Rotational Motion

Given, initial velocity of sphere, $u=2.8 \mathrm{~m} / \mathrm{s}$

Acceleration on the inclined plane,

$a=\frac{g \sin \theta}{1+\frac{k^2}{R^2}}$ .... (i)

where, $k$ and $R$ are the radius of gyration and radius of sphere, respectively.

Moment of inertia of sphere,

$\begin{aligned} I & =m k^2=\frac{2}{5} m R^2 \\ \frac{k^2}{R^2} & =\frac{2}{5} \quad \text{.... (ii)} \end{aligned}$

From Eqs. (i) and (ii), we have,

$\begin{aligned} a & =\frac{g \sin 30^{\circ}}{1+\frac{2}{5}} \quad\left[\because \theta=30^{\circ}, \text { given }\right] \\ & =\frac{5 g}{14}=\frac{25}{7} \mathrm{~m} / \mathrm{s}^2 \quad\left[\therefore g=10 \mathrm{~m} / \mathrm{s}^2\right] \end{aligned}$

If $s$ be the maximum distance travelled by the sphere, then at maximum distance, $v=0$.

$\therefore$ From equation of motion,

$\begin{aligned} v^2 & =u^2-2 a s \\ 0 & =u^2-2 a s \\ \Rightarrow \quad s & =\frac{u^2}{2 a}=\frac{(2.8)^2 \times 7}{2 \times 25}=1.09 \mathrm{~m} \simeq 1.38 \mathrm{~m} \end{aligned}$

Moment of inertia of the insect disc system,

$\mathrm{MI}=\frac{1}{2} M R^2+m x^2$

where, $m=$ mass of insect

and $\quad x=$ distance of insect from centre.

Clearly, as the insect moves along the diameter of the disc. Moment of inertia first decreases and then increases.

By conservation of angular momentum, angular speed first increases and then decreases.

$\begin{aligned} & \text { } \tau_{\mathrm{ext}}=\frac{d L}{d t}=0 \\ & \Rightarrow \frac{d L}{d t}=0 \Rightarrow L=\text { constant, i.e. } L_{\text {initial }}=L_{\text {final }} \end{aligned}$

Here, mass of ring $M=3 \mathrm{~kg}$.

radius of ring $r=0.6 \mathrm{~m}$

force applied $f=3 \mathrm{~N}$

acceleration $a=0.4 \mathrm{~m} / \mathrm{s}^2$

coefficient of friction between stick and ring $=\frac{F}{10}$

Now, according figure $f=M a$

$\begin{aligned} 3-f_s & =f=M a \\ \text{or} \quad f_s & =3-M a \\ & =3-3 \times 0.4=1.8 \mathrm{~N} \end{aligned}$

Now taking torque about $O$ We have

$f_s R-f_k R=I \cdot \alpha$

$=M R^2 \times \frac{a}{R}$

$\begin{array}{r} \left(\because \alpha=\text { angular acceleration }=\frac{a}{R}\right. \\ \text { and } \left.I=M R^2\right) \end{array}$

$\begin{gathered} \text { or } 1.8 \times 0.6-f_k \times 0.6=3 \times 0.6 \times 0.4 \\ f_k=1.8-1.2=0.6 \mathrm{~N} \end{gathered}$

If $\mu$ is the coefficient of kinetic friction, then

$\begin{aligned} f_k & =\mu \times 3 \\ \Rightarrow \quad \mu & =\frac{f_k}{2}=\frac{0.6}{3}=0.2=\frac{2}{10}=\frac{F}{10} \end{aligned}$

Hence on comparing, we get

$F=2 \mathrm{~N}$

The kinetic energy of a rolling solid sphere

$\begin{aligned} & =k_{\text {trans }}+k_{\text {rot }} \\ & =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{5} m R^2 \omega^2 \\ & =\frac{1}{2} m v^2+\frac{1}{5} m v^2 \\ & =7 / 10 ~m v^2 \end{aligned}$

Different solid bodies have different kinetic energy as for different solid bodies, $I$ is different.