Rotational Motion MCQ Questions & Answers in Basic Physics | Physics

Let $$v$$ be the linear velocity of centre of mass of the spherical body and $$w$$ its angular velocity about centre of mass. Then $$\omega = \frac{v}{{2R}}$$
KE of spherical body $${K_1} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$$
$${K_1} = \frac{1}{2}m{v^2} + \frac{1}{2}{\left( {2mR} \right)^2}\left( {\frac{{{v^2}}}{{4{R^2}}}} \right) = \frac{3}{4}m{v^2}\,......\left( {\text{i}} \right)$$
Speed of the block will be
$$\eqalign{ & v' = \left( \omega \right)\left( {3R} \right) = 3R\omega = \left( {3R} \right)\left( {\frac{v}{{2R}}} \right) = \frac{3}{2}v \cr & \therefore {\text{KE}}\,{\text{of}}\,{\text{block}}\,{K_2} = \frac{1}{2}m{v^2} \cr & = \frac{1}{2}m{\left( {\frac{3}{2}v} \right)^2} = \frac{9}{8}m{v^2}\,.....\left( {{\text{ii}}} \right) \cr} $$
From equations (i) and (ii),
$$\frac{{{K_1}}}{{{K_2}}} = \frac{2}{3}$$

By definition $$\alpha = \frac{{d\omega }}{{dt}}$$
i.e. $$d\omega = \alpha \,dt$$
So, if in time $$t$$ the angular speed of a body changes from $${\omega _0}$$ to $$\omega $$
$$\int_{{\omega _0}}^\omega {d\omega = \int_0^t {\alpha \,dt} } $$
If $$\alpha $$ is constant $$\omega - {\omega _0} = \alpha t\,\,{\text{or}}\,\,\omega = {\omega _0} + \alpha t\,......\left( {\text{i}} \right)$$
Now, as by definition $$\omega = \frac{{d\theta }}{{dt}}$$
Eq. (i) becomes $$\frac{{d\theta }}{{dt}} = {\omega _0} + \alpha t$$
i.e. $$d\theta = \left( {{\omega _0} + \alpha t} \right)dt$$
So, if in time $$t$$ angular displacement is $$\theta .$$
$$\int_0^\theta {d\theta = \int_0^1 {\left( {{\omega _0} + \alpha t} \right)dt} \,\,{\text{or}}\,\,\theta = } {\omega _0}t + \frac{1}{2}\alpha {t^2}\,......\left( {{\text{ii}}} \right)$$
Given $$\alpha = 3\,rad/{s^2}$$
$${\omega _0} = 2\,rad/s,\,t = 2s$$
Here, $$\theta = 2 \times 2 + \frac{1}{2} \times 3 \times {\left( 2 \right)^2}$$
or, $$\theta = 4 + 6 = 10\,rad$$
Alternative
As we know that equation of circular motion
$$\theta = \omega t + \frac{1}{2}\alpha {t^2}\,\,\left( {{\text{where symbols have their usual meaning}}} \right)$$
Putting the value of $$\omega ,t,\alpha $$  from question.
So, $$\theta = 2 \times 2 + \frac{1}{2} \times 3 \times 2 \times 2 = 10\,rad$$

Acceleration of an object rolling down an inclined plane is given by $$a = \frac{{g\sin \theta }}{{1 + \frac{I}{{m{r^2}}}}}$$
where, $$\theta =$$  angle of inclination of the inclined plane
$$m =$$  mass of the object
$$I =$$  moment of inertia about the axis through centre of mass
For disc, $$\frac{I}{{m{r^2}}} = \frac{{\frac{1}{2}m{r^2}}}{{m{r^2}}} = \frac{1}{2}$$
For solid sphere, $$\frac{I}{{m{r^2}}} = \frac{{\frac{2}{5}m{r^2}}}{{m{r^2}}} = \frac{2}{5}$$
For hollow sphere, $$\frac{I}{{m{r^2}}} = \frac{{\frac{2}{3}m{r^2}}}{{m{r^2}}} = \frac{2}{3}$$
$$\eqalign{ & \therefore {a_{{\text{disc}}}} = \frac{{g\sin \theta }}{{1 + \frac{1}{2}}} = \frac{2}{3}g\sin \theta = 0.66\,g\sin \theta \cr & {a_{{\text{solid sphere}}}} = \frac{{g\sin \theta }}{{1 + \frac{2}{5}}} = \frac{5}{7}g\sin \theta = 0.71\;g\sin \theta \cr & {a_{{\text{hollow sphere}}}} = \frac{{g\sin \theta }}{{1 + \frac{2}{3}}} = \frac{3}{5}g\sin \theta = 0.6g\sin \theta \cr} $$
Cleraly, $${a_{{\text{solid}}\,{\text{sphere}}}} > {a_{{\text{disk}}}} > {a_{{\text{hallow}}\,{\text{sphere}}}}$$
Type of sphere is not mentioned in the question. Therefore, we will assume the given sphere as solid sphere.
$$\therefore {a_{{\text{solid}}\,{\text{sphere}}}} = {a_{{\text{hallow}}\,{\text{sphere}}}} > {a_{{\text{disk}}}}$$

$$\eqalign{ & {X_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}} \cr & = \frac{{1 \times 0 + 1 \times 3 + 1 \times 0}}{{1 + 1 + 1}} = 1 \cr & {Y_{cm}} = \frac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}} \cr & = \frac{{1 \times 0 + 1 \times 0 + 1 \times 4}}{{1 + 1 + 1}} = \frac{4}{3} \cr} $$
Therefore the coordinates of centre of mass are $$\left( {1,\frac{4}{3}} \right).$$

Kinetic energy of rotation is $${K_{{\text{rot}}}} = \frac{1}{2}I{\omega ^2} = \frac{1}{2}M{k^2}\frac{{{v^2}}}{{{R^2}}}\,\,\left[ {K = \sqrt {\frac{I}{M}} } \right]$$
where, $$k$$ is radius of gyration.
Kinetic energy of translation is $${K_{{\text{trans}}}} = \frac{1}{2}M{v^2}$$
Thus, total energy $$E = {K_{{\text{rot}}}} + {K_{{\text{trans}}}}$$
$$\eqalign{ & = \frac{1}{2}M{k^2}\frac{{{v^2}}}{{{R^2}}} + \frac{1}{2}M{v^2} \cr & = \frac{1}{2}M{v^2}\left( {\frac{{{k^2}}}{{{R^2}}} + 1} \right) \cr & = \frac{1}{2}\frac{{M{v^2}}}{{{R^2}}}\left( {{k^2} + {R^2}} \right) \cr & {\text{Hence,}}\,\,\frac{{{K_{{\text{rot}}{\text{.}}}}}}{E} = \frac{{\frac{1}{2}M{k^2}\frac{{{v^2}}}{{{R^2}}}}}{{\frac{1}{2}\frac{{M{v^2}}}{{{R^2}}}\left( {{k^2} + {R^2}} \right)}} \cr & = \frac{{{k^2}}}{{{k^2} + {R^2}}} \cr} $$

Torque $$\vec \tau = \vec r \times \vec F = \left( {\hat j - \hat i} \right) \times \left( - {F\hat k} \right) = - F\left( {\hat i + \hat j} \right)$$

The moment of inertia of the rod about $$O$$ is $$\frac{1}{3}m{\ell ^2}.$$  The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of of the rod in this condition is $$\frac{1}{2}I{\omega ^2}$$   where $$I$$ is the moment of inertia of the rod about $$O.$$  When the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh where $$h$$ is the maximum height to which the centre of mass (C.M.) rises
$$\therefore mgh = \frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{1}{3}m{l^2}} \right){\omega ^2}\,\, \Rightarrow h = \frac{{{\ell ^2}{\omega ^2}}}{{6g}}$$

The moment of inertia of solid sphere $$A$$  about its diameter $${I_A} = \frac{2}{5}M{R^2}$$
The moment of inertia of a hollow sphere $$B$$  about its diameter $${I_B} = \frac{2}{3}M{R^2}$$
$$\therefore \,\,{I_A} < {I_B}$$

When sphere rolls, then it has both translational and rotational kinetic energy
$$\eqalign{ & \therefore K = {K_{{\text{rot}}}} + {K_{{\text{trans}}}} \cr & = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} \cr} $$
$$\because $$ Moment of inertia of the sphere about its diameter is $$I = \frac{2}{5}m{r^2}$$
$$\eqalign{ & \therefore K = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\omega ^2} + \frac{1}{2}m{v^2}\,\,\left( {{\text{as}}\,v = r\omega } \right) \cr & \therefore K = \frac{1}{5}m{v^2} + \frac{1}{2}m{v^2} = \frac{7}{{10}}m{v^2} \cr & \therefore \frac{{{K_t}}}{K} = \frac{{\frac{1}{2}m{v^2}}}{{\frac{7}{{10}}m{v^2}}} = \frac{5}{7} \cr} $$

A flywheel is a large heavy wheel with a long cylindrical axle supported on ball bearings. Its centre of mass lies on its axis of rotation, so that it remains at rest in any position. Rotational kinetic energy of flywheel is given by $${K_{{\text{rot}}}} = \frac{1}{2}I{\omega ^2}$$
where, $$I =$$  moment of inertia of the wheel about the axis of rotation
$$\omega =$$  angular velocity of flywheel
Given, Rotational kinetic energy $${K_r} = 360\,J$$
Angular velocity $$\omega = 30\,rad/s$$
$$\eqalign{ & \therefore I = \frac{{2{K_r}}}{{{\omega ^2}}} = \frac{{2 \times 360}}{{{{\left( {30} \right)}^2}}} \cr & = 0.8\,kg{\text{-}}{m^2} \cr} $$