Rotational Motion

Practice Questions:

Question 1: difficult

An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is :

1. mg/√3

2. mg/4

3. μmg/√3

4. 3mg/4

View Answer

Toppling will take place about right most point so,

F ×(√3a/2) = mg × a/2

⇒ F= mg/√3

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Question 2: difficult

A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre

1. m g a sin θ

2. (m g a sin θ)/3

3. (m g a sin θ)/4

4. (m g a sin θ)/2

View Answer

As the object slides with constant speed friction force is balancing mgsinθ. Torque of Normal reaction and friction force about centre of object will cancel each other.

τ=mg.sinθ × a/2 = (mg a sinθ)/2

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Question 3: moderate

A wheel having moment of inertia 2 kg-m² about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in one minute would be

1. (2π/15) N-m

2. (π/12) N-m

3. (π/15) N-m

4. (π/18) N-m

View Answer

ω = ω₀ + α.t

0= (60 × 2π /60) + α.60

⇒ α = - π/30

Torque = I α = (2× π/30) = (π/15) N-m

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Question 4: moderate

The figure shows a horizontal block of mass M suspended by two wires A and B. The centre of mass of the block is closer to B than A. (i) Is the magnitude of the torque due to wire A is greater, less or equal to that due to B w.r.t. centre of mass ? (ii) Which wire A or B exerts more force on the block ?

1. (i) greater (ii) B

2. (i) equal (ii) B

3. (i) less (ii) A

4. (i) greater (ii) A

View Answer

As the object is in rotational equilibrium, Net torque acting on the object is zero.

so, Torque of T_A = Torque of T_B

T_AX_A= T_BX_B

\[ \frac{T_{A}}{T_{B}}=\frac{X_{B}}{X_{A}} \]

\[ X_{B} < X_{A} \]

\[ T_{A} < T_{B} \]

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Question 5: moderate

A wire of mass m and length L is bent in the form of a circular ring. The moment of inertia of the ring about its axis is

1. mL²

2. mL²/ 4π²

3. mL²/ 2π²

4. mL²/ 8π²

View Answer

Let the radius of Ring formed is R. Then

2πR= L ⇒ R = L/2π

Moment of Interia= m (L/2π)²= mL²/ 4π²

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Question 6: moderate

Three rings each of mass m and radius r are so placed that they touch each other. The radius of gyration of the system about the axis as shown in the figure is

1. \[ \sqrt{\frac{5}{3}}r \]

2. \[ \sqrt{\frac{5}{6}}r \]

3. \[ \sqrt{\frac{7}{2}}r \]

4. \[ \sqrt{\frac{7}{6}}r \]

View Answer

Moment of Inertia of ring about it's diameter is mR²/2. ( Using Perpendicular axis theorem)

Moment of Inertia about tangential axis in plane of ring will be mR²/2 + mR²= 3/2 mR²

Total moment of inertia about the axis shown in figure

= (3/2 mR² )× 2 + 1/2 mR²= 7/2 mR²

Radius of Gyration is k then 3m×k²= 7/2 mR²

so,

\[ k= \sqrt{\frac{7}{6}}r \]

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Question 7: difficult

A uniform pole of length l and mass m is pivoted on the ground with a frictionless hinge O. The pole is free to rotate without friction about an horizontal axis axis passing through O and normal to plane of the page. The pole makes an angle θ with the horizontal. The pole is released from rest in the position shown, then linear acceleration of the free end (P) of the pole just after its release would be

1. (2g cosθ)/3

2. 2g/3

3. g

4. (3g cosθ)/2

View Answer

Torque of gravitational force mg about O is mgcosθ ×L/2

Torque = I × α

mgcosθ ×L/2 = (mL²/3) × α

α = (3gcosθ)/2

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Question 8: easy

A circular ring of wire of mass M and radius R is making n revolutions/sec about an axis passing through a point on its rim and perpendicular to its plane. The kinetic energy of rotation of the ring is given by

1. 4π²MR²n²

2. 2π²MR²n²

3. π²MR²n²/2

4. 8π²MR²n²

View Answer

Rotational Kinetic Energy = ½ I ω²= ½(2MR²)(2πn)²= 4π²MR²n²

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Question 9: easy

What is moment of inertia in terms of angular momentum (L) and kinetic energy (K)

1. \[ \frac{L²}{K} \]

2. \[ \frac{L²}{2K} \]

3. \[ \frac{L}{2K^{2}} \]

4. \[ \frac{L}{2K} \]

View Answer

\[ K = \frac{1}{2}I \omega^{2} \]

and

L = I ω ⇒ Substituting we get,

\[ I= \frac{L²}{2K} \]

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Question 10: easy

A disc is rolling (without slipping) on a horizontal surface C is its centre and Q and P are two points equidistant from C. Let v_P, v_Q and v_C be the magnitude of velocities of points P, Q and C respectively, then

1. \[ V_{Q}> V_{C}> V_{P} \]

2. \[ V_{Q}< V_{C}< V_{P} \]

3. \[ V_{Q}= V_{P }, V_{C}= \frac{1}{2} V_{P} \]

4. \[ V_{Q}< V_{C}> V_{P} \]

View Answer

During pure rolling the point on ground acts as instantaneous axis of rotation. Distance from the point of contact with ground determines speed of point. so,

\[ V_{Q}> V_{C}> V_{P} \]

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