(2)

[IMAGE 72]

For pure rolling $v_{cm}=r\omega$

Given, $v_{cm}=v$

$\therefore$   $v=r\omega$                               ...(i)

Net velocity is the vector sum of translational and rotational velocity and $\theta$ is the angle between translational and rotational velocity.

$\left|{\overrightarrow{v}}_{\text{net}}\right|$$=\sqrt{v^2+{\left(r\omega \right)}^2+2\left(v\right)\left(r\omega \right)\cos \theta }$                        ...(i)

For point Q, $\theta =180°$

$\left|{\overrightarrow{v}}_Q\right|$$=\sqrt{v^2+r^2{\omega }^2+2rv\omega \cos 180°}$

           $=\sqrt{{(v-r\omega )}^2}=v-r\omega$

$\left|{\overrightarrow{v}}_Q\right|$$=r\omega -r\omega =0 \text{(using (i))}$                            ...(ii)

Now for point P, $\theta =0°$

Using eq. (i), we have

$\left|{\overrightarrow{v}}_P\right|$$=\sqrt{v^2+{\left(r\omega \right)}^2+2\left(v\right)\left(r\omega \right)\cos 0°}$

       $=\sqrt{{(v+r\omega )}^2}=v+r\omega$

Using eq. (i), we have   

$\left|{\overrightarrow{v}}_P\right|$$=2r\omega$                                   ...(iii)

So, it is clear from equation (ii) and (iii) that point P moves faster than point Q.

(2)

Required work done $=-(K_f-K_i)=0+K_i=K_i$

$=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2=\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)\frac{v^2}{R^2}+\frac{1}{2}m{v}^2$

$=\frac{3}{4}m{v}^2=\frac{3}{4}\times 100\times {(20\times {10}^{-2})}^2=3 \text{J}$

(4)

Using law of conservation of energy,

$mgh=\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{4}m{r}^2{\omega }^2$

$\Rightarrow mg s \sin 30°=\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{4}m{v}_{\text{CM}}^2$

$\Rightarrow s=\frac{3}{4}$ $\frac{v_{\text{CM}}^2}{g \sin 30°}$$\Rightarrow s=\frac{3}{4}\times \frac{4^2}{10\times \frac{1}{2}}$

$\Rightarrow s=\frac{3\times 4\times 2}{10}=\frac{12}{5}=2.4 \text{m}$

(2)

Translational kinetic energy, $K_t=\frac{1}{2}m{v}^2$

Rotational kinetic energy, $K_r=\frac{1}{2}I{\omega }^2$

$\therefore$  $K_t+K_r=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{v}{r}\right)}^2$

$\therefore$  $K_t+K_r=\frac{7}{10}m{v}^2$                                                  $[\because I=\frac{2}{5}m{r}^2\text{ (for sphere)}]$

So,   $\frac{K_t}{K_t+K_r}=\frac{5}{7}$

(4)

Time taken by the body to reach the bottom when it rolls down on an inclined plane without slipping is given by

            $t=\sqrt{\frac{2l(1+\frac{k^2}{R^2})}{g\sin \theta }}$

$\text{Since }g\text{ is constant and }l,R\text{ and }\sin \theta \text{ are same for both}$

$\therefore$    $\frac{t_d}{t_s}=\frac{\sqrt{1+\frac{k_d^2}{R^2}}}{\sqrt{1+\frac{k_s^2}{R^2}}}=\sqrt{\frac{1+\frac{R^2}{2R^2}}{1+\frac{2R^2}{5R^2}}}$     $(\because k_d=\frac{R}{\sqrt{2}}, k_s=\sqrt{\frac{2}{5}}R)$

          $=\sqrt{\frac{3}{2}\times \frac{5}{7}}=\sqrt{\frac{15}{14}}\Rightarrow t_d>t_s$

Hence, the sphere gets to the bottom first.

(1)

Acceleration of the solid sphere slipping down the incline without rolling is

          $a_{\text{slipping}}=g\sin \theta$                                ...(i)

Acceleration of the solid sphere rolling down the incline without slipping is

          $a_{\text{rolling}}=\frac{g\sin \theta }{1+\frac{k^2}{R^2}}=\frac{g\sin \theta }{1+\frac{2}{5}}=\frac{5}{7}g\sin \theta$                     ..(ii)

                                                            $(\because \text{For solid sphere, }\frac{k^2}{R^2}=\frac{2}{5})$

Divide eqn. (ii) by eqn. (i), we get

             $\frac{a_{\text{rolling}}}{a_{\text{slipping}}}=\frac{5}{7}$