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Systems of Particles and Rotational Motion — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsSystems of Particles and Rotational Motion4 Marks
Question
Read the passage given below and answer the following questions from 1 to 5.

Rolling Motion
The rolling motion can be regarded as the combination of pure rotation and pure translation. It is also one of the most common motions observed in daily life.

Suppose the rolling motion (without slipping) of a circular disc on a level surface. At any instant, the point of contact P0 of the disc with the surface is at rest (as there is no slipping). If v CM is the velocity of centre of mass which is the geometric centre C of the disc, then the translational velocity of disc is v CM , which is parallel to the level surface. Velocity of centre of mass, $\text{V}_\text{cm}=\text{R}\omega$

  1. A solid cylinder is sliding on a smooth horizontal surface with velocity v0 without rotation. It enters on the rough surface. After that it has travelled some distance, the friction force increases its:
  1. translational kinetic energy
  2. rotational kinetic energy
  3. total mechanical energy
  4. angular momentum about an axis passing through point of contact of the cylinder and the surface
  1. A cylinder rolls down an inclined plane of inclination 30°, the acceleration of cylinder is:
  1. $\frac{\text{g}}{3}$
  2. g
  3. $\frac{\text{g}}{2}$
  4. $\frac{2g}{3}$
  1. Sphere is in pure accelerated rolling motion in the figure shown:

Choose the correct option.

  1. The direction of fs is upwards
  2. The direction of fs is downwards
  3. The direction of gravitational force is upwards
  4. The direction of normal reaction is downwards
  1. Kinetic energy of a rolling body will be:
  1. $\frac{1}{2}\text{mv}^2_\text{ cm}(\text{l}+\frac{\text{k}^2}{\text{R}^2}\big)$
  2. $\frac{1}{2}\text{I}\infty^2$
  3. $\frac{1}{2}\text{mv}^2\text{cm}$
  4. (d) None of the above
  1. A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a:
  1. solid sphere
  2. hollow sphere
  3. solid cylinder
  4. hollow cylinder
✓

Answer

  1. (b) rotational kinetic energy

Explanation:

The frictional force will reduce v 0, hence translational KE will also decrease. It will increases w, which increases its rotational kinetic energy. There is no torque about the line of contact, angular momentum will remain constant. The frictional force will decrease the mechanical energy.

  1. a$\frac{\text{g}}{3}$

Explanation:

$\text{a}=\frac{\text{g}\sin\theta}{1+\frac{\text{k}^2}{\text{R}^2}}=\frac{\text{g}\sin30^\circ}{1+\frac{1}{2}}\Rightarrow\text{a}\frac{\frac{\text{g}}{2}}{\frac{3}{2}}=\frac{\text{g}}{3}$

  1. (a) The direction of fs is upwards

Explanation:

As we know that,

The direction of fs will be upwards to provide torque for rolling of sphere.

  1. $\frac{1}{2}\text{mv}^2_\text{ cm}(\text{l}+\frac{\text{k}^2}{\text{R}^2}\big)$

Explanation:

KE of a rolling body = Rotational KE + Translational KE

$=\frac{1}{2}\text{l}\omega^2+\frac{1}{2}\text{mv}^2_\text{cm}(\because\text{I}=\text{mk}^2)$

and $\text{v}_\text{cm}=\text{R}\omega$

$=\frac{1}{2}\frac{\text{mk}^2\text{v}^2_\text{cm}}{\text{R}^2}+\frac{1}{2}\text{mv}^2_\text{cm}$

where, k is the corresponding radius of gyration of the body.

$\frac{1}{2}\text{mv}^2_\text{ cm}(\text{l}+\frac{\text{k}^2}{\text{R}^2}\big)$

  1. (d) hollow cylinder

Explanation:

When a body rolls down on inclined plane, it is accompanied by rotational and translational kinetic energies.

Rotational kinetic energy$=\frac{1}{2}\text{I}\omega^2=\text{K}_\text{R}$

where, I is the moment of inertia and w is the angular velocity. Translational kinetic energy for pure rolling,

$\text{v}_\text{cm}=\text{r}\omega$

$=\frac{1}{2}\text{mv}^2_\text{cm}=\text{K}_\text{r}=\frac{1}{2}\text{m}(r\omega)^2$

where, m is mass of the body, vCM is the velocity and $\omega$ is the angular velocity.
Given,
Translational KE = Rotational KE

$\therefore \frac{1}{2}\text{m}(\text{r}^2\omega^2)=\frac{1}{2}\text{I}\omega^2$

$\Rightarrow\text{I}=\text{mr}^2$

We know that, mr 2 is the moment of inertia of hollow cylinder about its axis, where m is the mass of hollow cylindrical body and r is the radius of the cylinder.

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