553 + Interview Questions in ROTATIONAL-MOTION IN GENERAL AWARENESS Page 8 InterviewSolution

351.	A spool has the shape shown in figure. Radii of inner and outer cylinders are R and 2 R respectively. Mass of the spool is `3 m` and its moment of inertia about the shown axis is `2 m R^(2)`. Light threads are tightly wrapped on both the cylindrical parts. The spool is placed on a rough surface with two masses `m_(1) = m` and `m_(2) = 2m` connected to the strings as shown. The string segment between spool and the pulleys `P_(1)` and `P_(2)` are horizontal. The centre of mass of the spool is at its geometrical centre. System is released from rest. (a) What is minimum value of coefficient of friction between the spool and the table so that it does not slip? (b) Find the speed of `m_(1)` when the spool completes one rotation about its centre.
Answer» Correct Answer - `mu_("min") = (1)/(8)`

Discussion

352.	Two cylinders having radii `2R` and `R` and moment of inertia `4I` and `I` about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity `omega_(0)`. The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this A. angular momentum of system is conservedB. kinetic energy is conservedC. neither the angular momentum nor the kinetic energy is conservedD. both the angular momentum and kinetic energy are conserved
Answer» Correct Answer - C

Discussion

353.	A ball is projected with velocityd of `20sqrt(2)` m/s at an angle of `45^(@)` with horizontal and at the same instant another plate is rotating with constant angular velocity `omega=(pi)/(4)` rad/sec in vertical plane as shown in the figure ( assume the length of plate is sufficient for collision to take place). If the mass of the plate is much larger than the mass of the ball, the plate is initially in horizontal position and collision is perfectly elastic, then choose the correct statement (s) A. Time when the ball collides with the plate is 2 (SI unit)B. velocity of the ball just after collision with the plate is `10(2+pi)`(SI unit)C. Distance of ball when it again collides on the ground surface from the projection point is `20pi`(SI unit)D. Maximum height achieved by the ball is 20 (SI unit)
Answer» Correct Answer - A::B::C::D

Discussion

354.	The particle of mass 1 kg is projected with velocity `20sqrt(2)`m/s at `45^(@)` with ground . When , the particle is at highest point `(g=10 m//s^(2))`,
Answer» Correct Answer - (A)Q,(B)Q,(C)S

Discussion

355.	A particle of mass `m` moves in the `XY` plane with a velocity `v` along the straight line `AB`. If the angular momentum of the particle with respect to origin `O` is `L_(A)` when it is at A and `L_(B)` when it is at B, then A. `L_(A) gt L_(B)`B. `L_(A)=L_(B)`C. the relationship between `L_(A)` and `L_(B)` depends upon the slope of the line `AB`D. `L_(A) lt L_(B)`
Answer» Correct Answer - b

Discussion

356.	A particle of mass `m` moves in the `XY` plane with a velocity `v` along the straight line `AB`. If the angular momentum of the particle with respect to origin `O` is `L_(A)` when it is at A and `L_(B)` when it is at B, then A. `L_(A) gt L_(B)`B. `L_(A) = L_(B)`C. The relationship between `L_(A)` and `L_(B)` depends upon the slop of the line `AB`D. `L_(A) lt L_(B)`
Answer» Drop a `bot^(ar)` from `O` to the line `AB`, `bot^(ar)` distance is same for positions `A` and `B`.

Discussion

357.	(i) A solid sphere of radius R is released on a rough horizontal surface with its top point having thrice the velocity of its bottom point A `(V_(A) = V_(0))` as shown in figure. Calculate the linear velocity of the centre of the sphere when it starts pure rolling. (ii) Solid sphere of radius R is placed on a rough horizontal surface with its centre having velocity `V_(0)` towards right and its angular velocity being `omega_(0)` (in anticlockwise sense). Find the required relationship between `V_(0)` and `omega_(0)` so that - (a) the slipping ceases before the sphere loses all its linear momentum. (b) the sphere comes to a permanent rest after some time. (c) the velocity of centre becomes zero before the spinning ceases.
Answer» Correct Answer - (i) `(12V_(0))/(7)` (ii) (a) `V_(0) gt (2)/(5) omega_(0) R` (b) `V_(0) = (2)/(5) omega_(0) R` (c) `V_(0) lt (2)/(5) omega_(0) R`

Discussion

358.	`O` is the centre of an equilateral triangle `ABC`. `F_(1), F_(2)` and `F_(3)` are the three forces acting along the sides `AB, BC` and `AC` respectively. What should be the value of `F_(3)` so that the total torque about `O` is zero? A. `F_(1)+F_(2)`B. `F_(1)-F_(2)`C. `(F_(1)+F_(2))/(2)`D. `2(F_(1)+F_(2))`
Answer» `tau_(0)=F_(1)r+F_(2)r-F_(3)r=0` `F_(3)=F_(1)+F_(2)`

Discussion

359.	Using theorem of parallel axes, calculate the moment of inertia of a disc of mass 400 g and radius 7 cm about an passing through its edge and perpendicular to its plane.
Answer» Correct Answer - `29400 g cm^(2)`

Discussion

360.	Four 2 kg masses are connecte dby `1.4` m spokes to an axle. A force of 24 N acts on a lever `1/2`m long to produce angular acceleration `alpha`. The magnitude of `alpha` (in rad `s^(-2)` is the angle between r and F is `30^(@)`.A. 24B. 12C. 6D. 3
Answer» b)As, force x distance = `rho = Ialpha` `rArr Fsin30^(@) xx 1/2 = 4[2 xx(1/4)^(2)]alpha` or `24 xx 1/2 xx 1/2 = alpha/2` `therefore` `alpha = 12rads^(-2)`

Discussion

361.	The angular displacement at any time t is given by `theta(t) = 2t^(3)-6^(2)`. The torque on the wheel will be zero atA. 1 sB. 0.1 sC. 2 sD. 0.2s
Answer» a) Given, `theta`(t) = 2t^(2)-6t^(2) rArr (dtheta)/(dt) = 6t^(2)-12t` `d^(2)theta)/(dt^(2)` = 12t -12 For torque, `rho = Ialpha`=0 `(d^(2)theta))/(dt^(2))` =0 rArr 12t^(2)-12 =0 `therefore` t=1s

Discussion

362.	If the equation for the displacement of a particle moving in a circular path is given by `(theta)=2t^(3)+0.5`, where `theta` is in radians and `t` in seconds, then the angular velocity of particle after `2 s` from its start isA. `8 rad//s`B. `12 rad//s`C. `24 rad//s`D. `36 rad//s`
Answer» `theta=2t^(3)+0.5` `omega=(d theta)/(dt)=6t^(2)` At `t=2 s, omega=6(2)^(2)=24 rad//s`

Discussion

363.	A couple produces.A. linear motionB. rotational motionC. translational motionD. oscillatory motion
Answer» Correct Answer - B

Discussion

364.	A couple produces.A. no motionB. linear and rorarional motionC. purely rotational motionD. purely linear motion
Answer» Correct Answer - c

Discussion

365.	A ball of mass `0.25kg` attached to the end of a string of length `1.96m` moving in a horizontal circle. The string will break if the tension is more than `25N`. What is the maximum speed with which the ball can be moved.A. `14 m//s`B. `3 m//s`C. `3.92 m//s`D. `5 m//s`
Answer» Correct Answer - a

Discussion

366.	A rod of negligble mass and length l is pioveted at its centre. A particle of mass m is fixed to its left end and another particle of mass 2 m is fixed to the right end. If the system is released from rest and after sometime becomes vericle, The speed v of the two mases and angular velocity at that instant are A. `sqrt(gl//3),sqrt(4g//3l`B. `sqrt(4gl//3),sqrt(4g//3l)`C. `sqrt(4gl//3),sqrt(4gl//3)`D. `sqrt(gl//3),sqrt(gl//3)`
Answer» Correct Answer - A

Discussion

367.	The M.I. of a thin uniform stick of mass 9 gm about an axis passing through one end perpendicular to the length of a meter stick isA. `90gm cm^(2)`B. `9 kg m^(2)`C. `3gm m^(2)`D. `9.8 kg m^(2)`
Answer» Correct Answer - C `I = (ml^(2))/(3)=(9xx1)/(3)=3 gm^(2)`.

Discussion

368.	When a perosn throws a meter stivk it is found that the centre of the stick is moving with speed `10 m//s` and left end stick with speed `20 m//s`. Both points move vertically upwards at that moment. Then angular speed the stick is :A. `20 rad//s`B. `10 rad//s`C. `30 rad//s`D. None of these
Answer» Correct Answer - A

Discussion

369.	A railway carriage has its CG at a height of 1 m above the rails , which are 1 m apart . Calculate the maximum safe speed at which it can travel round an unbanked curve of radius 80 m .
Answer» Correct Answer - `19.8 ms^(-1)`

Discussion

370.	A hexagonal pencil of mass M and sides length `a` has been placed on a rough incline having inclination `theta` Friction is large enough to prevent sliding. If at all the pencil moves, during one full rotation each of its 6 edges, in turn, serve as instantaneous axis of rotation. (a) Show that for `theta gt 30^(@)` the pencil cannot remain at rest. (b) For inclination of incline `theta lt 30^(@)` the pencil will not roll on its own. A sharp impulse J is given to the pencil parallel to the incline at its upper edge (see figure). Friction does not allow the pencil to slide but it begins to rotate about the edge through A with initial angular speed `omega_(0)`. Find `omega_(0)`. Moment of inertia of the pencil about its edge is I. (c) Find minimum value of J so that the pencil will turn about A, and B will land on the incline. (d) If kinetic energy acquired by the pencil just after the impulse is `K_(0)`, find its kinetic energy just before edge B lands on the incline
Answer» Correct Answer - (b) `omega_(0) = (sqrt(3)Ja)/(I)` (c) `omega = sqrt(2Mga)/(I)(1-cos(30^(@)-theta)))` (d) `K = K_(0) + Mga sin theta`

Discussion

371.	A small object of uniform density rolls up a curved surface with an initial velocity v. it reaches up to a maximum height of `(3v^2)/(4g) with respect to the initial position. The object isA. ringB. solid sphereC. hollow sphereD. disc
Answer» `(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))=mgh=mgxx(3v^(2))/(4g)=(3)/(4)mv^(2)` `1+(k^(2))/(R^(2))=(3)/(2)implies(k^(2))/(R^(2))=(1)/(2)`, Disc

Discussion

372.	A small object of uniform density rolls up a curved surface with an initial velocity v. it reaches up to a maximum height of `(3v^2)/(4g) with respect to the initial position. The object isA. ringB. solidC. hollow sphereD. disc
Answer» Correct Answer - D (d) By the concept of energy conservation `(1)/(2) mv^2 + (1)/(2)Iomega^2 = mg((3v^2)/(4g))` `:. (1)/(2)mv^2 + (1)/(2)I(v^2)/(R^2) = (3)/(4)mv^2 [:. V = R omega]` `:. (1)/(2)I (V^2)/(R^2) = (3)/(4)mv^2 - (1)/(2)mv^2 = (1)/(4)mv^2` `rArr I = (1)/(2)mR^2` This is the formula of the moment of inertia of the disc.

Discussion

373.	A small object of uniform density rolls up a curved surface with an initial velocity v. it reaches up to a maximum height of `(3v^2)/(4g) with respect to the initial position. The object isA. solid sphereB. hollow sphereC. discD. ring
Answer» Correct Answer - C By the law of conservation, `PE = KE_("rolling")` `mgh = (1)/(2)mv^(2)(1+(K^(2))/(R^(2)))` `g xx (3v^(2))/(4g)=(1)/(2)v^(2)(1+(K^(2))/(R^(2)))` `(3)/(2)=1+(K^(2))/(R^(2))` `therefore (K^(2))/(R^(2))=(3)/(2)-1=(1)/(2)` The value of `(K^(2))/(R^(2))` is `(1)/(2)` for disc.

Discussion

374.	Two discs of moments of inertia `I_(1) and I_(2)` about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds `omega_(1) and omega_(2)` are brought into contact face to face with their axes of rotation coincident. (a) Does the law of conservation of angular momentum apply to the situation ? Why ? (b) Find the angular speed of the two-disc system. (c ) Calculate the loss in kinetic energy of the system in the process. (d) Account for this loss.
Answer» (a) By the conservation of angular momentum `I_(1)omega_(1)+I_(2)omega_(2)=(I_(1)+I_(2))omega` `omega=(I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2))`, where `omega` is the final angular speed (b) Initial `K.E., K_(i)=(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2)omega_(2)^(2)` Final `K.E., K_(f)=(1)/(2)(I_(1)+I_(2))omega^(2)` `=(1)/(2)(I_(1)+I_(2))((I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2)))^(2)` `K_(i)=K_(f)=(1)/(2)(I_(1)I_(2))/(I_(1)+I_(2))(omega_(1)-omega_(2))^(2)=+ve, K_(i)gtK_(f)` `K_(i)-K_(f)=DeltaK:` loss in kinetic energy

Discussion

375.	Let `I_(A)` and `I_(B)` be moments of inertia of a body about two axes `A` and `B` respectively. The axis `A` passes through the centre of mass of the body but `B` does not. ThenA. `I_(A) lt I_(B)`B. If the axes are parallel, `I_(A) lt I_(B)`C. If the axes are parallel, `I_(A) = I_(B)`D. If the axes are not parallel, `I_(A) gt I_(B)`
Answer» The `M.I.` about an axis passing through the center of mass is maximum.

Discussion

376.	A semi-circular ring has mass m and radius R as shown in figure. Let `I_(1),I_(2) ,I_(3) " and "I_(4)` be the moments of inertia about the four axes as shown . Axis 1 passes through centre and is perpendicular to plane of ring. Then , match the following.
Answer» Correct Answer - (A)R,(B)P,(C)P,(D)Q

Discussion

377.	The moment of inertia of a thin uniform rod of mass `M` and length `L` about an axis passing through its mid-point and perpendicular to its length is`I_0`. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is.A. `I_(0)+ML^(2)//4`B. `I_(0)+2ML^(2)`C. `I_(0)+ML^(2)`D. `I_(0)+ML^(2)//2`
Answer» Correct Answer - a

Discussion

378.	The moment of inertia of a thin uniform rod of mass `M` and length `L` about an axis passing through its mid-point and perpendicular to its length is`I_0`. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is.A. `I_(0)+ML^(2)`B. `I_(0)+(ML^(2))/(2)`C. `I_(0)+(ML^(2))/(4)`D. `I_(0)+2ML^(2)`
Answer» `I_(end)=I_(c.m.)+Md^(2)=I_(0)+M((L)/(2))^(2)=I_(0)+(ML^(2))/(4)`

Discussion

379.	The moment of inertia of a thin rod of mass M and length L about an axis perpendicular to the rod at a distance L/4 from one end isA. `(19ML^(2))/(48)`B. `(38ML^(2))/(48)`C. `(7ML^(2))/(48)`D. `(ML^(2))/(12)`
Answer» Correct Answer - C `I_(0)=I_(c )+Mh^(2)=(ML^(2))/(12)+(ML^(2))/(16)` `=(4ML^(2)+3ML^(2))/(48)=(7ML^(2))/(48)`

Discussion

380.	If `I_(1),I_(2)` and `I_(3)` are the moments of inertia about the natural axies of solid sphere, hollow sphere and a spherical shell of same mass and radii, the correct result of the following isA. `I_(1)gt I_(2)gt I_(3)`B. `I_(3)gt I_(2) gt I_(1)`C. `I_(2)gt I_(1)gt I_(3)`D. `I_(2)=I_(3)gt I_(1)`
Answer» Correct Answer - D `I_(1)=(2)/(5)MR^(2), = 0.4` `I_(2)=(2)/(3)MR^(2),=0.6` `I_(3)=(2)/(3)MR^(2),=0.6`

Discussion

381.	A uniform wire has been bent in shape of a semi circle. The semicircle is suspended about a horizontal axis passing through one of its ends, so that the semicircular wire can swing in vertical plane. Find the angle `alpha` that the diameter of the semicircle makes with vertical in equilibrium.
Answer» Correct Answer - `alpha = tan^(-1) ((2)/(pi))`

Discussion

382.	Moment of inertia of a thin uniform rod about an axis passing through one end perpendicular to its length is I . Then moment of inertia the same rod about the central axis perpendicular to its plane isA. `I//4`B. 2IC. 4ID. 3 I
Answer» Correct Answer - A `I_(e )=I,I_(c )= ?` `(I_(e ))/(I_(c ))=(mL^(2))/(3)xx(12)/(ml^(2))=4` `I_(e )=4I_(c )` `therefore I_(c )=I//4`.

Discussion

383.	Two solid spheres of different materials have the same moments of inertia about their diameters. If `r_(1)` and `r_(2)` are their radii, ratio of their densities isA. `r_(1)^(3) : r_(2)^(3)`B. `r_(2)^(3):r_(2)^(3)`C. `r_(1)^(5):r_(2)^(5)`D. `r_(2)^(5):r_(1)^(5)`
Answer» Correct Answer - D `I_(1)=I_(2)` `(2)/(5)M_(1)R_(1)^(2)=(2)/(5)M_(2)R_(2)^(2)` `rho_(1)V_(1)R_(1)^(2)=rho_(2)V_(2)R_(2)^(2)` `rho_(1)(4pi)/(3)R_(1)^(3)R_(1)^(2)=rho (4pi)/(3)R_(1)^(3)R_(2)^(3)` `(rho_(1))/(rho_(2))=(R_(2)^(5))/(R_(1)^(5))`.

Discussion

384.	If a disc of mass m and radius r is reshaped into a ring a radius 2r,the mass remaining the same, the radius of gyration about centroidal axis perpendicular to plane goes up by a factor of `sqrt(x)` . Find the value of x.
Answer» Correct Answer - 8

Discussion

385.	A thin wire of length l and mass m is bent in the form of a semicircle as shown in the figure. Its moment of inertia about an axis joining its free ends will be A. `m l^(2)`B. zeroC. `m l^(2)// pi^(2)`D. `m l^(2)//2 pi^(2)`
Answer» Correct Answer - D `pi R = l therefore = (l)/(pi)` M..I. of ring about its diameter `=(1)/(2)MR^(2)` `therefore` MI of semicircle `=(1)/(2)m((l)/(pi))^(2)=(ml^(2))/(2pi^(2))`

Discussion

386.	A thin wire of mass m and length l is bent in the form of a ring . Moment of inertia of that ring about an axis passing through its centre and perpendicular to its through its centre and perpendicular to its plane isA. `ml^(2)//4pi`B. `ml^(2)//2pi`C. `ml^(2)//4pi^(2)`D. `ml^(2)//2pi^(2)`
Answer» Correct Answer - C `2pi r = l " " therefore r = (l)/(2 pi)` `I = mr^(2)` `= (ml^(2))/(4pi^(2))`. `I = mr^(2)` `= (ml^(2))/(4pi^(2))`.

Discussion

387.	The moment of inertia of a uniform rod about a perpendicular axis passing through one end is `I_(1)`. The same rod is bent into a ring and its moment of inertia about a diameter is `I_(2)`. Then `I_(1)//I_(2)` isA. `(pi)/(3)`B. `(8pi^(2))/(3)`C. `(5pi)/(3)`D. `(8pi^(2))/(5)`
Answer» Correct Answer - B `I=(mL^(2))/(3), I_(1)=(mR^(2))/(2)` `(I)/(I_(1))=(mL^(2))/(3)xx(2)/(MR^(2))` `= (2)/(3)xx(L^(2))/(R^(2))` `=(2)/(3)xx((2pi R^(2)))/(R^(2)) because L = 2 pi R` `= (2)/(3)xx (4pi^(2)R^(2))/(R^(2))` `(I)/(I_(1))=(8pi^(2))/(3)`.

Discussion

388.	The moment of inertia of a uniform rod about a perpendicular axis passing through one end is `I_(1)`. The same rod is bent into a ring and its moment of inertia about a diameter is `I_(2)`. Then `I_(1)//I_(2)` isA. `(pi^(2))/(3)`B. `(2pi^(2))/(3)`C. `(4pi^(2))/(3)`D. `(8pi^(2))/(3)`
Answer» `I_(1)=(ML^(2))/(3)` `L=2piRimplies R=(L)/(2pi)` `I_(2)=(1)/(2)MR^(2)=(1)/(2)M((L)/(2pi))^(2)=(ML^(2))/(8pi^(2))` `(I_(1))/(I_(2))=(8pi^(2))/(3)`

Discussion

389.	A thin wire of length L and uniform linear mass density `rho` is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX is` A. `(rhoL^3)/(8pi^2)`B. `rhoL^3)/(16pi^2)`C. `(5rhoL^3)/(16pi^2)D. `(3rhoL^3)/(8pi^2)`
Answer» Correct Answer - D (d) Moment of inertia about the diameter of the circular loop (ring) `= (1)/)2) MR^2` Using parallel axis theorem The moment of inertia of the loop about XX axis is `I_(XX) = (MR)^2/(2)+MR^2 = (3)/(2) MR^2` Where M = mass of the loop and R =radius of the loop Here `M = Lrho and R = (L)/(2pi),` `:.I_(XX) = (3)/(2)(Lrho)((L)/(2pi))^2 = (3L^3rho)/(8pi^2)`

Discussion

390.	A massless spool of inner radius `r` outer radius `R` is placed against a vertical wall and a titled split floor as shown. A light inextensible thread is tightly wound around the spool through which a mass `m` is hainging. There exists no friction at point `A`, while the coefficient of friction between the spool and point `B` is `mu`. The angle between the two surface is `theta` A. The magnitude of force on the spool at B in order ot maintain equilibrium is mg` sqrt(((r )/( R))^(2)+(1-( r)/( R))^(2)(1)/(tan^(2)theta))`B. The magnitude of force on the spool at B in order to maintain equilibrium is mg `(1-(r )/( R))(1)/( tan ^(2) theta)`C. The minimum value of `mu` for the system to ramain in equilibrium is `( cot theta)/((R//r)-1)`D. The minimum value of `mu` for the system to ramain in equilibrium is `( tan theta)/((R//r)-1)`
Answer» Correct Answer - A::D

Discussion

391.	A solid sphere is rolling without slipping on rough ground as shown in figure. If collides elastically with an identical another sphere at rest. There is no friction between the two spheres . Radius of each sphere is R and mass is m. What is the net angular impulse imparted to second sphere by the external forces?A. `(2)/(7)mRv`B. `(5)/(7)mRv`C. `(2)/(5)mRv`D. `(7)/(10)mRv`
Answer» Correct Answer - A

Discussion

392.	A plank `P` is placed on a solid cylinder `S`, which rolls on a horizontal surface. The two are of equal mass. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of `P` to the kinetic energy of `S` is: A. `1:1`B. `2:1`C. `8:3`D. `5:3`
Answer» Correct Answer - C

Discussion

393.	One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. it is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is A. `(1)/(2)MR^(2)`B. `(1)/(4)MR^(2)`C. `(1)/(8)MR^(2)`D. `sqrt2MR^(2)`
Answer» For the complete disc, `M.I.` about the axis `=(1)/(2)(4M)R^(2)` For the quarter disc, `M.I.=(1)/(4)[(1)/(2)(4M)R^(2)]` `=(1)/(2)MR^(2)`

Discussion

394.	A disc of mass 2 kg and diameter 40 cm is free to rotate about an axis passing through its centre and perpendicular to its plane. If a force of 50 N is applied to the disc tangentially Its angular acceleration will beA. `100 rad//s^(2)`B. `25 rad//s^(2)`C. `250 rad//s^(2)`D. `500 rad//s^(2)`
Answer» Correct Answer - C `tau = I alpha` `therefore alpha =(tau)/(I)=(rF)/(I)=(0.2xx50)/(mR^(2)/2)=(10xx2)/(2xx0.2xx0.2)` `= 250 rad//s^(2)`

Discussion

395.	The moment of inertia of the body about an axis is 1.2 kg `m^(2)`. Initially the body is at rest. In order to produce a rotational kinetic energy of 1500J, an angualr acceleration of 25 `rad/s^(2)` must be applied about the axis for the duration ofA. 2sB. 4sC. 8sD. 10 s
Answer» a) Rotational kinetic energy, `1/2 Iomega^(2)` According to question, `1/2Iomega^(2)`=1500 `1/2l(alphat)^(2) = 1500 ` `(1.2) xx (25)^(2) xx t^(2) = 3000` `1.2 xx 625 xx t^(2) = 3000` `t^(2) = (3000/(1.2 xx 625)` = 4 `t=2s`

Discussion

396.	If all a sudden the radius of the earth increases, thenA. the angular than that of the earth will be greater than of the sunB. the orbital speed of the earth will increaseC. the periodic time of the earth will increaseD. the energy and angular momentum will remain constant
Answer» Correct Answer - C

Discussion

397.	Three particles of masses 1 kg, 2 kg and 3 kg are at distance 1 m, 2 m and 3 m from the axis of rotation. The moment of intertia of the system isA. `24 kg m^(2)`B. `12 kg m^(2)`C. `36 kg m^(2)`D. `48 kg m^(2)`
Answer» Correct Answer - C `l = sum mr^(2)=1(1)^(2)+2(2)^(2)+3(3)^(2)` `= 1+8+27` `36 kg m^(2)`.

Discussion

398.	Mass of bigger disc having radius 2R is M. A disc of radius R is cut from bigger disc. Moment of intertia of disc about an axis passing through periphery and perpendicular to plane isA. `(27MR^(2))/8`B. `(29MR^(2))/8 `C. 3.5 MRD. 2 M`R^(2)`
Answer» b) Surface density of motional disc is `sigma = (M/(pi(2R)^(2)) = M/(4piR^(2))` Mass of cutting portion is `m_(1) = sigma xx piR^(2) = M/40` `I= I_(1)-I_(2)` Where, `I_(1)` = Moment of inertia of disc about given axis without cutting portion `I_(2)` = Moment of inertia due to cutting portion `I = (M(2R)^(2))/2 + M(2R)^(2) - [(m_(1)R^(2))/2 + m_(1)(3R)^(2)]` `6MR^(2) - (19MR^(2))/8 = (29MR^(2))/8`

Discussion

399.	A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing though O and P is `I_O and I_P` respectively. Both these axes are perpendiucalr to the plane of the lamina. The ratio `I_P/I_O` ot the nearest integer is
Answer» Correct Answer - C Let `sigma` be the surface mass density. Then `I_O = (1)/(2)sigma(2R)^2]xx(2R)^2 - ` `[(1)/(2)(sigmapiR^2)^2 + sigma(pi R^2)xxR^2]` `=(13)/(2) pi sigma^4` `I_P = 8pi sigma R^4 + sigma pi(2R)^2xx(2R)^2` `[(1)/(2) sigma(pi R^2)R^2+sigma(piR^2)(sqrt((2R)^2+R^2))^2]` `=24pi sigma R^4 - 5.5 sigma R^4 =18.5 pi sigma R^4` `:. (I_P)/(I_O) = (18.5pi sigmaR^4)/(13)/(2) pi sigmaR^4 = (37)/(13) ~~3`

Discussion

400.	Consider the situation as shown in the diagram. The pulley has radius `R` and moment of inertia `I`. The block is release when the spring (spring constant `k`) is unstretched. Find the speed of block when the vlock has fallen a distance `h`. Assume no slipping.
Answer» Decrease in the gravitation potential energy of block `=` gain in the kinetic energy of (block `+` pulley) `+` spring energy `mgh=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)+(1)/(2)kh^(2)` `omega=(v)/(R )` `mgh-(1)/(2)kh^(2)=(1)/(2)mv^(2)+(1)/(2)I((v)/(R))^(2)` `=(1)/(2)mv^(2)(1+(I)/(mR^(2)))` `v=sqrt((2mgh-kh^(2))/(m(1+(I)/(mR^(2)))))`

Discussion

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