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151.

A rod of mass `m` spins with an angular speed `omega=sqrt(g/l)`, Find its a. kinetic energy of rotation. b. kinetic energy of translation c. total kinetic energy.

Answer» Kinetic enegy of rotation `K_("rotational") = 1/2 I_(C)omega^(2)`
`=1/2 ((ml^(2))/12)(sqrt(g)/l)^(2)=1/24mlgl`
b Kinetic energy of translation, `K_("rotational")=1/2 mv_(C)^(2)`
`=1/2 m[(omegal/2)^(2)]=1/2 m[sqrt(g/l)xxl/2]^(2)=1/3mgl`
c. total kinetic energy
`K_("total")=K_("translational")+K_("rotational")`
`=1/6mgl=1/2I_(P)omega^(2)=1/2[(ml^(2))/2][sqrt(g/l)]^(2)=1/6mgl`
Discussion
152.

Calculate the moment of inertia of a uniform rod of mass `M` and length `l` about an axis passing through an end and perpendicular to the rod. The rod can be divided into a number of mass elements along the length of the rod.

Answer» Taking an element of mass `dm` at a distance `x` from the relation axis.
`I=int(dm)r^(2)-int_(0)^(1)(M/ldx)x^(2)=(Ml^(2))/3`
Discussion
153.

A circular disc `X` of radius `R` is made from an iron of thickness `t`, and another disc `Y` of radius `4R` is made from an iron plate of thickness `t//4`. Then the relation between the moment of mertia `I_(x)` and `I_(Y)` is :A. `I_(y)=32 I_(x)`B. `I_(y)=16 I_(x)`C. `I_(y)=I_(x)`D. `I_(y)=64I_(x)`

Answer» Correct Answer - D
Mass of disc (X), `mx=piR^(2)t rho`
Where `rho` =density of material of disc
`:. I_(x)=1/2 m_(x)R^(2)=1/2 R^(2)t rho R^(2)`
`I_(x)=1/2 pi rho R^(4)......(i)`
Again mass of disc (Y)
`m_(y)=pi=(4R)^(2)t/4 rho=4piR^(2) t rho`
and `I_(y)=1/2 m_(y)(4R^(2))=1/2 4piR^(2)t rho.16 R^(2)`
`rArr I_(y)=32pi t rho R^(4)......(ii)`
`:. (I_(y))/(I_(X))=(32pi t rho R^(4))/(1/2 pi rho t R^(4))`
`rArr =64`
`:. :. I_(Y)=64 I_(X)`
Discussion
154.

A conical pendulum consists of a simple pendulum moving in a horizontal circle as shown in the figure. `C` is the pivot, `O` the centre of the circle in which the pendulum bob moves and `omega` the constant angular velocity of the bob. If `vecL` is the angular momentum about point `C`, then A. `vecL` is constantB. only direction of `vecL` is constantC. only magnitude of `vecL` is constantD. none of the above

Answer» Correct Answer - C
The distance of `vecL` is perpendicular to the line joining the bob at point `C`. Since this line keeps changing its orientation is space, direction of `vecL` keeps changing, however, as `omega` is constant, magnitude of `vecL` remains constant.
Method 2 The torque about point is perpendicular to te angular momentum vector about point `C`. hence it can only change the directon of `L` and not its magnitude.
Discussion
155.

A man of mass `100 kg` stands at the rim of a turtable of radius `2 m` and moment of inertia `4000 kgm^(2)` mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man now walks along the outer edge of the turntable with a velocity of `1m//s` relative to the earth a. With what angular velocity and in what direction does the turntable rotate? b. Through what angle will it have rotated when the man reaches his initial position on the turntable? c. Through what angle will it have rotated when the man reaches his initial position relative to the earth?A. `36^@` in clockwise directionB. `36^@` in anticlockwise directionC. `72^@` in clockwise directionD. `72^@` in anticlockwise direction

Answer» Correct Answer - A
If the man completes one revolution relative to the earth, then
`theta_(m)=2pi`
`Time=(2pi)/theta_(m)=(2pi)/0.5`
During this time the angular displacement of the table
`theta_(t)=omega_(t)` (time)`=-0.05xx((2pi)/0.5)`
`=-pi/5`radians
`theta_(t)=36^@` in clockwise direction.
Discussion
156.

A man of mass `100 kg` stands at the rim of a turtable of radius `2 m` and moment of inertia `4000 kgm^(2)` mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man now walks along the outer edge of the turntable with a velocity of `1m//s` relative to the earth a. With what angular velocity and in what direction does the turntable rotate? b. Through what angle will it have rotated when the man reaches his initial position on the turntable? c. Through what angle will it have rotated when the man reaches his initial position relative to the earth?A. The table rotates anticlockwise (in the direction of the man motion) with angular velocity `0.05 rad//s`.B. The table rotates clockwise (opposite to the man) with angular velocity `0.1 rad//s`.C. The table rotates clockwise (opposite to the man) with angular velocity `0.05 rad//s`.D. The table rotates anticlockwise (in the direction of the man motion) with angular velocity `0.1 rad//s`

Answer» Correct Answer - C
By conservation of anglar mometum on the man table system,
`L_(i)=L_(f)`
`0+0=I_(m)omega_(m)=I_(t)omega_(t)`
`impliesomega_(t)=(I_(m)omega_(m))/I_(t)` where `omega_(m)=v/r=1/2rad//s`
`implies omega_(t)=-(100(2)^(2)xx(1/2))/4000`
`implies omega_(m)=v/r=1/2rad//s`
`implies omega_(t)=-1/20rad//s`
Thus, the table rotates clockwise (opposite to the man) with angular velocity `0.05 rad//s.`
Discussion
157.

A uniform rod of length `l` and mass 2 m rests on a smooth horizontal table. A point mass `m` moving horizontally at right angles to the rod with velocity `v` collides with one end of the rod and sticks it. ThenA. `1/6mv^(2)`B. `1/3mv^(2)`C. `mv^(2)`D. `(mv^(2))/5`

Answer» Correct Answer - A
Loss in kinetic energy of the system as a result of collision
`=K_("Initial")-K_("Final")=1/2mv^(2)-{1/2(m+2m)v_(CM)^(2)=1/2I_(CM)omega^(2)}`
`=1/2mv^(2)-{1/2(3m)(v/3)^(2)+1/2((ml^(2))/3)(v/l)}`
For the composite `I_(CM)=(ml^(2))/3=1/2mv^(2)-1/3m v^(2)=1/6mv^(2)`
Discussion
158.

`AB` is a horizontal diameter of a ball of mass `m=0.4 kg` and radius `R=0.10m`. At time `t=0`, a sharp impulse is applied at `B` at angle of `45^@` with the horizontal as shown in figure so that the ball immediately starts to move with velocity `v_(0)=10ms^(-1)` a. Calculate the impulse and angular velcity of ball just after impulse provided. If coefficient of kinetic friction between the floor and th ball is `mu=0.1`. Calculate. The b. velocity of ball when it stops sliding, c. time t at that instant d. horizontal distance travelled by the ball up to that instant. e. angular displacement of the ball about horizontal diameter perpendicular to `AB`, up to that instant, and f. energy lost due to friction (`g=10ms^(-2))`

Answer» Since, line of action of impulse does not pass through centre of mass of the sphere, therefore, just after application of impulse, the sphere starts to move, both translationally and rotationally. Translation motion is produced by horizontal component of the impulse, while rotational motion is produced by momnt of the impulse. Let the impulse applied be `J`.
Then its horizontal coponent provides horizontal motion.
`J.cos45^@` which gives `J=4sqrt(2)kgms^(-1)`
Moment of inertial of ball about centroidal axis is
`I=2/4mR^(2)=1.6xx10^(-30kgm^(2)`
Initial angular moment of ball (about centre) =`J(Rsin45^@)`
Angular impule will change angular momentum of the ball.
`J.R sin 45^@=/_L=(Iomega_(0)-0)`
Which gives `omega_(0)=250rads^(-1)`(clockwise)
Discussion
159.

A uniform disc of mass `m` and radius `R` has an additional rim of mass `m` as well as four symmetrically placed masses, each of mass `m//4` tied at positions `R//2` from the centre as shown in Fig. What is the total moment of inertia of the disc about an axis perpendicular to the disc through its centre?

Answer» `I=I_("disc")+I_("ring")+I_("point masses")`
`=(mR^(2))/2+mR^(2)+[m/4+(R/2)^(2)]=(7mR^(2))/4`
Discussion
160.

A particle performing uniform circular motion gas angular momentum `L`. If its angular frequency is double and its kinetic energy halved, then the new angular momentum is :A. 2LB. 4LC. L/2D. L/4

Answer» Correct Answer - D
Kinetic energy =`1/2 L omega`
`(K_(1))/(K_(2))=(L_(1)omega_(1))/(L_(2)omega_(2))`
`L_(2)=L//4`
Discussion
161.

A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The maximum value of `V_0` for whic the disk will roll without slipping is-A. `mugsqrt(M/k)`B. `mugsqrt(m/(2K)`C. `mugsqrt((3M)/k)`D. `mugsqrt((5M)/(2k))`

Answer» Correct Answer - C
From the previous question we can get
`f=(2kx)/3`
`f_("max")=muMg`
For disc not to slip `f_("max")gefimpliesxle3/2(muMg)/k`………i
From conservation of energy
`1/2(2k)x^(2)=1/2I_(CM)omega^(2)+1/2Mv_(0)^(2)`
`implies x=sqrt(3/4(MV_(0)^(2))/k)`
Putting the value of `x` in eqn i we get
`implies V_(0)lemugsqrt((3M)/k)`
Discussion
162.

A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The centre of mass of the disk undergoes simple harmonic motion with angular frequency `omega` equal to -A. `sqrt(k/M)`B. `sqrt((2k)/M)`C. `sqrt((2K)/(3M))`D. `sqrt((4K)/(3M))`

Answer» Correct Answer - D
From the previous question `a=-(4kx)/(3M)`
Comparing with `a=-omega^(2)x,` we get
`omega=sqrt((4k)/(3M))`
Discussion
163.

Two light vertical springs with equal natural length and spring constants `k_(1)` and `k_(2)` are separated by a distance `l`. Their upper end the ends `A` and `B` of a light horizontal rod `AB`. A vertical downwards force `F` is applied at point `C` on the rod. `AB` will remain horizontal in equilibrium if the distance `AC` is A. `l/2`B. `(lk_(1))/(k_(2)+k_(1))`C. `(lk_(2))/(k_(1))`D. `(lk_(2))/(k_(1)+k_(2))`

Answer» Correct Answer - D
AC=x
`x_(0)=ext^(0)`. In each spring
Torque about C
`K_(1)x_(0)x=K_(2)x_(0)(l-x)`
`x=(lk_(2))/(k_(1)+k_(2))`
Discussion
164.

A loop and a disc have same mass and roll without slipping with the same linear velocity `v`. If the total kinetic energy of the loop is `8 J`, the kinetic energy of the disc must be.

Answer» Correct Answer - 6
Total kinetic energy of loop `=1/2Iomega+1/2Mv^(2)`
`=1/2(MR^(2)omega^(2))+1/2Mv^(2)=Mv^(2)=8j(given)`
Total kinetic energy of disc `=1/2Iomega^(2)+1/2Mv^(2)`
`-=1/2(1/2MR^(2))((v^(2))/(R^(2)))+1/2Mv^(2)=3/4Mv^(2)=3/4(8)=6J`
Discussion
165.

A loop and a disc have same mass and roll without slipping with the same linear velocity `v`. If the total kinetic energy of the loop is `8 J`, the kinetic energy of the disc must be.A. `8J`B. `6J`C. `16J`D. `4J`

Answer» Correct Answer - B
Total kinetic energy of loop
`=1/2mv^(2)+1/2lomega^(2)`
`=1/2mv^(2)+1/2(mr^(2))(v^(2))/(r^(2))=mv^(2)=8`
Again total kinetic energy of disc
`=1/2mv^(2)+1/2lomega^(2)=1/2mv^(2)+1/2xx1/2mr^(2)xx(v^(2))/(r^(2))`
`=1/2mv^(2)+1/4mv^(2)=3/4mv^(2)=3/4xx8J=6J`
Discussion
166.

The disc of radius `r` is confined to roll without slipping at `A` and `B`. If the plates have the velocities shown, then A. angular velocity of the disc is `2V//r`B. linear velocity `V_(0)=V`C. angular velocity of the disc is `3V//2r`D. none of these

Answer» Correct Answer - A::B
`v_(0)=romega_(0)=3v`……i
`v_(0)+romega_(0)=-v`….ii
From eqn i and ii we get
`2v_(0)=2vimpliesv_(0)=v`
`v_(0)+v_(0)=-romega_(0)impliesomega=(2v)/r`
Discussion
167.

A uniform disc of mass `m` and radius `R` is rolling up a rough inclined plane which makes an angle of `30^@` with the horizontal. If the coefficients of static and kinetic friction are each equal to `mu` and the only force acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is and its direction is .(write up or down) the inclined plane.A. `(mg)/2`B. `(mg)/4`C. `(Mg)/6`D. `Mg`

Answer» Correct Answer - C
under the given condition only posibility is that friction is upward and it accelerated downwards as shown below:
the equation of motion are :
`a=(mg sin theta-1)/m=(mg sin 30^(@)-f)/m=g/2-1/m....(1)`
`alpha=(tau)/I=(fR)/I=(fR)/((mR^(2))/2)=(2f)/(mR)......(2)`
For rolling (no spelling) ltbrltgt `a=R alpha` or `g//2-f//m=2f//m`
`:. (3f)/m=g//2` or `f=mg//6`
Discussion
168.

Suppose you are standing on the edge of a spinning platform and step off at right angles to the edge (radially outward). Now consider it the other way. You are standing on the ground next to a spinning carousel and you step onto the platform at right angles to the edge (radially inward).A. There is no change in rotational speed of the carousel in either situation.B. There is a change in rotational speed in the first situation but not the second.C. There is a change in rotational speed in the second situation but not the first.D. There is a change in rotational speed in both instances

Answer» Correct Answer - C
In first case the impulse on the platform will act radially inward and it will not produce any angular impulse.
In second case, when man steps into platform, moment of inertia of system increases, so rotational speed decreases.
Discussion
169.

Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertias of the system about a line perpendicular to the plane of the square and passing through the centre of the square.

Answer» The perpendicular distance of every particle from the given line is `a//sqrt(2)`. The moment of inertia of one particle is therefore, `m(a//sqrt(2))^(2)=1/2ma^(2)`. The moment of inertia of the system is, therefore.
`4xx1/2ma^(2)=2ma^(2)`
Discussion