NEET Rotational Motion MCQs & PYQs

Q: 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

ω = ω0 + α.t 0= (60 × 2π /60) + α.60 ⇒ α = - π/30 Torque = I α = (2× π/30) = (π/15) N-m

Q: 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

Let the radius of Ring formed is R. Then 2πR= L ⇒ R = L/2π Moment of Interia= m (L/2π)²= mL²/ 4π²

Q: 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

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

Q: A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

\[ \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}= mgh =mg\frac{3v^{2}}{4g}= \frac{3}{4}mv^{2} \] \[ \frac{1}{2}I\omega^{2}= \frac{1}{4}mv^{2} \] Solving I = MR²/2 so, Object is a disc or hollow cylinder.

Q: A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be

\[ mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}= \frac{1}{2}mv^{2}+ \frac{1}{2}(\frac{2}{5}mR^{2})\frac{v^{2}}{R^{2}} \] Solving we get, \[ v= \sqrt[]{\frac{10}{7}gh} \]

Q: A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is

Given, rotational kinetic energy is 40% of total energy. so, \[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \] Solving , \[ I = \frac{2}{5}mR^{2} \] Object is Solid Sphere.

Q: A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity . Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

If We take Ring and 4 small blocks as one system net Torque will be zero, Using Principal of conservation of angular momentum. \[ I_{1}\omega= \left( I_{1}+ I_{2} \right)\omega_{1} \] \[ MR^{2}\omega= \left( MR^{2}+ 4mR^{2} \right)\omega_{1} \] \[ \omega_{1}= \left( \frac{M}{M+4m} \right)\omega \]

Q: A particle is projected with a speed v at 45° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

Angular Momentum = momentum × ( Perpendicular distance of momentum from axis of rotation ) Angular Momentum = mv cos (45º) × h = mvh/√2

Q: A particle of mass m = 5 units is moving with a uniform speed v = 3√2 m in the XOY plane along the line Y = X + 4. The magnitude of the angular momentum about origin is

Distance of line \[ ax+by+c=0 \] from point (x1,y1) is given by \[ d = \left( \frac{ax_{1}+ by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right) \] So, distance of direction of velocity from origin is d= 2√2 Angular momentum = Perpendicular distance of momentum × momentum = 2√2 × 5 ×3√2= 60 Unit

Practice NEET Rotational Motion Questions

Question 1:

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

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 2:

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 ?

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 3:

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

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 4:

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

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 5:

moderate

A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

\[ \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}= mgh =mg\frac{3v^{2}}{4g}= \frac{3}{4}mv^{2} \]

\[ \frac{1}{2}I\omega^{2}= \frac{1}{4}mv^{2} \]

Solving I = MR²/2 so, Object is a disc or hollow cylinder.

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Question 6:

moderate

A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be

\[ mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}= \frac{1}{2}mv^{2}+ \frac{1}{2}(\frac{2}{5}mR^{2})\frac{v^{2}}{R^{2}} \]

Solving we get,

\[ v= \sqrt[]{\frac{10}{7}gh} \]

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Question 7:

moderate

A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is

Given, rotational kinetic energy is 40% of total energy. so,

\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]

Solving ,

\[ I = \frac{2}{5}mR^{2} \]

Object is Solid Sphere.

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Question 8:

moderate

A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity . Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

If We take Ring and 4 small blocks as one system net Torque will be zero, Using Principal of conservation of angular momentum.

\[ I_{1}\omega= \left( I_{1}+ I_{2} \right)\omega_{1} \]

\[ MR^{2}\omega= \left( MR^{2}+ 4mR^{2} \right)\omega_{1} \]

\[ \omega_{1}= \left( \frac{M}{M+4m} \right)\omega \]

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Question 9:

moderate

A particle is projected with a speed v at 45° with the horizontal. The magnitude of angular
momentum of the projectile about the point of projection when the particle is at its maximum height h is

Angular Momentum = momentum × ( Perpendicular distance of momentum from axis of rotation )

Angular Momentum = mv cos (45º) × h = mvh/√2

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Question 10:

moderate

A particle of mass m = 5 units is moving with a uniform speed v = 3√2 m in the XOY plane along the line Y = X + 4. The magnitude of the angular momentum about origin is

Distance of line

\[ ax+by+c=0 \] from point (x1,y1) is given by

\[ d = \left( \frac{ax_{1}+ by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right) \]

So, distance of direction of velocity from origin is d= 2√2

Angular momentum = Perpendicular distance of momentum × momentum = 2√2 × 5 ×3√2= 60 Unit

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