iit advanced assignment with solutions on torque, angular

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Solutions to Assignment On Torque, Angular Momentum and Moment of Inertia; Part # 2

Q1. Calculate the kinetic energy of a tractor crawler belt (see figure) of mass m if the tractor moves with velocity v. There is no slipping. The dimensions of the wheels are as shown in the figure. Ans.1: The velocity of the CM of the tractor wheel is v. Therefore; vupper part = v+ v = 2v and vlower part = v− v = 0

Let mass of belt AB and CD is m1 each and BC and DA taken together is m2, then the total mass of the belt is (2m1 + m2). Now;

TKEBELT = LKEAB + LKECD + LKEBC+DA + RKEBC+DA =12m1 2v( )2 + 0+ 1

2m2 v( )

2+12Iω 2

⇒ = 2m1v2 +12m2v

2 +12m2R

2( ) vR⎛

⎝⎜

⎞

⎠⎟2

= 2m1v2 +m2v

2 = 2m1 +m2( )v2 =mv2

Q2. A cylinder of mass m is suspended through two strings wrapped around it as shown in figure. Find (a) the tension T in the string and (b) the speed of the cylinder as it falls through a distance h. Ans.2: Considering translational motion, the equation will be

mg− 2T =ma Considering rotational motion, the equation will be

2T ⋅R = Iα ⇒ 2T ⋅R = IaR ⇒ 2T = Ia

R2=Ma2

Solving we get; a = 23g and T = mg

g

Further; v2 − 02 = 2 23g

⎛

⎝⎜

⎞

⎠⎟h ⇒ v = 4gh

3

Q3. Two horizontal discs rotate freely about a common vertical axis passing through their centers. The moments of inertia of the discs relative to the axis are equal to I1 and I2, and the angular velocities are ω1 and ω2 . The upper disc falls on the lower disc and after some time, both discs begin to rotate as a single body due to friction. Find

(a) The steady state angular velocity of the discs, and (b) The work performed by the frictional forces in the process.

Ans.3: (a) By conservation of angular momentum; we get

I1ω1 + I2ω2 = I1 + I2( )ω f ⇒ ω f =I1ω1 + I2ω2

I1 + I2

(b) Work = ΔKE = KEfinal −KEinitial =12I1 + I2( )ω f

2 −12I1ω1

2 +12I2ω2

2⎡

⎣⎢⎤

⎦⎥= −

I1I22 I1 + I2( )

ω1 −ω2( )2

Q4. A uniform solid cylinder of mass m can rotate freely about its axis; which is kept horizontal. A particle of mass m0 hangs from the end of a light string wound round the cylinder. When the system is allowed to move, find the acceleration with which the particle descends. Ans.4: Considering translational motion, the equation will be

m0g−T =m0a Considering rotational motion, the equation will be

T ⋅R = Iα = 12mR2 a

R⎛

⎝⎜

⎞

⎠⎟ ⇒ T = ma

2

NARAYANA INSTITUTE OF CORRESPONDENCE COURSES

Physics : Rotational Motion

Total kinetic energy of the sphere at A

= K.E. of the rotation + K.E. of translation of sphere

2 21 1

2 2I m� v Z

2

2 22

1 2 12 5 2

vmr mv

r§ ·§ ·¨ ¸¨ ¸

© ¹© ¹ �

2mv … (ii) 7

10

For just completing the vertical circular loop

� �

2mvmg

R r

�

or v R . � �2 r � g

Now total kinetic energy of sphere at 27

10A m v

� �7

10m R r g � .

Loss of potential energy = gain of K.E.

� � � �72

10h R r m R r g� � �mg .

Solving we get,

27 17

10 10R§ · u �¨ ¸

© ¹h r

substituting the given values, we get

27 17

1 0.10 10§ ·� u¨ ¸© ¹

2 1.36m h u .

Problem 3. Calculate the kinetic energy of a tractor crawler belt (see figure) of

mass m if the tractor moves with velocity v. There is no slipping. The dimensions of the wheels are as shown in the figure. v

A B

CD Solution: The velocity of the centre of mass of the tractor wheel is v .

velocity of the lower part of the belt in contact with the ground is

� . 0v v

Velocity of the upper part between the two wheels is 2v v v �

Suppose that the mass of the upper portion of the belt AB between the two wheels is m1 and the total mass of the portion of the belt in contact with the wheels between BC and DA is m2. The total mass of the belt is:

� � . 1 22m m�

FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 z Ph.: (011) 32001131, 32001132, Fax : (011) 41828320

26

Physics : Rotational Motion NARAYANA INSTITUTE OF CORRESPONDENCE COURSES

The kinetic energy (KE) of the belt

= KE of AB + KE of CD + KE of BC + KE of DA

= � �2

2 2 21 2 2

1 1 12 0 .

2 2 2v

m v m RR

§ ·� � � ¨ ¸© ¹

m v

� 2 21 22m v m v

2mv .

Problem 4. A cylinder of mass m is suspended through two strings wrapped

around it as shown in figure. Find (a) the tension T in the string and (b) the speed of the cylinder as it falls through a distance h.

T T

mg Solution: The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of

the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. Suppose the centre of the cylinder falls with an acceleration a. The angular

acceleration of the cylinder about its axis is ar

D , as the cylinder does not slip over the

strings.

The equation of motion for the centre of mass of the cylinder is

mg � … (i) 2T ma

and for the motion about the centre of mass, it is

21 1

2 2Tr mr mra§ · ¨ ¸

© ¹2 D

or, 1

2T m 2 … (ii) a

From (i) and (ii),

2

3 a and g

6

mg T .

As the centre of the cylinder starts moving from rest, the velocity after it has fallen through a distance h is given by

2 223

§ · ¨ ¸© ¹

v g h

or, 43gh

v .

Problem 5. A solid spherical rotor of radius 30 cm and weight is connected to a link and freely rotates about with angular velocity 60 radian/second. If the rotor is suddenly allowed to rest on a vertical wall, what time will elapse before it comes to rest? The coefficient of friction between the wall and the rotor is 0.25 and inclination of link with respect to the vertical is 15º.

WAB B

r

15o

B CR

A

rB C

T

W+ RP


27


Solving we get; a = 2m0

m+ 2m0

⎛

⎝⎜

⎞

⎠⎟g

Q5. Two masses m1 = 12kg and m2 = 8kg are tied to the ends of a string which passes over a pulley of an Atwood’s machine. The mass of the pulley is M = 10kg and its radius R = 0.1m. Find the tensions in the string and the acceleration of the system.

Q6. A solid cylinder of diameter D is mounted on a frictionless horizontal axle. A string is wrapped around it and a heavy block is attached to the free end of the string. The block is allowed to fall freely. If the speed of the block just before striking the ground be v, then

(a) v is directly proportional to diameter D. (b) v is directly proportional to square of diameter D. (c) v is inversely proportional to diameter D. (d) v is independent of D. *

Q7. A uniform circular disc of radius R with a concentric circular hole of radius R/2 rolls down an inclined plane. The fraction of its total energy associated with its

rotational motion is: (a) 5/13 * (b) ¼ (c) ½ (d) 1

Q8. A ring starts from rest and acquires an angular speed of 10 rad/s in 2 s. The mass of the ring is 500 g and its radius is 20 cm. The torque on the ring is

(a) 0.02Nm (b) 0.20Nm (c) 0.10 Nm* (d) 0.01 Nm.

Q9. A loop and a disc have the 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) 6J* (b) 8J (c) 10J (d) 12J.

Q10. A wheel and an axle, having a total moment of inertia 0.002 kg-m2, is made to rotate about a horizontal axis by means of an 800 g mass attached to a cord (assumed massless) that is wound around its axle. The radius of the axle is 2 cm. Starting from rest, how far does the mass fall in order to give the wheel a speed of 3 rev/s?

(a) 2.25cm (b) 3.25cm (c) 4.5 cm* (d) 5. 25 cm.

Q11. A small meteorite of mass m travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass M and radius R. The length of the day was T before the meteorite struck. After the meteorite strikes the earth, find the new length of day on earth.



7. Two masses m 1 and are tied to the ends of a string which passes over a pulley of an Atwood’s machine. the mass of the pulley is and its radius R = 0.1m. Find the tensions in the string and the acceleration of the system.

1 2kg 2m 8k

M 1

g

0kg RM

T2

T1

a

m g1

m g2 m1

m2a

8. A uniform disc of mass M and radius R is initially at rest. Its axis is fixed through O. A block of mass m is moving with speed v1 on a frictionless surface passes over the disc to the dotted position. On coming in contact with the disc, it slips on it. the slipping ceases before the block loses contact with the disc, due to the high friction.

Now prove that � �

12

vv1 M / 2m

�

RO

v1 v2

m m

9. Find the angular momentum and rotational kinetic energy of earth about its own axis. Find the duration for which this amount of energy can supply 1 kilowatt power to each of 3.5u109 persons on earth ? Mass of earth = 6.0u1024., radius = 6.4u103km.

10. A uniform solid wheel of mass m and radius 15 cm is free to rotate without friction about a fixed horizontal axis passing through its centre. A particle of equal mass m strikes it at a pint P after falling vertically downward as shown in the figure, and sticks to the wheel.

60o P

Particle

37.5 cm

Find the maximum angular speed of the wheel. find the force on the axis when the angular speed of the wheel reaches a maximum.

11. A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle of 53º with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder.

53o

N1

N2


46


Q12. If earth suddenly shrinks to 1/64th of its original volume keeping mass same, the length of day on earth will become

(a) 2 hr (b) 1.5 hr (c) 3 hr (d) 6 hr. Q13. A uniform rod of length is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle θ, its angular velocity is given by

(a) 6glsinθ2

⎡⎣⎢

⎤⎦⎥*

(b) 6glcosθ

2⎡⎣⎢

⎤⎦⎥

(c) 6glsinθ[ ]

(d) 6glcosθ[ ]

Q14. A disc of mass M and radius R is pivoted about a horizontal axis through its centre and a particle of the same mass M is attached to the rim of the disc. If the disc is released from rest with the small body at the end of a horizontal radius, the angular speed when the small body is at the bottom is:

(a) g4R

(b) g2R

(c) 3g4R

(d) 4g3R

*

Q15. A uniform rod of length 6a and mass 8m lies on a smooth horizontal table. Two particles of mass m and 2m, moving in the same horizontal plane but in opposite directions with speeds 2v and v respectively strike the rod normally as shown in figure and stick to the rod. Denoting angular velocity (about the centre of mass), and translational velocity of centre of mass by ω and vc respectively after the collision, find the value of vc and ω. Q16. In pure rolling fraction of its total energy associated with rotation is a for α ring and β for a solid sphere. Then

(a) α = 1/2 (b) α = 1/4 (c) β = 2/5 (d) β = 2/7

Q17. A turntable turns about a fixed vertical axis, making one revolution in 10s. The moment of inertia of turntable about the axis is 1200 kgm2. A man of 80kg, initially standing at the centre of the turntable, runs out along the radius. What is the angular speed of the turntable, when the man is 2m from the axis of rotation?



13. A uniform rod of length A is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle T , its angular velocity is given by

(a) 6

sin2

g T© ¹A

(b) 6

cos2

g T© ¹A§ ·¨ ¸

(c) 6

sing

T© ¹A§ ·¨ ¸ (d)

6cos

gT

© ¹A§ ·¨ ¸ .

T

AA|

B

l

§ ·¨ ¸

14. A disc of mass M and radius R is pivoted about a horizontal axis through its centre and a particle of the same mass M is attached to the rim of the disc. If the disc is released from rest with the small body at the end of a horizontal radius, the angular speed when the small body is at the bottom is:

(a) 4gR© ¹

§¨ (b) ·

¸ 2gR© ¹

§ ·¨ ¸

(c) 34gR© ¹

§¨ (d) ·

¸43gR© ¹

§ ·¨ ¸ .

15. A symmetric lamina of mass M consists of a square shape with a semicircular section over each of the edge of the square as shown in figure. The side of the square is 2a. The moment of inertia of the lamina about an axis through its centre of mass and perpendicular to the plane is 1.6 . The moment of inertia of the lamina about the tangent AB in the plane of the lamina is:

2M a

(a) 4. (b) 3. 28Ma 22Ma

(c) 6. Ma (d) 1. Ma . 24 26

A

B

2a

16. A smooth sphere A is moving on a frictionless horizontal plane with angular speed Z and centre of mass velocity v . It collides elastically and head-on with an identical sphere B which is at rest. All surfaces are frictionless. After the collision, their angular speeds are Z and , respectively. Then, A BZ

(a) Z (b) Z 0B A BZ

(c) Z � (d) Z . A ZB B Z

17. A cubical block of side a is moving with a velocity v on a smooth horizontal plane

as shown in the figure. It hits a ridge at point O. The angular speed of the block after it hits O is:

(a) 3

4

va

(b) 3

2

va

(c) 3

(d) zero. 2

v

a

O

18. A solid uniform sphere, rotating about a horizontal axis (with rotational K.E. = ), is gently placed on a

rough horizontal plane. After some time the sphere begins pure rolling with total Kinetic Energy E. Then,

0E


52



SECTION–III

MULTIPLE CHOICE QUESTIONS 1. Which of the following statement(s) is/are correct for a spherical body rolling without slipping on a rough

horizontal ground at rest.

(a) The acceleration of a point in contact with ground is zero

(b) The speed of some of the point(s) is (are) zero

(c) Friction force may or may not be zero

(d) Work done by friction may or may not be zero

2. A uniform rod of length 6a and mass 8m lies on a smooth

horizontal table. Two particles of mass m and 2m, moving in the same horizontal plane but in opposite directions with speeds 2v and v respectively strke the rod normally as shown in figure and stick to the rod. Denoting angular velocity (about the centre of mass), total energy and translational velocity of centre of mass by Z, E and vc respectively after the collision.

m

2v

2m

v2a

a

(a) vc = 0 (b) Z = 53v

a

(c) Z = 5v

a (d) E =

235

mv

3. A particle moves in acircle of radius r with angular velocity ZZ. At some instant its velocity is v and

radius vector with respect to centre of the circle is . At this particular instant centripetal acceleration a of the particle would be

o o

o

ro

c

(a) Z × (b) v × Z o o

vo o

(c) Z × o o o

Z§ ·u© ¹r¨ ¸ (d) v × ( ) o o o

Zur

4. A particle of mass m is travelling with a constant velocity v = v along the line y = b, z = 0. Let dA be

the area swept out by the position vector from origin to the particle in time dt and L the magnituded angular momentum of particle about origin at any time t. Then

oˆ

0i

(a) L = constant (b) L z constant

(c) dA

dt =

2Lm

(d) dA

dt =

2L

m

5. In pure rolling fraction of its total energy associated with rotation is a for D ring and E for a solid sphere. Then

(a) D = 2

1 (b) D =

4

1

(c) E = 5

2 (d) E =

7

2


54


Q18. You are given tow discs A and B. A has twice the moment of inertia as compared to B. A is rotating but B is not. Now A is placed on B. As a result, 15000 J of energy is lost. Find initial K.E. of disc A. Q19. Consider a disc of mass M and radius R rotating with an angular velocity of ω about its geometrical axis. A small object of mass m gently falls on the edge of the disc

and sticks to it. Find the new angular velocity of disc. M

M + 2m⎡⎣⎢

⎤⎦⎥ω⎧

⎨⎩

⎫⎬⎭

Q20. A uniform rod AB of length L, mass M is hinged at one end A. The rod is kept in the horizontal position by a massless string tied to the point B as shown. Find the reaction of the hinge on the end A at the instant string is cut. Ans.: When the string is cut, the weight of rod will form torque about the hinge, so

τ A = mgL2

⎛⎝⎜

⎞⎠⎟ ⇒ Iaα = mg L

2⎛⎝⎜

⎞⎠⎟

⇒ ML2

3α = mg L

2⎛⎝⎜

⎞⎠⎟ ⇒ α = 3

2gL

The acceleration of CM of rod will be;

a = rα = L2α = 3

4g

Hence, considering translatory motion of the rod, we get;

mg − Ra = macm ⇒ mg − Ra =3mg4

⇒ Ra =mg4

Q21. A string is wrapped around a solid cylinder and then the end of string is held stationary while the cylinder is released from rest. Find the acceleration of the cylinder and the tension in the string. [2g/3, mg/3] Q22. In figure, the steel balls A and B have mass of 500g each, and are rotating about the vertical axis with an angular velocity of 4 rad/s at a distance of 15 cm from the axis. Collar C is now forced down until the balls are at a distance of 5 cm from the axis. How much work should be done to move the collar down? [1.44 J]

Q23. A bullet of mass m moving with velocity u just grazes the top of a solid cylinder of mass M and radius R resting on a rough horizontal surface as shown. Assuming the cylinder rolls without slipping, find the angular velocity of the cylinder and the final velocity of bullet. Ans.: Considering pure rotation about the point of contact O, and using conservation of angular momentum we get,

mu( ) 2R( ) = mv( ) 2R( )+ Iω

Also; v = 2R( )ω and I = MR2

2+MR2


Therefore, on solving we get;

ω = 4mu8m + 3M( )R and v = 8mu

8m + 3M( )

Q24. A circular wooden hoop of mass m and radius R rests on a smooth frictionless surface. A bullet, also of mass m, and moving with velocity v strikes the hoop and gets embedded in it. The thickness of hoop is much smaller than R. Find the angular velocity with which the system rotates after the bullet strikes the hoop. Ans.: Let the velocity of CM of system after the bullet strikes is Vcm. Therefore; Velocity of CM after strike will be

Vcm =m v( )+m 0( )

m +m= v2

Also, position of CM of system will be

xcm =m 0( )+m R( )

m +m= R2

Using conservation of angular momentum about CM, we get

mv( ) R2= Iω ⇒ mv( ) R

2= m R

2⎛⎝⎜

⎞⎠⎟2

+ mR2 +m R2

⎛⎝⎜

⎞⎠⎟2⎧

⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

⎛

⎝⎜

⎞

⎠⎟ω

After solving we get; ω = v3R

Q25. A stick of length L and mass M lies on a frictionless horizontal surface on which it is free to move in anyway. A ball of mass m moving with speed v (as shown) strikes elastically with the stick. What must be the mass of ball so that it remains at rest immediately after collision? Ans.: By conservation of linear momentum

mv +M 0( ) = m 0( )+MV ⇒ V = mvM

By conservation of angular momentum

mvd = Iω ⇒ ω = mvdI

Since collision is perfectly elastic, therefore, K.E. of system also remain conserved i.e. 12mv2 = 1

2MV 2 + 1

2Iω 2

Substituting the values of v and ω, we get;

12mv2 = 1

2M mv

M⎛⎝⎜

⎞⎠⎟2

+ 12

ML2

12⎡

⎣⎢

⎤

⎦⎥

mvdmL2 /12

⎡⎣⎢

⎤⎦⎥

2

⇒ m = ML2

L2 +12d 2

Q26. Four forces of same magnitude act on a square as shown in figure. The square can rotate about point O; which is mid point of one of the side of square. The force which can produce the maximum torque is:

(a) F1. (b) F2. (c) F3. (d) F4.


Q27. A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane:

(a) The hollow sphere reaches the bottom first. (b) The solid sphere reaches the bottom with greater speed. * (c) The solid sphere reaches the bottom with greater kinetic energy. (d) The two spheres will reach the bottom with equal linear momentum.

Q28. Figure shows a particle moving at constant velocity v and four points with their xy coordinates. If L1, L2, L3 and L4 are the angular momentum about the points a, b, c and d respectively, then:

(a) L1 < L3. (b) L1 > L4. (c) L1 = L3 = L4. (d) (L1 = L3) < L4 < L2. *

Q29. Minimum coefficient of friction required to cause pure rolling of a cylinder down an inclined plane of inclination θ is

(a) sin θ. (b) sin θ/3 (c)tanθ/3 (d)tanθ/2. Q30. A rod of mass m and length L is hinged to its one end and held vertical. A point mass m is attached to other end, is allowed to rotate about the hinge. The velocity of the rod when it becomes horizontal is

(a) 3gL

(b) 32

gL

(c) 34

gL *

(d) None of these Q31. A mass is moving with constant velocity parallel to x-axis. Its angular momentum wrt origin:

(a) Remain constant * (b) Goes on increasing (c) Goes on decreasing (d) Zero.

Q32. A body slides down on an incline plane and reaches the bottom with a velocity v. If the same body were in the form of a ring, its velocity at the bottom would have been:

(a) v (b) 2v (c) v / 2

(d) 25v

Q33. A ring rolls on a plane surface. The fraction of its total energy associated with its rotation is:

(a) ½ (b) 1 (c) 1/3 (d) 2 *

Q34. A solid cylinder of mass M and radius R rolls down an inclined plane with height h without slipping. The speed of its centre of mass when it reaches the bottom is

(a) 2gh


(b) 43gh *

(c) 34gh

(d) 4gh

Q35. The moment of the force about the origin, as shown in the figure is

(a) 10 k Nm (b) 10 j Nm (c) 5 k Nm * (d) 5 j Nm

Q36. A uniform cylinder has a length L and radius R. If MI of this cylinder about its geometrical axis is equal to MI of the same cylinder about an axis passing through centre and perpendicular to its length is:

(a) L = R. (b) L = 3R. (c) L = R / 3 . (d) L = 3R.

Q37. A hole of radius R/2 from edge is cut from a thin circular plate of radius R. The MI of the plate about an axis passing through centre of main plate and perpendicular to plate is

(a) 5MR2/7. (b) 7 MR2/12. (c) 11 MR2/24 (d) 13 MR2/24.*

Q38. A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of the mass m are gently attached to the opposite ends of diameter of the ring. The wheel now rotates with an angular velocity:

(a) ωMM +m

(b) ω M − 2m( )M +m

(c) ωMM + 2m

*

(d) ω M − 2m( )M + 2m

Q39. ABC is a rectangular plate of uniform thickness. The sides AB, BC and AC are in the ratio 4, 3, 5 as shown in figure. IAB, IBC and ICA are MI of the plate about AB, BC and CA respectively. Which of the following is correct?

(a) ICA is maximum. (b) IAB > IBC (c) IBC > IAB * (d) IAB + IBC = ICA

Q40. A child is standing with folded hands at the centre of a rotating platform, rotating about its centre. The kinetic energy of the system is K. The child now stretches his arms so that the MI of the system doubles. The KE of the system now becomes:


(a) 2K (b) K/2 * (c) K/4 (d) 4K. Q41. The MI of a semicircular disc of mass M and radius R about a line perpendicular to the plane of disc through the centre is

(a) 2 MR2/5 (b) MR2/4 (c) MR2/2* (d) MR2 Q42. A disc of radius R/3 is removed from edge from a thin circular plate of radius R and mass 9M. The MI of the plate about an axis passing through centre of main plate and perpendicular to plate is

(a) 8MR2. (b) 4MR2. * (c) 40MR2/9 (d) 37MR2/9.

Q43. A circular disc X of radius R is made from an iron plate of thickness t and another plate Y of radius 4R is made from an iron plate of thickness t/4. The ratio between MI IY/IX is:

(a) 1 (b) 16 (c) 32 (d) 64*

Q44. A solid sphere is rotating in free space. If the radius of sphere is increased keeping mass same, which of the following will not be affected?

(a) Moment of inertia (b) Angular momentum* (c) Angular velocity (d) Rotational kinetic energy.

Q45. A uniform solid sphere of radius R having MI equal to I about its diameter is melted to form a disc of thickness t and radius r. The moment of inertia of disc about an axis passing through its edge and perpendicular to the disc is I, then radius of disc is

(a) 2R/ 15 (b) 2R/ 5

(c) 25

R

(d) 215

R.

Q46. Four point masses each of value m are placed at the corners of a square ABCD of side L. The MI of system about an axis parallel to diagonal will be

(a) 2mL2. (b) 3mL2.* (c) 4mL2. (d) mL2.

Q47. A horizontal platform is rotating with an angular velocity around its vertical axis passing through centre. At the same instant of time a viscous fluid of mass m is dropped at the centre and allowed to spread out and finally fall. The angular velocity during this period

(a) Decreases continuously. (b) Decreases initially and increases again.* (c) Remain unaltered. (d) Increases continuously.


Q48. The angular momentum of a body changes from A to 4A in 4 seconds. The torque acting on the body is

(a) 3A/4 * (b) 4A (c) 3A (d) 3A/2

Q49. A wooden log of mass M and length L is hinged by a frictionless nail at point O. A bullet of mass m strikes it with velocity v and sticks to it. Find the angular velocity of system immediately after collision.

3mvM + 3m( )L

⎡

⎣⎢

⎤

⎦⎥

Q50. An electric motor exerts a constant torque of 10Nm on a grindstone mounted on a shaft. MI of system about the shaft is 10 kgm2. If the system starts from rest, find the work done in 8s.

(a) 1600J* (b) 1200J (c) 800J (d) 600J

Q51. The power of an automobile engine is advertised to be 200hp at 600rpm. Find the corresponding torque.

(a) 137 Nm (b) 237 Nm* (c) 337 Nm (d) 287 Nm

Q52. A cable is wrapped several times around a uniform solid cylinder that can rotate about its geometric axis. The cylinder has diameter of 12 cm and mass of 50kg. The cable is pulled with a force of 9N. Assuming cable unwound without stretching or slipping, find the acceleration of cylinder.

(a) 0.3 m/s2. (b) 0.32 m/s2. (c) 0.36 m/s2.* (d) 0.4 m/s2.

Q53. A bicycle wheel has a rim of mass 1kg and ahs 50 spokes each of mass 5g. If the radius of wheel is 40cm, then MI of wheel is

(a) 0.16 kgm2. (b) 0.174 kgm2.* (c) 0.18 kgm2. (d) 0.196 kgm2.

Q54. A 2kg rock has a velocity of 12 m/s when at point P as shown. Find the angular momentum of body about point D.

(a) 115.2 kgm2s– 1.* (b) 125.2 kgm2s– 1. (c) 135 kgm2s– 1. (d) None of these

Q55. A solid cylinder of mass M and radius 2R is connected to a string and a block of mass M as shown. The pulley has mass M and radius R. The friction is sufficient to cause rolling without sliding. Find the acceleration of block after the system is released.


(a) 2g/5 (b) g/2 (c) g/3 * (d) g/2.

Q56. A uniform disc of radius a has a hole of radius b at a distance c from the centre as shown/ If the disc is free to rotate about an axis passing through centre of hole A and perpendicular to the plane of disc. Find the MI of this disc about

this axis. M2

a2 + b2 + 2c2a2

a2 − b2⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

Q57. Two spheres each of M, Rare in contact as shown. Find the MI if they are rotated about a common tangent.

(a) 7MR2/5 (b) 2MR2/5 (c) 14MR2/5* (d) 4MR2/5

Q58. A boy of mass M stands on a platform of radius R capable to rotate freely about its axis. The moment of inertia of axis is I. The system is at rest. A friend of boy throws a ball of mass m with a velocity v (horizontally) towards him. The boy on the platform

catches the ball. Find the angular velocity of the system in this process. mvR

I + M +m( )R2⎡

⎣⎢

⎤

⎦⎥

Q48. A solid sphere is rolling on a frictionless surface as shown in figure with a translational velocity v m/s. If sphere climbs upto height h then value of v should be

(a) 107gh

(b) 2gh (c) 2gh

(d) 107gh *

Q49. A truck, initially at rest with a solid cylindrical paper roll, moves forward with a constant acceleration a. The cylinder roll is lying parallel to the forward edge of the truck at a distance d from the rear edge of the truck. Find the distance ‘s’ which the truck travels before the paper roll moves off the edge of its horizontal surface. Friction is sufficient to prevent slipping between the paper and the truck. Ans.49: Consider translational motion of cylinder, we get

ma− f =mar Consider translational motion of cylinder, we get

f ⋅ r = Iα = 12mr2α ⇒ f = mr

2

2rarr

⎛

⎝⎜

⎞

⎠⎟=

mar2

Solving we get; ar =2a3


Hence; d = 0+ 122a3

⎛

⎝⎜

⎞

⎠⎟t2 ⇒ t = 3d

a

The distance traveled by the truck relative to ground before the paper rolls off is given

by S = 12at2 = 3d

2

Q50. A metal ball of mass m is put at the point A of a loop track and the vertical distance of A from the lower most point of track is 8 times the radius R of the circular part. The linear velocity of ball when it rolls of the point B to a height R in the circular track will be

(a) 10gR

(b) 7gR5

(c) 7 gR10

(d) 5gR

Ans.50: Applying law of conservation of energy at points A and B, we get;

mg 8R( ) = 12mv2 + 1

2Iω 2 +mgR ⇒ mg 7R( ) = 1

2mv2 + 1

225mR2

⎛

⎝⎜

⎞

⎠⎟vR⎛

⎝⎜

⎞

⎠⎟2

⇒ v = 10gR

Q51. The torque !τ on a body about a given point is found to be

!A x !L where

!A is a

constant vector and !L is angular momentum of the body about that point. From this it

follows that

(a) d!Ldtis perpendicular to

!L at all times. *

(b) The component of !L in the direction of does not change with time. *

(c) The magnitude of !L does not change with time. *

(d) !L does not change with time.

Ans.51: Based on theoretical concept. Answers are (a), (b) and (c). Q52. A spherical body of radius R rolls on a horizontal surface with linear velocity v. Let L1 and L2 be the magnitudes of angular momenta of the body about centre of mass and point of contact P. Then

(a) L2 = 2L1 if radius of gyration K = R. * (b) L2 = 2L1 for all cases (c) L2 > 2L1 if radius of gyration K < R (d) L2 > 2L1 if radius of gyration K > R. *

Ans.52: L1 = Iω =MK 2ω and L2 = Iω +MRv =MK2ω +MR Rω( ) =Mω K 2 + R2( )

Therefore; If K = R; then L2 = 2L1 and If K > R; then L2 > 2L1 Hence answers are (a) and (d) Q53. A hollow sphere of radius R rests on a horizontal surface of finite coefficient of friction. A point object of mass m moved horizontally and hits the sphere at a height of R/2 above its center. The collision is instantaneous and completely inelastic. Which of the following is/are correct?


(a) Total linear momentum of the system is not conserved. * (b) Total angular momentum about center of mass of the system remains conserved* (c) The space gets finite angular velocity immediately after collision* (d) The sphere moves with finite speed immediately after collision*

Ans.53: Impulse due to normal reaction is finite. So friction force gives finite impulse. Therefore frictional torque causes a finite angular impulse about center of mass of system so angular momentum about center of mass of system will change. All other options are correct.

iit advanced assignment with solutions on torque, angular

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