Rotational Motion MCQ Questions & Answers in Basic Physics | Physics

This is the case of vertical motion when the body just completes the circle. Here
$$v = \sqrt {5gL} \,\,.....(i)$$

Applying energy conservation,
Total energy at $$A=$$  Total energy at $$P$$
$$\eqalign{ & \frac{1}{2}m{v^2} = \frac{1}{2}m{\left( {\frac{v}{2}} \right)^2} + mgh \cr & \Rightarrow h = \frac{{3{v^2}}}{{8g}} \cr & = \frac{3}{{8g}} \times 5gL = \frac{{15L}}{8}\,\,.....(ii) \cr} $$
In $$\Delta OPM,\,\,\cos \theta = \frac{{L - h}}{L} = \frac{{L - \frac{{15L}}{8}}}{L} = \frac{{ - 7}}{8}$$
Therefore, the value of $$\theta $$ lies in the range $$\frac{{3\pi }}{4} < \theta < \pi $$

Let the $$C.M.$$  be $$'b'.$$ Then,
$$\frac{{\left( {n - 1} \right)mb + ma}}{{mn}} = 0 \Rightarrow b = - \frac{1}{{n - 1}}a$$

The coordinates of C.M of three particle are
$$\eqalign{ & x = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{\;{m_1} + {m_2} + {m_3}}} \cr & \& \,\,y = \frac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{{\;{m_1} + {m_2} + {m_3}}} \cr & {\text{here}}\,{m_1} = {m_2} = {m_3} = m \cr & {\text{so}}\,x = \frac{{\left( {{x_1} + {x_2} + {x_3}} \right)m}}{{m + m + m}} = 2, \cr & y = \frac{{\left( {{y_1} + {y_2} + {y_3}} \right)m}}{{m + m + m}} = 2 \cr} $$

Moment of inertia about $$z$$-axis,
$${I_z} = m{r^2}\,\left( {{\text{about centre of mass}}} \right)$$

Applying parallel axes these,
$$\eqalign{ & {I_z} = {I_{cm}} + m{k^2} \cr & {I_{cm}} = {I_z} - m{\left( {\frac{2}{{\;m}}r} \right)^2} \cr & = m{r^2} - \frac{{m4{r^2}}}{{{\pi ^2}}} = m{r^2}\left( {1 - \frac{4}{{{\pi ^2}}}} \right)\,\,{\text{i}}{\text{.e}}{\text{.,}}\,k = 4 \cr} $$

$$M.I.$$  of a circular wire about an axis $$nn '$$ passing through the centre of the circle and perpendicular to the plane of the circle $$ = M{R^2}$$

As shown in the figure, X-axis and Y-axis lie in the plane of the ring . Then by perpendicular axis theorem
$$\eqalign{ & {I_X} + {I_Y} = {I_Z} \cr & \Rightarrow 2{I_X} = M{R^2}\left[ {\because {I_X} = {I_Y}\left( {{\text{by symmetry}}} \right){I_Z} = M{R^2}} \right] \cr & \therefore {I_X} = \frac{1}{2}M{R^2} \cr} $$

Angular velocity of particle is given by $$\omega = \frac{{2\pi }}{T}$$
$${\text{or}}\,\,\omega \propto \frac{1}{T}\,\,\,\left[ {T = {\text{time period of the particle}}} \right]$$
It simply implies that $$\omega $$ does not depend on mass of the body and radius of the circle.
$$\therefore \frac{{{\omega _1}}}{{{\omega _2}}} = \frac{{{T_2}}}{{{T_1}}}$$
but given time period is same, i.e. $${T_1} = {T_2}$$
Hence, $$\frac{{{\omega _1}}}{{{\omega _2}}} = \frac{1}{1}$$

Angular momentum will remain the same since external torque is zero.

$$\eqalign{ & {\text{For}}\,{\text{rolling, }}{v_A} = 2\,m/s \cr & {\text{or}}\,4 - 1.\omega = 2 \cr & {\text{or}}\,\omega = 2\,rad/\sec \,\left( {{\text{clockwise}}} \right) \cr} $$

The distribution of mass about axis $$EF$$  is minimum so radius of gyration is minimum and therefore moment of inertia is minimum about $$EF.$$

Moment of inertia of a hollow cylinder of mass $$M$$ and radius $$r$$ about its own axis is $$M{r^2}.$$