Rotational Motion MCQ Questions & Answers in Basic Physics | Physics

$$I = \frac{{\frac{m}{2}{{\left( {\frac{\ell }{2}} \right)}^2}}}{3} + \frac{m}{2}{\left( {\frac{\ell }{2}} \right)^2} = \frac{{m{\ell ^2}}}{6}$$

$$\mu = \frac{F}{R} = \frac{{mg\sin \alpha }}{{mg\cos \alpha }} = \tan \alpha $$

Applying conservation of angular momentum $$I'\omega ' = I\omega $$
$$\eqalign{ & \left( {m{R^2} + 2M{R^2}} \right)\omega ' = m{R^2}\omega \cr & \Rightarrow \omega ' = \omega \left[ {\frac{m}{{m + 2M}}} \right] \cr} $$

angular acceleration $$\alpha = \frac{{3g}}{{2L}}$$

$$\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2} \Rightarrow \theta = 100\,rad$$
∴ Number of revolutions $$ = \frac{{100}}{{2\pi }} = 16\left( {{\text{approx}}{\text{.}}} \right)$$

Angular velocity of flywheel is given by $$\omega = 2\,\pi \nu $$
where, $$\nu $$ is number of revolutions per second or frequency of revolution
Here, $$\nu = 120\,rev/\min $$
$$\eqalign{ & \therefore \omega = \frac{{2\pi \times 120}}{{60}} \cr & = 4\,\pi \,rad/s \cr} $$

Change in angular momentum of the system $$=$$  angular impulse given to the system about the centre of mass $${\left( {{\text{Angular momentum}}} \right)_f} - {\left( {{\text{Angular momentum}}} \right)_i}$$
$$ = Mv \times \frac{L}{2}.....(i)$$
Let the system starts rotating with the angular velocity $$\omega .$$
Moment of Inertia of the system about its axis of rotation [centre of mass of the system]
$$\eqalign{ & = M{\left( {\frac{L}{2}} \right)^2} + M{\left( {\frac{L}{2}} \right)^2} = \frac{{2M{L^2}}}{4} = \frac{{M{L^2}}}{2} \cr & {\text{From, }}(i)\,\,\,\,I\omega - 0 = Mv\frac{L}{2} \cr & \Rightarrow \omega = \frac{{Mv}}{I} \times \frac{L}{2} = \frac{{Mv}}{{\frac{{M{L^2}}}{2}}} \times \frac{L}{2} = \frac{v}{L} \cr} $$

$$\eqalign{ & {I_{AX}} = m{\left( {AB} \right)^2} + m{\left( {OC} \right)^2} \cr & = m{\ell ^2} + m{\left( {{\ell ^2}\cos {{60}^ \circ }} \right)^2} \cr & = m{\ell ^2} + \frac{{m{\ell ^2}}}{4} \cr & = \frac{5}{4}m{\ell ^2} \cr} $$

Let the distance of the centre of mass from the carbon atom be $${x_{{\text{cm}}}}.$$
The mass of carbon, $${m_1} = 12\,amu$$
The mass of oxygen, $${m_2} = 16\,amu\,\,\left[ {{\text{atomic mass unit}}} \right]$$

From definition of centre of mass $${x_{CM}} = \frac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$$
$$\eqalign{ & = \frac{{\left( {12\,amu} \right) \times 0 + \left( {16\,amu} \right) \times r}}{{12\,amu + 16\,amu}} \cr & = \frac{{16}}{{28}}r = \frac{{16}}{{28}} \times 1.12 \times {10^{ - 10}}m \cr & = 0.64 \times {10^{ - 10}}m \cr} $$

Let the mass per unit area be $$\sigma .$$
Then the mass of the complete disc $$ = \sigma \left[ {\pi {{\left( {2R} \right)}^2}} \right] = 4\pi \sigma {R^2}$$
The mass of the removed disc $$ = \sigma {\left( {\pi R} \right)^2} = \pi \sigma {R^2}$$
Let us consider the above situation to be a complete disc of radius $$2R$$  on which a disc of radius $$R$$  of negative mass is superimposed. Let $$O$$  be the origin. Then the above figure can be redrawn keeping in mind the concept of centre of mass as :

$$\eqalign{ & {x_{c.\,m}} = \frac{{\left( {4\pi \sigma {R^2}} \right) \times 0 + \left( { - \pi \sigma {R^2}} \right)R}}{{4\pi \sigma {R^2} - \pi \sigma {R^2}}} \cr & \therefore \,{x_{c.\,m}} = \frac{{ - \pi \sigma {R^2} \times R}}{{3\pi \sigma {R^2}}} \cr & \therefore \,{x_{c.\,m}} = - \frac{R}{3}\,\,\,\, \Rightarrow \alpha = \frac{1}{3} \cr} $$