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Rigid Bodies: Translational Motion And Rotational Motion - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • 12 Questions around this concept.

Solve by difficulty

A heavy spool of inner radius ' 10 \mathrm{~cm} ' and outer radius ' 20 \mathrm{~cm} ' is lying on a rough horizontal plane. Thread of negligible mass is wound over it and is being pulled with a constant force at an angle ' \alpha ' from the vertical as shown in the figure. Find the angle ' \alpha ' for which the centre of the spool remains at rest.

Two balls of mass \mathrm{m_1}=1.0 \mathrm{~kg} and m_2=0.5 \mathrm{~kg} are suspended on two strings of length l_1=10 \mathrm{~cm} and l_2=20 \mathrm{~cm} at the end of a freely hanging rod.

The rod is rotating with an angular velocity of 15 \mathrm{rad} / \mathrm{s} about the vertical axle such that it remains in the vertical position. If the tension in the strings are T_1 and T_2 respectively, then find the sum of T_1 and T_2.

A rod with linear mass density given, as -

        \lambda=\frac{\lambda_0 x^2}{l^2}

where, x is the distance from its left most end and ' l ' is the length of the rod. If the rod is placed on a smooth horizontal plane and a force of constant magnitude F is applied on its rightmost end then the angular acceleration of the rod will be.

Two hollow cylinders of radii ' 1 m ' and ' 2 m ' respectively have common geometrical aixs as shown in the figure.

A fly starts with a constant speed ' v ' and hits the outer cylinder at ' X ' from inside. If both the cylinder starts rotating with an angular velocity of '1 rad/s'. The fly strikes another point ' Y ' where the distance between ' X ' and ' Y ' is 10 \mathrm{~cm}. Assuming the fly moves radially, find the speed of the fly.

A rope is wrapped around the rim of a wheel of moment of inertia 0.10 Kg-m2 and radius  10 cm. The wheel is fixed such that it can freely rotate about its axis. If a person is pulling the rope with the constant force of 10 N. Find the length of the rope detached from the pulley for t=0 to t=10 s

Concepts Covered - 2

Combined rotation and translation motion
  1. Pure Translational motion-

If each particle of it has same velocity/acceleration at a particular instant of time then A body is said to have pure translational motion.

  • Slipping-

It is motion in which  the body slides on a surface without rotation.

Example- Motion of a wheel on a frictionless surface.

Here  friction between the body and surface =f= 0 

Wheel possess only translatory kinetic energy

i.e., - K_T=\frac{1}{2}mv^2

 

  1. Pure rotational motion-

When a body rotates such that its axis of rotation does not move then that body is said to have pure rotational motion.

In pure rotational motion each particle of body has same angular velocity/acceleration about its axis of rotation at a particular instant  of time.

Example- Spinning  of wheel about fixed axis

Here  axis of rotation of a wheel  is fixed.

Here  body possess only rotary kinetic energy.

I.e K_R=\frac{1}{2}I\omega^2

Here Rotational angular momentum = \vec{L}= I\vec{w}

Where I = Moment of inertia about fixed axis of rotation

\omega = angular velocity of rotation

 

Another example- Motion of blades of a fan

 

  1. Combined rotation and translation motion

In this type of motion body is having both rotation and translation motion.

  • Rolling

 In case of rolling motion a body rotates about a fixed axis, and the axis of rotation also moves.

Example- Rolling of football on ground

Here friction between the body and surface = f\neq0

  1. Kinetic energy-

Total kinetic energy of body is sum of both translational and rotational kinetic energy.

K_{net}=K_T+K_R=\frac{1}{2}mV^2+\frac{1}{2}I\omega ^2

Using V=\omega R \ and \ I=mK^2 

K_{net}=K_T+K_R=\frac{1}{2}mV^2(1+\frac{K^2}{R^2})

 

  1. Net Velocity at a point-

\vec{V}_{net}= \vec{V}_{translation }+\vec{V}_{rotation }

                     Where, \vec{V}_{rot }= r\: w

 

Angular momentum in case of Combined rotation and translation motion

           

Angular momentum is always calculated about a particular point.

Net Angular momentum of body is sum of angular momentum due to both translational and rotational motion.

      \vec{L}=\vec{L}_{com}+(m \vec {r} \times \vec {v}_{com})

\vec{L}={I_{com} \omega}+(m \vec {r} \times \vec {v}_{com})

Where L_{com} represents the angular momentum of the body about center of mass and r is the position vector about which we have to calculate the angular momentum.

Study it with Videos

Combined rotation and translation motion
Angular momentum in case of Combined rotation and translation motion

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