Rotational Motion

The given situation is as shown below,

Here, $O$ is the point where centre of mass of whole strip exist.

The point $O^{\prime}$ is 100 cm or 1 m away from one end. So, its distance from centre of mass or from point $O$ is 1.5 m .

∴ Moment of inertia about an axis perpendicular to strip at point $O^{\prime}$ can be calculated using parallel axis theorem as

$ \begin{aligned} I^{\prime} & =I_{C M}+M R^2=\frac{M L^2}{12}+M R^2 \\ & =\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5 \\ & =2.08+2.25=4.33 \mathrm{~kg} \mathrm{~m}^2 \end{aligned} $

According to the question, given situation is shown below

In $\triangle A B C$,

$ \sin \theta=\frac{B C}{A C} \Rightarrow B C=A C \sin \theta=l \sin \theta $

From law of conservation of mechanical energy,

$ \begin{aligned} m g(l \sin \theta)= & \frac{1}{2} I \omega^2+\frac{1}{2} m v^2 \\ m g(l \sin \theta)= & \frac{1}{2}\left(\frac{1}{2} m r^2\right)\left(\frac{v}{r}\right)^2+\frac{1}{2} m v^2 \\ & {\left[\because I=\frac{1}{2} m r^2 \text { and } \omega=\frac{v}{r}\right] } \end{aligned} $

$ \begin{aligned} m g l \sin \theta & =\left(\frac{1}{4}+\frac{1}{2}\right) m v^2 \\ g l \sin \theta & =\frac{3}{4} v^2 \Rightarrow \quad v^2=\frac{4 g l \sin \theta}{3} \\ \Rightarrow \quad v & =\sqrt{\frac{4 g l \sin \theta}{3}} \\ & =\sqrt{\frac{4 \times 10 \times 0.6 \times \sin 30^{\circ}}{3}} \end{aligned} $

$ \begin{aligned} & \quad[\because l=60 \mathrm{~cm}=0.6 \mathrm{~m}] \\ &=\sqrt{\frac{12}{3}}=\sqrt{4}=2 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \begin{aligned} &\text { Mass per unit length, }\\ &\frac{d M}{d x}=\lambda|x| \Rightarrow d M=\lambda|x| d x \end{aligned} $

$ \begin{aligned} &\text { Moment of inertia about axis } A A\\ &I_{A A}=\int x^2 d M \end{aligned} $

$ \begin{aligned} &\text { Put the value of } d M\\ &\begin{aligned} I_{A A}= & \int_{-L / 2}^{+L / 2} x^2 \times \lambda|x| d x \\ = & \int_{-L / 2}^{+L / 2} x^2 \times 16|x| d x \\ & \quad\left[\because \lambda=16 \mathrm{~kg} / \mathrm{m}^2 \text { (given) }\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =16 \int_{-L / 2}^{+L / 2} x^2|x| d x \\ & =16\left[\int_{-L / 2}^0 x^2(-x) d x+\int_0^{+L / 2} x^2(+x) d x\right] \\ & \quad\left[\because|x|=\left\{\begin{array}{ll} -x & ; \\ +x & \text { if } x<0 \\ +x & \text { if } x \geq 0 \end{array}\right]\right. \end{aligned} $

$ \begin{aligned} &\text { So, }\\ &\begin{aligned} I_{A A} & =16\left[\int_{-L / 2}^0\left(-x^3\right) d x+\int_0^{+L / 2}\left(x^3\right) d x\right] \\ & =16\left[-\left[\frac{x^4}{4}\right]_{-L / 2}^0+\left[\frac{x^4}{4}\right]_0^{+L / 2}\right] \\ & =4\left[-\left[x^4\right]_{-L / 2}^0+\left[x^4\right]_0^{+L / 2}\right] \\ & =4\left[\begin{array}{l} \left.-\left\{(0)^4-\left(-\frac{L}{2}\right)^4\right\}\right] \\ \left.+\left\{\left(+\frac{L}{2}\right)^4-(0)^4\right\}\right] \end{array}\right. \end{aligned} \end{aligned} $

$ \begin{aligned} & =4\left[\left\{0-\frac{L^4}{16}\right\}+\left\{\frac{L^4}{16}-0\right\}\right] \\ & =4\left[\frac{L^4}{16}+\frac{L^4}{16}\right]=4\left[\frac{L^4}{8}\right]=\frac{L^4}{2} \end{aligned} $

Now by the theorem of parallel axes, the moment of inertia about axis $B B$,

$ \begin{aligned} I_{B B} & =I_{\mathrm{COM}}+\int\left(\frac{L}{2}\right)^2 d M \\ & =I_{\mathrm{COM}}+\frac{L^2}{4} \int d M \end{aligned} $

$ \begin{aligned} &\text { Put the value of } d M \text { and } I_{\text {COM }}\\ &\begin{aligned} I_{B B} & =I_{A A}+\frac{L^2}{4} \int_0^{L / 2} \lambda|x| d x \\ & =\frac{L^4}{2}+\frac{L^2}{4} \lambda \int_0^{L / 2}|x| d x \\ & =\frac{L^4}{2}+\frac{L^2}{4} \times 16 \int_0^{L / 2}(+x) d x \\ & \left.\quad [\because \lambda=16 \mathrm{~kg} / \mathrm{m}^2 \text { (given) }\right] \\ & =\frac{L^4}{2}+4 L^2\left[\frac{x^2}{2}\right]_0^{L / 2} \\ & =\frac{L^4}{2}+2 L^2\left[x^2\right]_0^{L / 2} \\ & =\frac{L^4}{2}+2 L^2\left[\left(\frac{L}{2}\right)^2-(0)^2\right] \\ & =\frac{L^4}{2}+2 L^2\left[\frac{L^2}{4}-0\right] \\ & =\frac{L^4}{2}+\frac{L^4}{2}=L^4=1^4 \\ & =1 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} \end{aligned} $

Given, mass of solid cylinder, $M=10 \mathrm{~kg}$

Radius, $R=40 \mathrm{~cm}=0.4 \mathrm{~m}$

$ L=9 R=9 \times 0.4=3.6 \mathrm{~m} $

Moment of inertia of the solid cylinder about the line passing through its centre and perpendicular to its axis is given as

$ \begin{aligned} I_{\mathrm{COM}} & =M\left[\frac{L^2}{12}+\frac{R^2}{4}\right] \\ & =10\left[\frac{(3.6)^2}{12}+\frac{(0.4)^2}{4}\right] \\ & =10[1.08+0.04]=10 \times 1.12 \\ \Rightarrow \quad I_{\mathrm{COM}} & =11.2 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} $

According to parallel axis theorem, moment of inertia of cylinder about a line passing through its edge and perpendicular to its axis is given as

$ \begin{aligned} I^{\prime} & =I_{\mathrm{COM}}+M\left(\frac{L}{2}\right)^2 \\ & =11.2+10\left(\frac{3.6}{2}\right)^2 \\ & =11.2+10(3.24) \\ & =11.2+324=43.6 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} $

$ \text { We take origin as fixed point. } $

Now, angular momentum about origin $O=$ angular momentum due to linear motion of centre of mass + angular momentum due to rotation of body about its' centre of mass

$ \begin{aligned} & \\ & \begin{aligned}& \\ \therefore L_O=L_{\text {linear motion }} & +L_{\text {rotational motion }} \\ & =m v R+I \omega \\ & =m R^2 \omega+I \omega \quad[\therefore v=R \omega] \\ & =\left(m R^2+I\right) \omega \end{aligned} \end{aligned} $

For solid sphere, $I=\frac{2}{5} m R^2$ and for solid cylinder, $I=\frac{1}{2} m R^2$

$ \begin{aligned} \text { So, }\left(L_{\text {sphere }}\right)_{\text {about ' } O^{\prime}} & =\left(m R^2+\frac{2}{5} m R^2\right) \omega \\ & =\frac{7}{5} m R^2 \omega \\ \text { and }\left(L_{\text {sphere }}\right)_{\text {about ' } O^{\prime}} & =\left(m R^2+\frac{1}{2} m R^2\right) \omega \\ & =\frac{3}{2} m R^2 \omega \end{aligned} $

$ \begin{aligned} & \text { So, ratio of } \frac{L_{\text {sphere }}}{L_{\text {cylinder }}}=\frac{\frac{7}{5} m R^2 \omega}{\frac{3}{2} m R^2 \omega} \\ & \Rightarrow \quad \frac{\frac{L_1}{L_2}}{2}=\frac{14}{15} \end{aligned} $

Let block falls through a height $h$, then loss of PE of block appears in form of KE of block and KE of rotation of pulley.

$ \Rightarrow \quad m g h=\frac{1}{2} m v^2+\frac{1}{2} \frac{M R^2}{2} \cdot \omega^2 $

Here, $\quad m=M=2 \mathrm{~kg}$,

So, $\quad g h=\frac{v^2}{2}+\frac{R^2}{4} \omega^2$

As $v=R \omega, g h=\frac{3}{4} R^2 \omega^2$

Differentiating with respect to time,

$ \begin{gathered} g \cdot \frac{d h}{d t}=\frac{3}{4} R^2 \cdot 2 \omega \cdot \frac{d \omega}{d t} \\ g v=\frac{3}{4} R^2 \times 2 \times \frac{v}{R} \cdot \alpha \\ \Rightarrow g=\frac{6}{4} R \alpha \Rightarrow \frac{4 g}{6}=R \alpha \Rightarrow R \alpha=\frac{2 g}{3} \end{gathered} $

But $R \alpha=a_t=$ tangential acceleration of pulley

= acceleration of block

So, $a_t=R \alpha=\frac{2}{3} g=\frac{20}{3} \mathrm{~m} / \mathrm{s}^2$

From 2nd equation of rotational motion,

$\theta = {\omega _0}t + {1 \over 2}\alpha {t^2}$

Given, $\theta$ = 5 rad, t = 1s, $\omega$₀ = 0

Substituting all these values in above equation, we get

$5 = 0 + {1 \over 2}\alpha {(1)^2} \Rightarrow \alpha = 0$ rad/s²

The angle rotated by wheel of car during first two seconds,

${\theta _2} = 0 + {1 \over 2} + \alpha {t^2} = {1 \over 2} \times 10 \times {2^2} = 20$ radians

Hence, angle rotated during the 2nd second

= $\theta$₂ = 0

= 2 $-$ 5 = 15 radians