Rotation- IIT JEE Exam

Rotational Mechanics Page 1ROTATIONAL MECHANICS

Rigid BodyRIGID BODY: is an idealization of a solid body of finite size in which deformation is

neglected. In other words, the distance between any two given points of a rigid body

remains constant in time regardless of external forces exerted on it.

All bodies will be henceforth assumed to be rigid, if not mentioned.

RIGID BODY ROTATING ABOUT A FIXED AXIS:In this section, we treat the rotation of a rigid object about a fixed axis, which is commonly referred to as pure rotational motionWe know from our study of circular motion:Δθ=ωo t+

12α t2

ω=ωo+αt

ω2=ωo2+2αΔθ

v⃗=d r⃗dt

=Rω e⃗t∨¿ v∨¿ Rω

a⃗=d v⃗dt

=−ω2R e⃗r+Rα e⃗ t

Centre of mass of rigid body

For a continuous(discrete) distribution of mass, the expression for the centre of mass of a

collection of particles is expressed in the form of an integral as

xCM=∫0

M

xdm

M

or xCM=∑mi xiM

For the case of a uniform rod this becomes

xCM=∫0

L

xMLdx

M= L

2

This example of a uniform rod previews some common features about the process of finding

the centre of mass of a continuous body. Continuous mass distributions require calculus

methods involving an integral over the mass of the object. Such integrals are typically

transformed into spatial integrals by relating the mass to a distance, as with the linear EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

dxy

xO

ry

rxO

Rotational Mechanics Page 2density M/L of the rod. Exploiting symmetry can give much information: e.g., the centre of

mass will be on any rotational symmetry axis. The use of symmetry would tell you that the

centre of mass is at the geometric centre of the rod without calculation. (For details, refer to

the previous material)The integration is to be performed under proper limits so that, as the integration variable goes through the limits, the smaller elements cover the entire body.For rigid bodies having regular geometrical shapes and uniform distribution of mass, the centre of mass is at their geometrical centre. E.g. Uniform hollow sphere, uniform solid sphere, uniform circular ring, uniform circular disc and uniform rod have centre of mass at their centre.For a plane lamina, it is the point of intersection of the diagonalsFor triangular plane lamina, it is the point of intersection of medians of the triangle.For rectangular or cubical block, it is the point of intersection of diagonals.

Example-1: Find the position of the centre of mass of a uniform semi-circular lamina, radius

r.

Solution We know that the centre of mass lies somewhere on the axis of symmetry of

the semi-circle, although where on the axis of symmetry we do not know.

We therefore divide the semi-circle into many tiny strips, each with

thickness dx.

Each strip is a distance of x from the y-axis. The mass of each strip is its

volume × density. The volume of each strip is just its area, since it is a

lamina and so has no depth. The area of each strip is dx × 2y .

We are told that the body is uniform, which means that the density is

constant. It doesn't matter what the density is, lets call it ρ. So the mass

of each strip is 2y ρ dx.

Now the total area of the semicircle is ½ πr2. So the total mass is ½ πr2 ρ.

(distance of centre of mass from O) × (weight of body) = the sum of: (the mass of each

particle) × (the distance of each particle from O)

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Rotational Mechanics Page 3Therefore: (centre of mass)= ∫0

r

2 yρdx ×x

½ π r2ρ

.

The integral is with respect to x, so we must replace the y by a function of x.

By thinking about similar triangles, we can deduce that y/r = x/r, so y = x

Therefore:

(Centre of mass)= ∫0r

2 yxρdx

½ π r2 ρ

(Centre of mass) =∫

0

r

2x2 ρdx

½ π r2 ρ

(Centre of mass) = [ 2r3 ρ3 ]

½π r2 ρ= 4 r

3π

So centre of mass is a distance of 4r/3π from O, on the axis of

symmetry.

Try this! Figure shows a circular metal plate of radius 2R from which a disk

if radius R has been removed. Let us call it object X. Locate the center of

mass of object X. Ans: R/3

TORQUE:Torque is a measure of how much a force acting on an object causes that object to rotate.Consider a particle P at a position vector r from a fixed point O. Let F be the force acting on the particle P. Then we define ‘the TORQUE of the force F about O’ asΓ⃗=r⃗ × F⃗Torque Γ is a vector quantity having its direction perpendicular to both r and F.EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 4TORQUE EQUATION:Let us consider a rigid object of arbitrary shape rotating about a fixed axis, as shown in the above Figure. The object can be regarded as an infinite number of mass elements dm of infinitesimal size. Each mass element dm rotates in a circle about the origin O, and each has a tangential acceleration at produced by an external tangentialForce dFt . From Newton’s 2nd lawdFt = (dm)atThe torque dΓ associated with the force dFt acts about the origin and is given bydΓ= r dFt = (r dm)atBecause at = rα, the expression for dΓ becomesdΓ=(r dm)rα= (r2 dm)αTo get the total Torque on the rigid body about the axis O we have to integrate the above equation.∫dΓ= ∫ (r2 dm) αSo, Γ = Iα Where I is called the moment of inertia of the rigid body about the axis O.CALCULATION OF MOMENT OF INERTIA OF DIFFERENT OBJECTS:

Moment of Inertia of a Rod about the central axis perpendicular to it! : Let λ = m/l be the mass per unit length of the rod mass m and length l.This integral will be given by ∫dm.x2 over the whole mass. dm = λ dx. Hence the integral becomes ∫ λ dx.x2 Which is given by λx3/3 limit –l/2 to l/2 = λ(l/2)3/3 - λ(-l/2)3/3 = λl3/12 .(λ l=m.)EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 5 Thus, moment of Inertia of a rod about its central axis is given by ml2/12

Moment of Inertia of a ring about the central axis

perpendicular to it:Elementary mass is dm dx can vary from 0 to 2π This integral will be given by ∫dm.x2 over the whole mass. x is constant. Since x=r .∫dm.x2 = x2∫dm = mx2 Thus, moment of Inertia of a ring about its central axis is given by mr2 Moment of Inertia of a Disc about the central axis perpendicular to it : An elementary mass of radius r, thickness dr is taken Elementary mass is dm dm=(2πr dr). λ .This integral will be given by ∫dm.r2 over the whole mass∫dm.r2 = ∫2πrdrλr2 = ∫2πλr3 dr = 2πλR4 /4 (Limit was from 0 to R) πλR2 =M Thus, moment of Inertia of a disc about its central axis is given by MR2/2

IMPORTANT THEOREMS ON MOMENT OF INERTIA:

PARALLEL AXIS THEOREM: Suppose we have to obtain the moment of inertia ‘I’ of a body of mass M about a given axis z.Let CM be the centre of mass of the body and I0 is the moment of inertia about the axis passing through CM and parallel to EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 6z. Let the distance between the two axes OCM be ‘d’. The parallel axis theorem states that I=I0+Md2

PERPENDICULAR AXIS THEOREM:Perpendicular axis theorem is applicable only for plane bodies.Let X and Y-axes be chosen in the plane of the body and Z axis perpendicular to this plane as shown in the above figure.Then the theorem states that Iz=Ix+Iy EXAMPLE 2 :Two blocks having masses 2kg and 3kg are connected

to each other by a light cord that passes over a Disc, having a ,mass

M=2 kg and and radius R=1m,as shown in Figure. Find the

acceleration of each block and the tensions T1 and T2?

SOLUTION: Here note that the tensions T1 and T2 are not same as the pulley is not massless. The pulley is rotating with the torque produced by the difference in T1 and T2. For 3Kg block 30 – T2 = 3a For 2Kg block T1 – 20 = 2a For Disc (T2 – T1)R = Iα (T2 – T1) = (2/2)(a/R) as I = MR2/2 and α = a/R T2 – T1 = a Solving the three equations we get a = 5/3; T2 = 25N; T1 = 70/3 N


Rotational Mechanics Page 7 MOMENT OF INERTIA OF HOMOGENEOUS RIGID BODIES WITH DIFFERENT GEOMETRIES:

ROTATIONAL ENERGY: Let us now look at the kinetic energy of a rotating rigid object, considering the object as a collection of particles and assuming it rotates about a fixed z axis with an angular speed ω. Each particle has kinetic energy determined by its mass and linear speed. If the mass of the ith particle is mi and its linear speed is vi its kinetic energy is Ki = (1/2) mivi2 = (1/2)miri2 ω2


Rotational Mechanics Page 8 The total kinetic energy of the rotating rigid object is the sum of the kinetic energies of the individual particles: K = (1/2) (Σmiri2) ω2 So, K = (1/2) Iω2

EXAMPLE 3:A Disc of radius 1m and mass 2kg is free to rotate about its centre as shown in figure. A string is wrapped over its rim and a block of mass 2Kg is attached to its free end. The system is released from rest. Find the speed of the block when it hits the ground 7.5m below it.SOLUTION:Let the speed of the block when it hits the ground be V. At this instant the angular velocity of the Disk is ω = V/R = V where R is Radius of the disk.The moment of Inertia of the disc I = MR2/2 = 1 By the principle of conversation of energy, the gravitational energy gained by the block is equal to the total kinetic energy gained by the system.As we know the Kinetic energy of the Disk is given by Iω2 Hence, 2(10)(7.5) = (1/2)(2)V2 + (1/2) I ω2 = (1/2)(2)V2 + (1/2)(1)(V)2 This gives V = 10 m/sANGULAR MOMENTUM:The instantaneous angular momentum L of the particle relative to the origin O is defined as the cross product of the particle’s instantaneous position vector r and its instantaneous linear momentum p: L⃗=r⃗ × p⃗Angular momentum L is a vector whose direction is perpendicular to both r and p.EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 9Now, let us find the angular momentum of a rigid body rotating about a fixed axis OO’.Let the angular velocity of the body be ω. The angular momentum of this rigid body is the sum of angular momentum of all the particles in the body. Let the angular velocity of the rigid body be ω about OO’.Then the linear velocity of the ith particle is vi=ri ω. The angular velocity of this particle is rxp = rx (mIvI) = miviri = miri2ω .So, the angular momentum of the whole body is L⃗ = Σmiri2ω⃗ = Iω⃗ So, L⃗=I ω⃗ Notice that the above equation is similar to the linear momentum equation P = mv. Now by differentiating the above equation L = Iω with respect to t, we get dL⃗/dt = I(dω⃗/dt) = Iα⃗ = τ⃗ total So, dL⃗/dt = τ⃗ total

CONSERVATION OF ANGULAR MOMENTUM: The law of conservation of angular momentum states that: If the total external torque on a system is zero, its angular momentum remains constant.EXAMPLE 3: Two similar spheres A and B of Mass M= 5kg and Radius r= 1m are attached to two ends of a rod of length L= 4m and mass 1 kg. The system rotates about a frictionless pivot O with a constant angular velocity ω = 6 rad/s. Find a) The angular momentum of the system about point O. b) Now if the sphere B is removed from the system find the final angular velocity of the system about O? SOLUTIONa) The moment of Inertia of the system about point O = the sum of moment of inertia of the two spheres and the rod about O. Moment of inertia of each sphere about an axis passing through centre of the sphere isEDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 10 (2/5)MR2 From parallel axis theorem, the moment of inertia of a sphere about an axis passing through O and perpendicular to the plane of figure is (2/5)MR2 + M ((L/2) +r) 2 = (2/5) (5) (1)2 + 5(2+1)2 = 47Moment of Inertia of the rod about O is (1/12)ML2 = 4/3Total moment of inertia about O = 47 + (4/3) + 47 = 95.33 Total angular momentum about O is Iω = (95.33) (6) = 572 kg.m2.s-1

b) The net external torque on the system is zero about O. So, the angular momentum of the system remains constant.Moment of inertia after removing sphere B is I = 47 + (4/3) = 48.33Let the final Angular velocity be ωU, then by conservation theorem Initial angular momentum = final angular momentum 572 = 48.33 ωU ωU = 11.83 rad/sec

ROLLING: UNIFORM ROLLING: In this section we treat the motion of a circular rigid object whose centre moves in a straight line with a constant linear velocity vc and it rotates in


Rotational Mechanics Page 11its plane about its centre with an angular velocity ω such that vc = Rω, where r is the radius of the circle. This is called Uniform Pure Rolling.This motion can be represented as the sum of pure translation and pure rotational motions. Note that while the centre moves in a straight line, a point on the rim moves in a path called the cycloid as shown in the following figure. The velocities and accelerations of each point on the rigid body in pure rolling motion can be obtained by adding the ‘pure translation’ and ‘pure rotational’ component at that point.An illustration on calculating velocities of four points P1, P2, P3 and P4 on a rolling disc moving with a horizontal velocity V is shown in the following figure:

Here we note that the velocity at point P4 which touches the ground is ZERO and the velocity at the topmost point P1 is TWICE the velocity of centre of mass. ACCELERATED ROLLING :Recall that rolling requires that linear and angular velocities are tied together by the equation of rolling motion: Vc = Rω In case of Uniform pure rolling there is no linear and angular accelerations as the body moves with uniform velocity. In accelerated rolling there is an external force and torque on the body. Differentiating the above equation of rolling motion gives the equation of accelerated rolling ac = Rα


Rotational Mechanics Page 12 FRICTION ON ROLLING BODY: In Uniform pure rolling there is no frictional force acting on the body while in accelerated pure rolling there will be frictional force acting on the body. In both these cases the work done by friction on the body is zero .This is because the point where the frictional force is acting on the body has zero instantaneous velocity and thus zero instantaneous displacement(recall, Work = Force x displacement). KINETIC ENERGY OF A ROLLING BODY: As we know that rolling motion can be divided into ‘Pure translational’ and ‘Pure rotational’ motions, the total kinetic energy of the rolling body is the sum of rotational and translational kinetic energies. Consider a rigid body rolling with an angular velocity ω on a flat surface, the K.E of the body is Translational K.E + Rotational K.E = (1/2) Mv2+ (1/2) Iω2

In pure rolling v = Rω. K.E = (1/2) (I + MR2) ω2 EXAMPLE 5A disc of mass M and radius R rolls without slipping on an inclined plane of inclination θ.The sphere starts from rest. Find a) the linear acceleration of the sphere and the force of friction acting on it. B) The kinetic energy of the disc at the end of t sec. c) The work done by friction on the disc in t sec.SOLUTION

a)

Linear motion Equation: Mg sin θ - F = Ma ....... (i)Rotational motion equation: FR = (1/2) (MR2)(a/R) F = (1/2) Ma ...... (ii)EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 13From (i) and (ii), F = (1/3) Mg sin θ N And linear acceleration a = (2/3) g sin θ m/s2b) Velocity of the disc at the end of t sec is V = 0 + at = (2/3) g t sin θ So angular velocity at the end of t sec is ω = V/R ω = (2/3R) g t sin θAs we recall, K.E = (1/2) (I + MR2) ω2 K.E = (1/2)(3/2)MR2ω2 = (3/4) MR2(4/9)R-2gt2 sin2 θ = (1/3) Mgt2 sin2 θ Ja) We recall that work done by the frictional force on the rolling body is ZERO.Let us prove it for a better understanding,The bottom most point is at rest at all times as we know from pure rotation. And work done is force times displacement at the application of the point of force. The displacement of the point of application is zero and hence the work done is zero.ANGULAR MOMENTUM OF A BODY IN COMBINED ROTATION AND TRANSLATION:Let us consider a rigid body shown in the figure. This is a case of combined rotation and translation with Velocity of centre of mass V and angular velocity about centre of mass ω. Now let us find the angular momentum L of the body about an arbitrary point O. L = Σ mi ri x vi = Σ mi (ri/cm + r0) x (vi/cm + v) EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 14Where rI and vI are the position and velocity vector of a mass mi on the body. Thus L = Σ mi (ri/cm x vi/cm) + [Σ mi ri/cm] x v + r0 x [Σ mi vi/cm] + [Σmi] r0 x v Since Σ mi ri/cm = M Rcm/cm =0 Σ mi vi/cm = M vcm/cm =0 Here note that ri , ri/cm , vi , vi/cm are all vectors.Thus, L = Σ mi (ri/cm x vi/cm) + M r0 x v0 L = Lcm + M (r0 x v0)Where Lcm is the angular momentum of the body around the centre of mass.EXAMPLE 6: A disc of mass M and radius R is initially rotating about its centre of mass at an angular velocity of ω and NO linear velocity. This disc is kept on a rough ground where it slips for some time and starts pure rolling. Find the translational velocity of the disc after it starts pure rolling.SOLUTION:When the disc is kept on rough ground friction starts acting on the body. The angular velocity decreases and the linear velocity (which is initially zero) increases until its starts pure rolling.Let V be the velocity of the disc when it starts pure rolling. Then final angular velocity will be V/R.The net Torque on the body about contact point O is zero as the frictional force acting on the body passes through O. So angular momentum of the body is conserved about O.INITIAL ANGULAR MOMENTUM: L = Lcm + M (r0 x v0) = Iω + 0 as initial linear velocity v 0 = 0 = (1/2) MR2ωFINAL ANGULAR MOMENTUM: L = Lcm + M (r0 x v0) = I (V/R) + MRVEDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 15 = (1/2) MR2 (V/R) + MRV = (3/2) MRVINITIAL ANGULAR MOMENTUM = FINAL ANGULAR MOMENTUM (1/2)MR2ω = (3/2) MRV V = (1/3) RωProblems

1. The moment of inertia of the sector of a disc, having mass M and radius r with sector angle 450 isa) Mr2/2 b) Mr2/8 c) Mr2/16 d) Mr2/42. A wheel having a moment of inertia 2 kg.m2 about its axis rotates at 50rpm about its axis. The torque that can stop the wheel in one minute isa) π/3 b) π/9 c) π/12 d) π/183. The moment of inertia of disc A is I. The moment of inertia of disc B with same mass and radius, but twice the thickness of disc A isa) I b) 2I c) 4I d) I/24. A force F acts on a cube of mass M and side L at the top edge as shown in the figure. What is the minimum force F required to topple the cube?a) 0 b) Mg c) Mg/2 d) Mg/35. A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What is the initial linear acceleration of its centre of mass?a) 3g b) 3g/2 c) 3g/4 d) g


Rotational Mechanics Page 166. A hole of radius R/2 is removed from a thin circular plate of radius R and mass M, What is the moment of inertia of the remaining plate about an axis through centre and perpendicular to the plane of the plate.a) 3MR2/32 b) 7MR2/32c) 13MR2/32 d) 9MR2/327. Which of the following is correct for a sphere of radius r with forward velocity of centre of mass v and clockwise angular velocity about centre ω, moving on a rough horizontal plane?a) If v > rω friction is backwards b) If v < rω friction is forward.c) Both a and b d) None 8. Which of the following statements is correct for a uniform sphere on an inclined plane, when a constant force is acting tangentially on its top most point, parallel to the inclined plane in the direction of motion?a) Frictional force is always upwardsb) Frictional force is always downwardc) Frictional force can either be upward or downward depending on the magnitude of force. d) Frictional force is Zero9. When a body starts from rest and rotates about a fixed axis with constant angular acceleration ,the radial acceleration of a point in the body is directly proportional to itsa) Tangential velocity b) Angular displacement squarec) Angular displacement d) Radial distance10. The centre of a wheel rolling on a plane surface moves with a speed v. A particle on the rim of the wheel at the same level as the centre will be moving at speeda)Zero b)v c)√2 v d)2v11. At an instant, a sphere of radius R is rolling without slipping on a horizontal surface with a linear acceleration ‘a’ and linear velocity ‘v’. the magnitude of acceleration of the particle on the sphere in contact with the ground isa) a b) (a2+ (v4/R2)) 0.5 c) √2.a d) a+ (v2/R)


Rotational Mechanics Page 1712. A disc rolls without slipping on a horizontal surface with a constant velocity V. What is the magnitude of velocity of a particle on the disc, whose velocity vector makes maximum angle with the horizontal?a) V b) √2.V c) √3V d) 2V13. A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end .The rod is released from rest in the horizontal position. What is its angular speed at the instant when its angular acceleration is zero?a) (g/L)0.5 b) (2g/L) 0.5 c) (3g/L) 0.5 d) 2(g/L)0.5

14. A square plate of mass 6m and side L is free to rotate about a frictionless pivot O.A small particle of mass m moving with a velocity V comes and sticks to the plate which is initially at rest. What is the final angular velocity of the system?a) V/3L b) 3V/L c) 3V/2L d) 2V/3L15. A wheel of radius R, mass M, and moment of inertia I, is mounted on a frictionless pivot at its centre, as shown in Figure. A light cord wrapped around the wheel supports an object of mass m. What is the linear acceleration of the object if g=10;I=30;m=10;R=1 all in MKS a) 0 b) 1.5 c) 2 d) 2.516. A uniform rod of mass m and length l = 0.6m is kept vertical with lower end clamped on a frictionless pivot. It is slightly pushed to let it fall down under gravity. Find the angular speed of the rod when it is passing through the lowest position. g=10m.s-1

a) 10 b) 5 c) 12 d) 100EDUDIGM 1B Panditya Road, Kolkata 29 www.edudigm.in 40034819

Rotational Mechanics Page 1817. A force F acts tangentially at the highest point of a sphere of mass m kept on a rough horizontal plane, If the sphere rolls without slipping, then the acceleration of the centre of the sphere isa) 10F/7M b) 5F/7M c) 3F/5M d) 5F/7M18. A sphere starts rolling down an incline of inclination 300.The speed of its centre when it has covered 14m isa) 2 b) 5 c) 7 d) 1019. Two uniform identical rods each of mass M and length l are joined to form a cross as shown. The moment of inertia of the cross about a bisector shown in dotted line isa) Ml2/3 b) Ml2/6 c) Ml2/12 d) Ml2/2420. A turn table of mass M and radius R is rotating with angular velocity ωU on frictionless bearing. A spider of mass M falls vertically onto the rim of the turn table and then walks in slowly towards the centre of the table. The angular velocity of the system when spider is at a distance r from the centre is proportional toa) R2/ (2R2+r2) b) R2/ (R2+r2) c) R2/ (R2+2r2) d) R2/ (2R2+2r2)21. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed on the top of an incline and released. The frictional coefficient between the objects and the incline is same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom bya) Solid sphere b)Hollow sphere c) Discd) All will take same time.


Rotational Mechanics Page 1922. The coefficient of friction on the ground is μ. The maximum value of F so that sphere will not slip is a) μmg b) 5μmg/2 c) 7μmg/5 d) 7μmg/223. One end of a uniform rod of mass M and length l is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity ω. The force exerted by the clamp on the rod has a horizontal componenta) Mω2l b) Zero c) Mg d) 0.5 Mω2l24. A uniform thin rod of mass M is bent into a square of side a. The moment of inertia of the square about an axis through its centre and perpendicular to the plane isa) Ma2/3 b) 7Ma2/3 c) Ma2/6 d) Ma2/1225. A uniform rod of mass M and length L is hanged by two strings as shown in figure. Suddenly one of the string is cut, the tension in other string at this instant is a) Mg b) Mg/2 c) Mg/3 d) Mg/426. 1. A shaft is turning at 25 rad/s at time zero. Thereafter, its angular acceleration is given byα = -10 - 15t rad.sec-2.The magnitude of angular velocity of the shaft at time t= 3sec is.a) 25 b) 185 c) 145 d) none of these27. A spool of wire of mass M and radius R is unwound under a constant force F. Assuming that the spool is a uniform solid cylinder that does not slip, the acceleration of the centre of mass is a) 2F/3M b) F/3M c) 4F/3M d) 5F/3M


Rotational Mechanics Page 2028. A triangular section of mass M with the dimensions shown in the figure, is free to rotate around a frictionless axis .The maximum angular acceleration that can be produced by a force F isa) 3F/2M b) 3F/4M c) 3F/M d) None of The Above29. A sphere of radius R is left from rest at point A. AO is rough and OB is frictionless .The sphere rolls without slipping in AO. What is the maximum height the sphere rises above O in track OB. a) R b) R/2 c) 3R/5 d) 5R/730. A sphere initially slides with a linear velocity of V on the smooth part of track. It enters the rough part of the track as shown in the figure. The linear velocity of the sphere when it starts pure rolling on the rough track isa) V b) V/2 c) 3V/5 d) 5V/731. A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed , smooth vertical wall. The speed of the sphere after it has started pure rolling in the backward direction isa) v b) v/7 c) 3v/7 d) v/232. An impulse J acts horizontally on a sphere of radius R at height h above its centre as shown in the figure. The rotational and translational kinetic energies of the sphere are equal after the impulse. Then h isa) √(2/5). Rb) R c) √2 .R d) R/233. Two particles of mass m each are attached to a light rod of length d, one at its centre and other at a free end.The rod is foxed at the other end and is rotated in a plane at an angular velocity ω. The angular momentum of the particle at the end with respect to the particle at the centre is


Rotational Mechanics Page 21a) mωd2 b) 0.5 mωd2 c) 0.25 mωd2 d) 0.75 mωd2

34. A cubical block of mass M and side L slides down a rough inclined plane of inclination 300 with a uniform velocity. The magnitude of torque of the normal force on the block about its centre isa) 0 MgL 0.5 MgL 0.25 MgL35. A hollow sphere of radius r lies on a smooth horizontal surface. It is pulled by a horizontal force acting tangentially from its highest point. The distance travelled by the sphere during the time it takes 1 full revolution isa) 2πr b) 2πr/3 c) 4πr d) 4πr/3

36. A rotating ball hits a rough horizontal plate with a vertical velocity v and angular velocity ω. The coefficient of friction is μ and the vertical velocity after collision is v/2.The reduction in the angular velocity after collision isa) μv/4R b) 5μv/4R c) 10μv/4R d) 15μv/4R37. A hollow sphere of mass 2kg is kept on a rough horizontal surface. A force of 10√3 N is applied as shown in the figure. Find the minimum value of μ so that the sphere starts pure rolling. a) 0.3 b) 0.2 c) 0.1 d) The body does not roll for all values of μ.38. A uniform circular disc of radius r is placed on a rough horizontal surface and given a linear velocity v rightward and angular velocity ω in the anticlockwise direction. The disc comes to rest after moving some distance to the right. Then a) 3v=2ω b) 2v=ωr c) v=ωr d) 2v=3ωr


Rotational Mechanics Page 2239. A cylinder of mass m rotating in the anticlockwise direction is put at the corner of two perpendicular rough walls with coefficient of friction μ. The normal reaction on the cylinder offered by the vertical wall isa) μmg/(1+μ2) b) μ2mg/ (1+μ2) c) μmg d) Zero40. A hollow sphere of mass √2 kg is released from the top of an inclined plane of inclination 45o and coefficient of friction is 0.2. Its kinetic energy after moving a distance of 8m isa) 7g b) 5g c) 3g d) None

Answers1) a 2) d 3) a 4)c 5)c 6)c 7) c 8) c 9) c 10) c 11) b 12) b13) c 14) a15) d 16) a 17) a 18) d 19) c 20 c 21) d 22) d 23) d 24) a 25) d 26) c27) c 28) c29) c 30 d 31) d 32) a 33) c 34) d 35) d 36) d 37) b 38) b 39) b 40) a


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