, `v=|vecv|={:(v_(0)(1-(t)/(tau))),(v_(0)((t)/(tau)-1)):}}{:("for t" le tau),("for t" gt tau):}` So `s=underset0oversettintv_0(1-t/tau)dt` for `tletau=v_0t(1-t//2tau)` and `s=underset0oversettauintv_0(1-t/2)dt+undersettauoversettintv_0(t/tau-1)dt` for `tgttau` `=v_0tau[1+(1-t//tau)^2]//2` for `tgttau` (A) `s=underset0overset4intv_0(1-t/tau)dt=underset0overset4int10(1-t/5)dt=24cm`. And for `t=8s` `s=underset0overset5int10(1-t/5)dt+underset5overset8int10(t/5-1)dt` On integrating and simplifying, we get `s=34cm`. On the basis of Eqs. (3) and (4), `x(t)` and `s(t)` plots can be drawn as shown in the answer sheet. |
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| 12. |
A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to `r_1` and `r_2` respectively. Find the angular momentum M of this planet relative to the centre of the Sun. |
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Answer» As the planet is under central force (gravitational interaction), its angular momentum is conserved about the Sun (which is situated at one of the focii of the ellipse) So, `mv_1r_1=mv_2r_2` or, `v_1^2=(v_2^2r_2^2)/(r_1^2)` (1) From the conservation of mechanical energy of the system (Sun+planet), `-(gammam_sm)/(r_1)+1/2mv_1^2=-(gammam_sm)/(r_2)+1/2mv_2^2` or, `-(gammam_s)/(r_1)+1/2v_2^2(r_2^2)/(r_1^2)=-((gammam_s)/(r_2))+1/2v_2^2` [Using (1)] Thus, `v_2=sqrt(2gammam_sr_1//r_2(r_1+r_2))` (2) Hence `M=mv_2r_2=msqrt(2gammam_sr_1r_2//(r_1+r_2))` |
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| 13. |
A balloon starts rising from the surface of the Earth. The ascertion rate is constant and equal to `v_0`. Due to the wind the balloon |
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Answer» Correct Answer - (a) `x=(a//2v_0)y^2`; (b) `w=av_0`, `w_tau=a^2y//sqrt(1+(ay//v_0)^2)`, `w_n=av_0//sqrt(1+(ay//v_0)^2)`. According to the problem (a) `(dy)/(dt)=v_0` or `dy=v_0dt` Integrating `underset0oversetyintdy=v_0underset0overset tint dt` or `y=v_0t` (1) And also we have `(dx)/(dt)=ay` or `dx=aydt=av_0tdt` (using 1) So, `underset(0)overset(x)intdx=av_0underset0oversettint t dt`, or, `x=1/2av_0t^2=1/2(ay^2)/(v_0)` (using 1) (b) According to the problem `v_y=v_0` and `v_x=ay` (2) So, `v=sqrt(v_x^2+v_y^2)=sqrt(v_0^2+a^2y^2)` Therefore `w_t=(dv)/(dt)=(a^2y)/(sqrt(v_0^2+ay^2))(dy)/(dt)=(a^2y)/(sqrt(1+(ay//v_0)^2)` Diff. Eq. (2) with respect to time. `(dv_y)/(dt)=w_y=0` and `(dv_x)/(dt)=w_x=a(dy)/(dt)=av_0` So, `w=|w_x|=av_0` Hence `w_n=sqrt(w^2-w_t^2)=sqrt(a^2v_0^2-(a^4y^2)/(1+(ay//v_0)^2))=(av_0)/(sqrt(1+(ay//v_0)^2)` |
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| 14. |
A planet A moves along an elliptical orbit around the Sun. At the moment when it was at the distance `r_0` from the Sun its velocity was equal to `v_0` and the angle between the radius vector `r_0` and the velocity vector `v_0` was equal to `alpha`. Find the maximum and minimum distances that will separate this planet from the Sun during its orbital motion. |
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Answer» From the conservation of angular momentum about the Sun. `mv_0r_0sin alpha=mv_1r_1=mv_2r_2` or, `v_1r_1=v_2r_2=v_0r_2sinalpha` (1) From conservation of mechanical energy, `1/2mv_0^2-(gammam_sm)/(r_0)=1/2mv_1^2-(gammam_sm)/(r_1)` or, `v_0^2/2-(gammam_s)/(r_0)=(v_0^2r_0^2sin^2alpha)/(2r_1^2)-(gammam_s)/(r_1)` (Using 1) or, `(v_0^2-(2gammam_s)/(r_0))r_1^2+2gammam_sr_1-v_0^2r_0^2sin alpha=0` So, `r_(1) =(-2gammam_(s)+-sqrt(4gamma^(2)m_(s)^(2)+4(v_(0)^(2)r_(0)^(2)sin^(2)alpha)(v_(0)^(2)-(2gammam_(s))/(r_(0)))))/(2(v_(0)^(2)-(2gammam_(s))/(r_(0))))` `=(1+-sqrt(1-(v_0^2r_0^2sin^2alpha)/(gammam_s)(2/r_0-(v_0^2)/(rm_s))))/((2/r_0-(v_0)/(gammam_s)))=(r_0[1+-sqrt(1-(2-eta)etasin^2alpha)])/((2-eta))` where `eta=v_0^2r_0//gammam_s`, (`m_s` is the mass of the Sun). |
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| 15. |
The density of a stationary body is equal to `rho_0`. Find the velocity (relative to the body) of the reference frame in which the density of the body is `eta=25%` greater than `rho_0`. |
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Answer» We define the density `rho` in the frame K in such a way that `rhodxdydz` is the rest mass `dm_0` of the element. That is `rhodxdydz=rho_0dx_0dy_0dz_0`, where `rho_0` is the proper density `dx_0`, `dy_0`, `dz_0` are the dimensions of the element in the rest frame `K_0`. Now `dy=dy_0`, `dz=dz_0`, `dx=dx_0sqrt(1-v^2/c^2)` if the frame K is moving with velocity, v relative to the frame `K_0`. Thus `rho=(rho_0)/(sqrt(1-v^2/c^2))` Defining `eta` by `rho=rho_0(1+eta)` We get `1+eta=(1)/(sqrt(1-v^2/c^2))` or , `v^2/c^2=1-(1)/((1+eta)^2)=(eta(2+eta))/((1+eta)^2)` or `v=csqrt((eta(2+eta))/((1+eta)^2))=(csqrt(eta(2+eta)))/(1+eta)` |
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| 16. |
A shell flying with velocity `v=500m//s` burts into three identical fragements so that the kinetic energy of the system increases `eta=1.5` times. What maximum velocity can one of the fragments obtain? |
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Answer» From the conservation of linear momentum of the shell just before and after its fragementation `3vecv=vecv_1+vecv_2+vecv_3` (1) where `vecv_1`, `vecv_2` and `vecv_3` are the velocities of its fragments. From the energy conservation `3etav^2=v_1^2+v_2^2+v_3^2` (2) Now `overset~vecv_i` or `vecv_(iC)=vecv_i-vecv_C=vecv_i-vecv` (3) where `vecv_C=vecv`=velocity of the C.M. of the fragements the velocity of the shell. Obviously in the C.M. frame the linear momentum of a system is equal to zero, so `overset~vecv_1+overset~vecv_2+overset~vecv_3=0` (4) Using (3) and (4) in (2), we get `3etav^2=(vecv+overset~vecv_1)^2+(vecv+overset~vecv_2)^2+(vecv-overset~vecv_1-overset~vecv_2)^2=3v^2+2overset~v_1^2+2overset~v_2^2+2overset~vecv_1*overset~vecv_2` or, `2overset~v_1^2+2overset~v_1overset~v_2costheta+2overset~v_2^2+3(1-eta)v^2=0` (5) If we have had used `overset~vecv_2=-overset~vecv_1-overset~vecv_3`, then Eq. 5 were contain `overset~v_3` instead of `overset~v_2` and so on. The problem being symmetrical we can look for the maximum of any one. Obviously it will be the same for each. For `overset~v_1` to be real in Eq. (5) `4overset~v_2^2cos^2thetage8(2oversetv_2^2+3(1-eta)v^2)` or `6(eta-1)v^2ge(4-cos^2theta)overset~v_2^2` So, `overset~v_2levsqrt((6(eta-1))/(4-cos^2theta))` or `overset~v_(2(max))=sqrt(2(eta-1))v` Hence `v_(2(max))=|vecv+overset~vecv_2|_(max)=v+sqrt(2(eta-1))v=v(1+sqrt(2(eta-1)))=1km//s` Thus owing to the symmetry `v_(1(max))=v_(2(max))=v_(3(max))=v(1+sqrt(2(eta-1)))=1km//s` |
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| 17. |
On the pole of the Earth a body is imparted velocity `v_0` directed vertically up. Knowing the radius of the Earth and the free-fall acceleration on its surface, find the height to which the body will ascend. The air drag is to be neglected. |
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Answer» Let the mass of the body be m and let it go upto a height h. From conservation of mechanical energy of the system `-(gammaMm)/(R)+1/2mv_0^2=(-gammaMm)/((R+h))=0` Using `(gammaM)/(R^2)=g`, in above equation and on solving we get, `h=(Rv_0^2)/(2gR-v_0^2)` |
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| 18. |
A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion. What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to? |
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Answer» Velocity of the body at height h, `v_h=sqrt(2g(H-h))`, horizontally (from the figure given in the problem). Time taken in falling through the distance h. `t=sqrt((2h)/(g))` (as initial vertical component of the velocity is zero.) Now `s=v_ht=sqrt(2g(H+h))xxsqrt((2h)/(g))=sqrt(4(Hh-h^2))` For `s_(max)`, `(d)/(ds)(Hh-h^2)=0`, which yields `h=H/2` Putting this value of h in the expression obtained for s, we get, `s_(max)=H` |
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| 19. |
A body of mass m is pushed with the initial velocity `v_0` up an inclined plane set at an angle `alpha` to the horizontal. The friction coefficient is equal to k. What distance will the body cover before it stops and what work do the friction forces perform over this distance? |
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Answer» Let s be the sought distance, then from the equation of increment of M.E. `DeltaT+DeltaU=A_(f r)` `(0-1/2mv_0^2)+(+mgssin alpha)=-kmg cos alphas` or, `s=(v_0^2)/(2g)//(sin alpha+kcos alpha)` Hence `A_(f r)=-kmgcos alphas=(-kmv_0^2)/(2(k+tan alpha))` |
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| 20. |
At the equator a stationary (relative to the Earth) body falls down from the height `h=500m`. Assuming the air drag to be negligible, find how much off the vertical, and in what direction, the body will deviate when it hits the ground. |
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Answer» Here `v_y=0` so we can take `y=0`, thus we get for the motion in the x-y plane. `underset(..)x=-2omegav_zcostheta` and `overset(..)z=-g` Integrating, `z=-1/2g t^2` `overset(.)x=omegagcosvarphit^2` So `x=1/3omegagcosvarphit^3=1/3omegagcosvarphi((2h)/(g))^(3//2)` `=(2omegah)/(3)cosvarphisqrt((2h)/(g))` There is thus a displacement to the east of `2/3xx(2pi)/(8)64xx500xx1xxsqrt((400)/(9*8))~~26cm`. |
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| 21. |
A wheel rotates around a stationary axis so that the rotation angle `theta` varies with time as `theta=at^(2)` where `a=0.2rad//s^(2)`. Find the magnitude of net acceleration of the point A at the rim at the moment `t=2.5s` if the linear velocity of the point A at this moment is `v=0.65m//s`. |
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Answer» Differentiating `varphi(t)` with respect to time `(dvarphi)/(dt)=omega_z=2at` (1) For fixed axis rotation, the speed of the point A: `v=omegaR=2a tR` or `R=(v)/(2at)` (2) Differenciating with respect to time `w_t=(dv)/(dt)=2aR=v/t`, (using 1) But `w_t=v^2/R=(v^2)/(v//2at)=2atv` (using 2) So, `w=sqrt(w_t^2+w_n^2)=sqrt((v//t)^2+(2atv)^2)` `=v/tsqrt(1+4a^2t^4)` |
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| 22. |
A horizontal smooth rod AB rotates with a constant angular velocity `omega=2.00rad//s` about a vertical axis passing through its end A. A freely sliding sleeve of mass `m=0.50g` moves along the rod from the point A with the initial velocity `v_0=1.00m//s`. Find the Coriolis force acting on the sleeve (in the reference frame fixed to rotating rod) at the moment when the sleeve is located at the distance `r=50cm` from the rotation axis. |
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Answer» The sleeve is free to slide along the rod AB. Thus only the centrifugal force acts on it. The equation is, `moverset(.)v=momega^2r` where `v=(dr)/(dt)`. But `overset(.)v=v(dv)/(dr)=(d)/(dr)(1/2v^2)` so, `1/2v^2=1/2omega^2r^2+const ant` or, `v^2=v_0^2+omega^2r^2` `v_0` being the initial velocity when `r=0`. The Coriolis force is then, `2momegasqrt(v_0^2+omega^2r^2)=2momega^2rsqrt(1+v_0^2//omega^2r^2)` `=2*83N` on putting the values.s |
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| 23. |
A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle `varphi` as `omega=omega_0-avarphi`, where `omega_0` and a are positive constants. At the moment `t=0` the angle `varphi=0`. Find the time dependence of (a) the rotation angle, (b) the angular velocity. |
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Answer» We have `omega=omega_0-avarphi=(dvarphi)/(dt)` Integration this Eq. within its limit for `(varphi)t` `underset(0)overset(varphi)int(dvarphi)/(omega_0-kvarphi)=underset0oversett int dt` or, `1n(omega_0-k varphi)/(omega_0)=-kt` Hence `varphi=(omega_0)/(k)(1-e^(-kt))` (1) (b) From the Eq., `omega=omega_0-kvarphi` and Eq. (1) or by differentiating Eq. (1) `omega=omega_0e^(-kt)` |
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| 24. |
A horizontally oriented uniform disc of mass `M` and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which can slide without friction a small body of mass `m`. A light thread running down through the hollow axle of the disc is tied to the body initially the body was located at the edge of the disc and the whole system rotated with and angular velocity `omega_(0)`. Then by means of a force `F` applied to the lower and of the thread the body was slowly pulled to the rotation axis. find: (a). The angular velocity of the system in its final state. (b). The work performed by the force F. |
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Answer» (a) As force F on the body is radial so its angular momentum about the axis becomes zero and the angular momentum of the system about the given axis is conserved. Thus `(MR^2)/(2)omega_0+momega_0R^2=(MR^2)/(2)omega` or `omega=omega_0(1+(2m)/(M))` (b) From the equation of the increment of the mechanical energy of the system: `DeltaT=A_(ext)` `1/2(MR^2)/(2)omega^2-1/2((MR^2)/(2)+mR^2)omega_0^2=A_(ext)` Putting the value of `omega` from part (a) and solving we get `A_(ext)=(momega_0^2R^2)/(2)(1+(2m)/(M))` |
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| 25. |
Find how the momentum of a particle of rest mass `m_0` depends on its kinetic energy. Calculate the momentum of a proton whose kinetic energy equals `500MeV`. |
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Answer» From the formula `E=(m_0c^2)/(sqrt(1-v^2/c^2)), p=(m_0v)/(sqrt(1-v^2//c^2))` we find `E^2=c^2p^2+m_0^2c^4` or `(m_0c^2+T)^2=c^2p^2+m_0^2c^4` or `T(2m_0c^2+T)=c^2p^2` i.e. `p=1/2sqrt(T(2m_0c^2+T))` |
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| 26. |
What work has to be performed in order to increase the velocity of a particle of rest mass `m_0` from `0.60c` to `0.80c`? Compare the result obtained with the value calculated from the classical formula. |
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Answer» The work done is equal to change in kinetic energy which is different in the two cases Classically i.e. in nonrelativistic mechanics, the change in kinetic energy is `1/2m_0c^2((0*8)^2-(0*6)^2)=1/2m_0c^20*28=1*14m_0c^2` Relativistically it is, `(m_0c^2)/(sqrt(1-(0*8)^2))-(m_0c^2)/(sqrt(1-(0*6)^2))=(m_0c^2)/(0*6)-(m_0c^2)/(0*8)=m_0c^2 (1*666-1*250)` `=0*416m_0c^2=0*42m_0c^2` |
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| 27. |
A neutron with kinetic energy `T=2m_0c^2`, where `m_0` is its rest mass, strikes another, stationary, neutron. Determine: (a) the combined kinetic energy `overset~T` of both neutrons in the frame of their centre of inertia and the momentum `overset~p` of each neutron in that frame, (b) the velocity of the centre of inertia of this system of particles. Instruction. Make use of the invariant `E^2-p^2c^2` remaining constant on transition from one inertial reference frame to another (E is the total energy of the system, p is its composite momentum). |
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Answer» (b) & (a) In the CM frame, the total momentum is zero. Thus `V/c=(cp_(1x))/(E_1+E_2)=(sqrt(T(T+2m_0c^2)))/(T+2m_0c^2)=sqrt((T)/(T+2m_0c^2))` where we have used the result of problem Then `(1)/(sqrt(1-V^2//c^2))=(1)/(sqrt(1-(T)/(T+2m_0c^2)))=sqrt((T+2m_0c^2)/(2m_0c^2))` Total energy in the CM frame is `(2m_0c^2)/(sqrt(1-V^2//c^2))=2m_0c^2sqrt((T+2m_0c^2)/(2m_0c^2))=sqrt(2m_0c^2(T+2m_0c^2))=overset~T+2m_0c^2` So `overset~T=2m_0c^2(sqrt(1+(T)/(2m_0c^2))-1)` Also `2sqrt(c^2overset~p^2+m_0^2c^4)=sqrt(2m_0c^2(T+2m_0c^2)), 4c^2overset~p^2=2m_0c^2T`, or `overset~p=sqrt(1/2m_0T)` |
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| 28. |
A thin uniform rod AB of mass `m=1.0kg` moves translationally with acceleration `w=2.0m//s^2` due to two antiparallel forces `F_1` and `F_2` (figure). The distance between the points at which these forces are applied is equal to `a=20cm`. Besides, it is known that `F_2=5.0N`. Find the length of the rod. |
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Answer» Since, motion of the rod is purely translational, net torque about the C.M. of the rod should be equal to zero. Thus `F_1l/2=F_2(l/1-a)` or, `F_1/F_2=1-(a)/(l//2)` (1) For the translational motion of rod. `F_2-F_1=mv_e` or `1-F_1/F_2=(mw_e)/(F_2)`(2) From (1) and (2) `(a)/(l//2)=(mw_c)/(F_2)` or, `l=(2aF_2)/(mw_c)=1m` |
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| 29. |
A spaceship of mass `m_0` moves in the absence of external forces with a constant velocity `v_0`. To change the motion direction, a jet engine is switched on. It starts ejecting a gas jet with velocity u which is constant relative to the spaceship and directed at right angles to the spaceship motion. The engine is shut down when the mass of the spaceship decreases to m. Through what angle `alpha` did the motion direction of the spaceship deviate due to the jet engine operation? |
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Answer» As `vecF=0`, from the equation of dynamics of a body with variable mass, `m(dvecv)/(dt)=vecu(dm)/(dt)` or, `dvecv=vecu(dm)/(m)` (1) Now `dvecvuarrdarrvecu` and since `vecu_|_vecv`, we must have `|dvecv|=v_0dalpha` (because `v_0` is constant) where `dalpha` is the angle by which the spaceship turns in time `dt`. So, `-u(dm)/(m)=v_0dalpha` or, `dalpha=-(u)/(v_0)(dm)/(m)` or, `alpha=-(u)/(v_0)underset(m_0)overset(m)int(dm)/(m)=u/v_(0) 1n(m_0/m)` |
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| 30. |
A ship moves with velocity `v=36km` per hour along an arc of a circle of radius `R=200m`. Find the moment of the gyroscopic forces exerted on the bearings by the shaft with a flywheel whose moment of inertia relative to the rotation axis equals `I=3.8*10^3 kg*m^2` and whose rotation velocity `n=300rpm`. The rotation axis is oriented along the length of the ship. |
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Answer» The revolutions per minute of the flywheel being n, the angular momentum of the flywheel is `lxx2pin`. The rate of precession is `v/R` Thus `N=2piINV//R=5*97kN.m`. |
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| 31. |
The side wall of a wide vertical vessel of height `h=75 cm` has a narrow slit (vertical) running all the way down to the bottom of the vessel. The length of the slit is `l=50 cm` and the width is `b=1 mm`. With the slit closed, water is filled to the top. Find the resultant reaction force of water coming out as the slit is opened. |
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Answer» Consider an element of height `dy` at a distance `y` from the top. The velocity of the fluid coming out of the element is `v=sqrt(2gy)` The force of reaction `dF` due to this is `dF=rhodAv^2`, as in the previous problem, `=rho(bdy)2gy` Integrating `F=rhogbunderset(h-1)overset(h)int2ydy` `=rhogb[h^2-(h-l)^2]=rhogbl(2h-l)` (The slit runs from a depth `h-l` to a depth h from the top.) |
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| 32. |
A cylindrical vessel of height h and base area S is filled with water. An orifice of area `s lt lt S` is opened in the bottom of the vessel. Neglecting the viscosity of water, determine how soon all the water will pour out of the vessel. |
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Answer» Let at any moment of time, water level in the vessel be H then speed of flow of water through the orifice, at that moment will be `v=sqrt(2gH)` (1) In the time interval `dt`, the volume of water ejected through orifice, `dV=svdt` (2) On the other hand, the volume of water in the vessel at time t equals `V=SH` Differentiating (3) with respect to time, `(dV)/(dt)=S(dH)/(dt)` or `dV=SdH` (4) Eqs. (2) and (4) `SdH=svdt` or `dt=S/s(dH)/(sqrt(2gH))` from (2) Integrating `underset(0)overset(t)intdt=(S)/(ssqrt(2h))underset(h)overset(0)int(dh)/(sqrtH)` Thus, `t=S/ssqrt((2h)/(g))` |
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| 33. |
Find the maximum power which can be transmitted by means of a steel shaft rotating about its axis with an angular velocity `omega=120rad//s`, if its length `l=200cm`, radius `r=1.50cm`, and the permissible torsion angle `varphi=2.5^@`. |
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Answer» The maximum power that can be transmitted by means of a shaft rotating about its axis is clearly `Nomega` where N is the moment of the couple producing the maximum permissible torsion `varphi`. Thus `P=(pir^4Gvarphi)/(2l)*omega=16*9kw` |
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| 34. |
Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance r from its axis. The length of the rod is equal to l, the torsion angle to `varphi`. |
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Answer» The energy between radii r and `r+dr` is, by differentiation, `(pir^3dr)/(l)Gvarphi^2` Its density is `(pir^3dr)/(2pirdrl)(Gvarphi^2)/(l)=1/2(Gvarphi^2r^2)/(l^2)` |
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| 35. |
Find the elastic deformation energy of a steel rod of mass `m=3.1kg` stretched to a tensile strain `epsilon=1.0*10^-3`. |
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Answer» We assume that the deformation is wholly due to external load, neglecting the effect of the weight of the rod. Then a well known formula says, elastic energy per unit volume `=1/2stressxxstrai n=1/2sigmaepsilon` This gives `1/2m/rhoEepsilon^2~~0*04kJ` for the total deformation energy. |
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| 36. |
Find the volume density of the elastic deformation energy in fresh water at the depth of `h=1000m`. |
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Answer» The energy density is as usual `1//2 stressxxstrai n`. Stress is the pressure `rhogh`. Strain is `betaxxrhogh` by defination of `beta`. Thus `u=1/2beta(rhogh)^2=23*5kJ//m^3` on putting the values. |
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| 37. |
Calculate the torque N twisting a steel tube of length `l=3.0m` through an angle `varphi=2.0^@` about its axis, if the inside and outside diameters of the tube are equal to `d_1=30mm` and `d_2=50mm`. |
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Answer» Clearly `N=underset(d_1//2)overset(d_z//2)int(2pir^3drvarphiG)/(l)=(pi)/(32l)Gvarphi(d_2^4-d_1^4)` using `G=81GPa=8*1xx10^(10)N/m^2` `d_2=5xx10^-2m, d_1=3xx10^-2m` `varphi=2*0^@=(pi)/(90)` radius, `l=3m` `N=(pixx8*1xxpi)/(32xx3xx90)(625-81)xx10^2N*m` `=0*5033xx10^3N*m~~0*5k N*m` |
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| 38. |
Find the elastic deformation energy of a steel rod whose one end is fixed and the other is twisted through an angle `varphi=6.0^@`. The length of the rod is equal to `l=1.0m`, and the radius to `r=10mm`. |
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Answer» When the rod is twisted through an angle `theta`, a couple `N(theta)=(pir^4G)/(2l)theta` appears to resist this. Work done in twisting the rod by an angle `varphi` is then `underset(0)overset(varphi)intN(theta)d theta=(pir^4G)/(4l)varphi^2=7J` on putting the values. |
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| 39. |
A small body of mass m is located on a horizontal plane at the point O. The body of mass m is located on a horizontal plane at the point O. The body acquires a horizontal velocity `v_0`. Find, (a) the mean power developed by the friction force during the whole time of motion, if the friction coefficient `k=0.27`, `m=1.0kg`, and `v_0=1.5m//s`, (b) the maximum instantaneous power developed by the friction force, if the friction coefficient varies as `k=alphax`, where `alpha` is a constant, and x is the distance from the point O. |
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Answer» Let the body m acquire the horizontal velocity `v_0` along positive x-axis at the point O. (a) Velocity of the body t seconds after the beginning of the motion, `vecv=vecv_0+vecwt=(v_0-kg t)veci` (1) Instantaneous power `P=vecF*vecv=(-kmgveci)*(v_0-kgt)veci=-kmg(v_0-kgt)` From Eq. (1), the time of motion `tau=v_0//kg` Hence sought average power during the time of motion `lt P gt =(underset(0)overset(tau)int-kmg(v_0-kgt)dt)/(tau)=-(kmgv_0)/(2)=-2W` (On substitution) From `F_x=mw_x` `-kmg=mw_x=mw_x(dv_x)/(dx)` or, `v_xdv_x=-kgdx=-agxdx` To find `v(x)`, let us integrate the above equation `underset(v_0)overset(v)intv_xdv_x=-alphagunderset(0)overset(x)intxdx` or, `v^2=v_0^2-alphagx^2` (1) Now, `vecP=vecF*vecv=-malphaxgsqrt(v_0^2-alphagx^2)` (2) For maximum power, `(d)/(dt)(sqrt(v_0^2x^2-lambdagx^4))=0` which yields `x=(v_0)/(sqrt(2alphag))` Putting this value of x, in Eq. (2) we get, `P_(max)=-1/2mv_0^2sqrt(alphag)` |
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| 40. |
A steel plate of thickness h has the shape of a square whose side equals l, with `h lt lt l`. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration `beta` (figure). Find the deflection `lambda`, assuming the sagging to be small. |
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Answer» The deflection of the plate can be noticed by going to a co-rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance `xi` from the axis is `a=xibeta` and the moment of the force exerted by the section between x and l is `N=rholhbetaunderset(x)overset(l)intxi^2dxi=1/3rholhbeta(l^3-x^3)`. From the fundamental equation `EI(d^2y)/(dx^2)=1/3rholhbeta(l^3-x^3)`. The moment of inertia `I=underset(-h//2)overset(+h//2)intz^2ldz=(lh^3)/(12)`. Note that the natural surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and z is perpendicular to it. `(d^2y)/(dx^2)=(4rhobeta)/(Eh^2)(l^3-x^3)`. Integrating `(dy)/(dx)=(4rhobeta)/(Eh^2)(l^3x-x^4/4)+c_1` Since `(dy)/(dx)=0`, for `x=0`, `c_1=0`. Integrating again, `y=(4rhobeta)/(Eh^2)((l^3x^2)/(2)-(x^5)/(20))+c_2` `c_2=0` because `y=0` for `x=0` Thus `lambda=y(x=l)=(9rhobetal^5)/(5Eh^2)` |
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| 41. |
A solid copper cylinder of length `l=65cm` is placed on a horizontal surface and subjected to a vertical compressive force `F=1000N` directed downward and distributed uniformly over the end face. What will be the resulting change of the volume of the cylinder in cubic millimeters? |
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Answer» Volume of a solid cylinder `V=pir^2l` So, `(DeltaV)/(V)=(pi2rDeltarl)/(pir^2l)+(pir^2Deltal)/(pir^2l)=(2Deltar)/(r)+(Deltal)/(l)` (1) But longitudinal strain `Deltal//l` and accompanying lateral strain `Deltar//r` are related as `(Deltar)/(r)=-mu(Deltal)/(l)` (2) Using (2) in (1), we get: `(DeltaV)/(V)=(Deltal)/(l)(1-2mu)` (3) But `(Deltal)/(l)=(-F//pir^2)/(E)` (Because the increment in the length of cylinder `Deltal` is negative) So, `(DeltaV)/(V)=(-F)/(pir^2E)(1-2mu)` Thus, `DeltaV=(-Fl)/(E)(1-2mu)` Negative sign means that the volume of the cylinder has decreased. |
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| 42. |
A paricle of mass m moves along a circle of radius R with a normal acceleration varying with time as `w_n=at^2`, where a is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power averaged over the first t seconds after the beginning of motion. |
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Answer» We have `w_n=v^2/R=at^2`, or, `v=sqrt(aR)t`, t is defined to start from the beginning of motion from rest. So, `w_t=(dv)/(dt)=sqrt(aR)` Instantaneous power, `P=vecF*vecv=m(w_thatu_t+w_nhatu_t)*(sqrt(aR)thatu_t)`, (where `hatu_t` and `hatu_t` are unit vectors along the direction of tangent (velocity) and normal respectively) So, `P=mw_tsqrt(aR)t=maRt` Hence the sought average power `lt P gt =(underset(0)overset(t)intPdt)/(underset(0)overset(t)intdt)=(underset(0)overset(t)intmaRtdt)/(t)` Hence `lt P ge(maRt^2)/(2t)=(maRt)/(2)` |
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| 43. |
A particle moves in the xy-plane with constant acceleration `a` directed along the negative y-axis. The equation of path of the particle has the form `y= bx - cx^2`, where b and c are positive constants. Find the velocity of the particle at the origin of coordinates. |
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Answer» Correct Answer - `v_0=sqrt((1+a^2)w//2b)`. According to the problem `vecw=w(-vecj)` So, `w_x=(dv_x)/(dt)=0` and `w_y=(dv_y)/(dt)=-w` (1) Differentiating Eq. of trajectory, `y=ax-bx^2`, with respect to time `(dy)/(dt)=(adx)/(dt)-2bx(dx)/(dt)` (2) `(dy)/(dt):|:_(x=0)=a(dx)/(dt):|_(x=0)` Again differentiating with respect to time `(d^2y)/(dt^2)=(ad^2x)/(dt^2)-2b((dx)/(dt))^2-2bx(d^2x)/(dt^2)` or, `-w=a(0)-2b((dx)/(dt))^2-2bx(0)` (using 1) or, `(dx)/(dt)=sqrt((w)/(2b))` (using 1) Using (3) in (2) `(dy)/(dt):|_(x=0)=asqrt((w)/(2b))` Hence, the velocity of the particle at the origin `v=sqrt(((dx)/(dt))_(x=0)^(2)+((dy)/(dt))_(x=0)^2)=sqrt((w)/(2b)+a^2(w)/(2b))` (using Eqns (3) and (4)) Hence, `v=sqrt((w)/(2b)(1+a^2))` |
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| 44. |
A light non-stretchable thread is wound on a massive fixed pulley of radius R. A small body of mass m is tied to the free end of the thread. At a moment `t=0` the system is released and starts moving. Find its angular momentum relative to the pulley axle as a funciton of time t. |
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Answer» On the given system the weight of the body m is the only force whose moment is effective about the axis of pulley. Let us take the sense of `omega` of the pulley at an arbitrary instant as the positive sense of axis of rotation (z-axis) As `M_z(0)=0`, so, `DeltaM_z=M_z(t)=intN_zdt` So, `M_z(t)=underset(0)overset(t)intmgRdt=mgRt` |
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| 45. |
Having gone through a plank of thickness h, a bullet changed its velocity from `v_0` to `v`. Find the time of motion of the bullet in the plank, assuming the resistance force to be proportional to the square of the velocity. |
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Answer» According to the problem `m(dv)/(dt)=-kv^2`or, `m(dv)/(v^2)=-kdt` Integrating, within the limits, `underset(v_0)overset(v)int(dv)/(v^2)=-k/m underset(0)overset(t)intdt` or, `t=m/k((v_0-v))/(v_0v)` (1) To find the value of k, rewrite `mv(dv)/(ds)=-kv^2`, or, `(dv)/(v)=-k/mds` On integrating `underset(v_0)overset(v)int(dv)/(v)=-k/m underset(0)overset(h)intds` So, `k=m/h1n(v_0)/(v)` (2) Putting the value of k from (2) and (1), we get `t=(h(v_0-v))/(v_0v1nv_0/v)` |
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| 46. |
Suppose we have made a model of the Solar system scaled down in the ratio `eta` but of materials of the same mean density as the actual materials of the planets and the Sun. How will the orbital periods of revolution of planetary models change in this case? |
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Answer» `T=2piR^(3//2)//sqrt(gammam_s)` If the distance are scaled down, `R^(3//2)` decreases by a factor `eta^(3//2)` and so does `m_s`. Hence T does not change. |
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| 47. |
Three forces are applied to a square plate as shown in Figure. Find the modulus, direction, and the point of application of the resultant force, if this point is taken on the side BC. |
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Answer» For coplanar forces, about any point in the same plane, `sumvecr_ixxvecF_i=vecrxxvecF_(n et)` (where `vecF_(n et)=sum vecF_i` = resultant force) or, `vecN_(n et)=vecrxxvecF_(n et)` Thus length of the arm, `l=(N_(n et))/(F_(n et))` Here obviously `|vecF_(n et)|=2F` and it is directed toward right along AC. Take the origin at C. Then about C, `vecN=(sqrt2aF+a/sqrt2F-sqrt2aF)` directed normally into the plane of figure. Here `a=` side of the square.) Thus `vecN=F(a)/(sqrt2)` directed into the plane of the figure. Hence `l=(F(a//sqrt2))/(2F)=(a)/(2sqrt2)=a/2sin45^@` Thus the point of application of force is at the mid point of the side BC. |
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| 48. |
A system consists of two springs connected in series and having the stiffness coefficients `k_1` and `k_2`. Find the minimum work to be performed in order to stretch this system by `Deltal`. |
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Answer» Since the springs are connected in series, the combination may be treated as a single spring of spring constant. `k=(k_1k_2)/(k_1+k_2)` From the equation of increment of `M.E.`, `DeltaT+DeltaU=A_(ext)` `0+1/2kDeltat^2=A`, or, `A=1/2((k_1k_2)/(k_1+k_2))Deltal^2` |
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| 49. |
At the moment `t=0` a particle of mass m starts moving due to a force `F=F_0 cos omegat`, where `F_0` and `omega` are constants. How long will it be moving until it stops for the first time? What distance will it traverse during that time? What is the maximum velocity of the particle over this distance? |
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Answer» According to the problem, the force acting on the particle of mass m is, `vecF=vecF_0 cos omegat` So, `m(dvecv)/(dt)=vecF_0cos omegat` or `dvecv=(vecF_0)/(m)cos omega tdt` Integrating, within the limits. `underset(0)overset(vecv_0)int dvecv=(vecF_0)/(m)underset(0)overset(t)intcos omegadt` or `vecv=(vecF_0)/(momega)sin omegat` It is clear from equation (1), that after starting at `t=0`, the particle comes to rest from the first time at `t=pi/omega`. From Eqs. (1), `v=|vecv|=(F_0)/(momega)sin omegat` for `tlepi/omega` (2) Thus during the time interval `t=pi//omega`, the sought distance `s=(F_0)/(mw)underset(0)overset(pi//omega)int sin omega t dt=(2F)/(momega^2)` From Eq. (1) `v_(max)=(F_0)/(momega)` as `|sin omega t|le1` |
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| 50. |
The potential energy of a particle in a certain field has the form `U=a//r^2-b//r`, where a and b are positive constants, r is the distance from the centre of the field. Find: (a) the value of `r_0` corresponding to the equilibrium position of the particle, examine where this position is steady, (b) the maximum magnitude of the attraction force, draw the plots `U(r)` and `F_r(r)` (the projections of the force on the radius vector r). |
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Answer» From the equation `F_r=-(dU)/(dr)` we get `F_r=[-(2a)/(r^3)+(b)/(r^2)]` (a) we have at `r=r_0`, the particle is in equilibrium position. i.e. `F_r=0` so, `r_0=(2a)/(b)` To check, whether the position is steady (the position of stable equilibrium), we have to satisfy `(d^2U)/(dr^2)gt0` We have `(d^2U)/(dr^2)=[(6a)/(r^4)-(2b)/(r^3)]` Putting the value or `r=r_0=(2a)/(b)`, we get `(d^2U)/(dr^2)=(b^4)/(8a^3)`, (as a and b are positive constant) So, `(d^2U)/(dr^2)=(b^2)/(8a^3)gt0`, which indicates that the potential energy of the system is minimum, hence this position is steady: (b) We have `F_r=-(dU)/(dr)=[-(2a)/(r^3)+(b)/(r^2)]` For F, to be maximum, `(dF_r)/(dr)=0` So, `r=(3a)/(b)` and then `F_(r(max))=(-b^3)/(27a^2)`, As `F_r` is negative, the force is attractive. |
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