Rotational Motion MCQ Questions & Answers in Basic Physics | Physics

Velocity at the bottom and top of the circle is $$\sqrt {5gr} $$  and $$\sqrt {gr} .$$  Therefore $$\left( {\frac{1}{2}} \right)M\left( {5gr} \right) = MgH\,{\text{and}}\,\left( {\frac{1}{2}} \right)M\left( {gr} \right) = Mgh.$$

Angular momentum about the point of contact, for solid homogeneous sphere of mass $$M$$ and radius $$R$$ is conserved.

Possible when $${m_2} = {m_4}$$

$$\eqalign{ & {X_{cm}} = \frac{{m \times 5.5 + 1 \times m}}{{2m}} = \frac{{6.5}}{2} = \frac{{13}}{4}cm. \cr & {Y_{cm}} = \frac{{m \times 3.5 + 1 \times m}}{{2m}} = \frac{{4.5}}{2} = \frac{9}{4}cm. \cr} $$

As insect moves along a diameter, the effective mass and hence the M.I. first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.

Similarly, velocity of point $$A$$ is given by
$${v_{AB}} = {\text{velocity of centre of mass}}\left( {{v_{CM}}} \right) + {\text{Linear velocity of point }}A\left( {R\omega } \right){\text{ }}$$
$$\eqalign{ & = {v_{CM}} + {v_{CM}}\,\,\left( {\because {v_{CM}} = R\omega } \right) \cr & = 2{v_{CM}} \cr} $$
Velocity of point $$B$$ is,
$$\eqalign{ & {v_B} = {v_{CM}} - R\omega \cr & = {v_{CM}} - {v_{CM}} \cr & = 0 \cr} $$
Thus, the velocity of point $$A$$ is $$2{v_{CM}}$$  and velocity of point $$B$$ is zero.

As for linear motion $$KE = \frac{{{p^2}}}{{2m}}$$
Similarly, for rotational motion $$K{E_{rot}} = \frac{{{L^2}}}{{2I}}$$
As said, $${\left( {KE} \right)_{rot}}$$   remains same.
i.e. $$\frac{1}{2}{I_1}\omega _1^2 = \frac{1}{2}{I_2}\omega _2^2$$
$$\eqalign{ & \Rightarrow \frac{1}{{2{I_1}}}{\left( {{I_1}{\omega _1}} \right)^2} = \frac{1}{{2{I_2}}}{\left( {{I_2}{\omega _2}} \right)^2} \cr & \Rightarrow \frac{{L_1^2}}{{{I_1}}} = \frac{{L_2^2}}{{{I_2}}} \cr & \Rightarrow \frac{{{L_1}}}{{{L_2}}} = \sqrt {\frac{{{I_1}}}{{{I_2}}}} \cr & {\text{but}}\,{I_1} = I,\,{I_2} = 2I \cr & \therefore \frac{{{L_1}}}{{{L_2}}} = \sqrt {\frac{I}{{2I}}} = \frac{1}{{\sqrt 2 }} \cr & {\text{or}}\,\,{L_1}:{L_2} = 1:\sqrt 2 \cr} $$

Apply parallel axes theorem of moment of inertia.
According to parallel axes theorem of moment of inertia,
$$\eqalign{ & I = {I_{CM}} + M{h^2} \cr & {\text{So,}}\,\,I = {I_0} + M{\left( {\frac{L}{2}} \right)^2} \cr & \Rightarrow I = {I_0} + \frac{{M{L^2}}}{4} \cr} $$

$$\eqalign{ & {I_{nn'}} = \frac{1}{{12}}m\left( {{a^2} + {a^2}} \right) = \frac{{m{a^2}}}{6} \cr & {\text{Also, }}DO = \frac{{DB}}{2} = \frac{{\sqrt 2 a}}{2} = \frac{a}{{\sqrt 2 }} \cr} $$
According to parallel axis theorem
$$\eqalign{ & {I_{mm'}} = {I_{nn'}} + m{\left( {\frac{a}{{\sqrt 2 }}} \right)^2} \cr & = \frac{{m{a^2}}}{6} + \frac{{m{a^2}}}{2} \cr & = \frac{{m{a^2} + 3m{a^2}}}{6} \cr & = \frac{2}{3}m{a^2} \cr} $$

A solid sphere rolling without slipping down an inclined plane

In this case, $${a_1} = \frac{{g\sin \theta }}{{1 + \frac{{{k^2}}}{{{R^2}}}}} = \frac{{g\sin \theta }}{{1 + \frac{{\left( {\frac{2}{5}} \right){R^2}}}{{{R^2}}}}}\,\,\left[ {\therefore {\text{for solid sphere,}}\,\,{k^2} = \frac{2}{5}{R^2}} \right]$$
$$ = \frac{{g\sin \theta }}{{\frac{7}{5}}} \Rightarrow {a_1} = \frac{5}{7}g\sin \theta $$
For a sphere slipping down an inclined plane
$$\eqalign{ & \Rightarrow {a_2} = g\sin \theta \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{\frac{5}{7}g\sin \theta }}{{g\sin \theta }} \cr & \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{5}{7} \cr} $$