Rotational Motion

Given, mass of solid sphere = 80 kg

radius of solid sphere, R_s = 15 m

radius of circular disc, R_c = 20 m

and time = 1 hour = 60 minute = 60 $\times$ 60 sec

$\therefore$ Moment of inertia of solid sphere, I_s = $\frac{2}{5}$ MR²

= $\frac{2}{5}$ $\times$ 80 $\times$ (15)²

= 7200 kg-m²

Similarly,

moment of inertia of the disc, I_c = $\frac{1}{2}$ MR$_c^2$

= $\frac{1}{2}$ $\times$ 80 $\times$ (20)²

= 16000 kg-m²

Rate of change of moment of Inertia $ = {{{I_c} - {I_s}} \over t}$

$ = {{16000 - 7200} \over {60 \times 60}}$

= $\frac{22}{9}$ kg-m² s^$-$1

Given, mass of solid sphere $=M$

Radius of solid sphere $=R$

Angular frequency, $\omega=600 \mathrm{rpm}=20 \pi \mathrm{~rad} / \mathrm{s}$

As we know that,

Kinetic energy, $E=\frac{1}{2} I \omega^2$

$\begin{aligned} & =\frac{1}{2} \times \frac{2}{5} M R^2(20 \pi)^2 \\ & =\frac{M R^2}{5}\left(400 \pi^2\right)=80 \pi^2 M R^2\end{aligned}$

We know the moment of inertia is a measure of how hard it is to spin an object. For wheel $A$, it is $5~\mathrm{kg}-\mathrm{m}^2$. For wheel $B$, it is $20~\mathrm{kg}-\mathrm{m}^2$.

Wheel $A$ is spinning at $10$ revolutions per second ($10~\mathrm{rps}$), and wheel $B$ is not moving at all at the start ($0~\mathrm{rps}$).

Let’s call the speed that both wheels spin at after they are connected $\omega$.

When the wheels are joined, the total rotational momentum (angular momentum) stays the same, because there is no friction or any loss.

The total rotational momentum before the clutch is connected is:
$I_1 \omega_1 + I_2 \omega_2$

After they are joined, the total moment of inertia is $I_1 + I_2$ and both wheels move at the new speed $\omega$:

$ I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega $

Now substitute the numbers:
$ 5 \times 10 + 20 \times 0 = (5+20) \omega $

$ 50 = 25\omega $

$ \omega = \frac{50}{25} = 2~\mathrm{rps} $

This means after the clutch is connected, both wheels will rotate together at $2~\mathrm{rps}$.

Moment of inertia of sphere and hollow cylinder are,

$I_{sphere}=\frac{2}{5}MR^2$

and $I_{cylinder}=MR^2$

Angle of inclination, $\theta=30\Upsilon$

Let, acceleration of body = a

Friction force = $f$

Mass of body = $m$

Radius of body = R

$mg\sin\theta-f=ma$ ..... (i)

Torque, $\tau=Rf=I\alpha$

$\Rightarrow f=I\alpha/R$ ..... (ii)

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

$m g \sin \theta-\frac{I \alpha}{R}=m a$

Angular acceleration, $\alpha=a / R$

$\begin{array}{ll} \Rightarrow & m g \sin \theta-\frac{I a}{R^2}=m a \\ \Rightarrow & m g \sin \theta=a\left(m+\frac{I}{R^2}\right) \\ \Rightarrow & a=\frac{m g \sin \theta}{m+\frac{I}{R^2}}=\frac{g \sin \theta}{1+\frac{I}{m R}} \end{array}$

$\because$ Distance covered by body, $s=\frac{1}{2} a t^2$

$\begin{aligned} & \because \quad s_{\text {sphere }}=s_{\text {cylinder }} \\ & \frac{1}{2} a_{\text {sphere }} \times t^2=\frac{1}{2} \times s_{\text {cylinder }} \times t^2 \\ & \Rightarrow \quad a_{\text {sphere }}=a_{\text {cylinder }} \\ & \Rightarrow \quad \frac{g \sin 30 \Upsilon}{1+\frac{\frac{2}{5} m R^2}{m R^2}}=\frac{g \sin \theta}{1+\frac{m R^2}{m R^2}} \\ & \Rightarrow \quad \frac{1 / 2}{7 / 5}=\frac{\sin \theta}{2} \Rightarrow \sin \theta=\frac{5}{7} \\ & \Rightarrow \quad \theta=\sin ^{-1}\left(\frac{5}{7}\right)=45.6 \Upsilon \approx 45 \Upsilon \\ & \end{aligned}$

Given,

Diameter of sphere = 2a

Radius of sphere = a

Mass of sphere = m

Side of square = b

$\because$ Moment of inertia of sphere, $I=\frac{2}{5} m a^2$

From parallel axis theorem, $I_p=I_{\mathrm{CM}}+m R^2$

So, net moment of inertia on sphere

$\begin{aligned} 1 & =2 \times\left(\frac{2}{5} m a^2+m b^2\right)+2 \times\left(\frac{2}{5} m a^2\right) \\ & =\frac{8 m a^2}{5}+2 m b^2 \end{aligned}$

Given, Energy, $E=684 \mathrm{~J}$

Initial angular frequency, $\omega_i=180 \mathrm{~rpm}=6 \pi$

Final angular frequency, $\omega_f=360 \mathrm{~rpm}=12 \pi$

Let, $I$ be the moment of inertia.

As we know that,

$\begin{aligned} E & =\frac{1}{2} I\left(\omega_f^2-\omega_i^2\right) \\ 684 & =\frac{1}{2} I\left[(12 \pi)^2-(6 \pi)^2\right] \\ \Rightarrow \quad I & =\frac{684 \times 2}{\left(144 \pi^2-36 \pi^2\right)}=\frac{684 \times 2}{1065.9}=1.28 \mathrm{~kg}-\mathrm{m}^2 \end{aligned}$

Given, mass of sphere $=m$

Spring constant $=k$

Angle of inclined $=\theta$

By using law of conservation of energy,

$\begin{aligned} \frac{1}{2} k x^2 & =m g x \sin \theta \\ \Rightarrow \quad x^2 & =\frac{2 m g x \sin \theta}{k} \\ x & =\frac{2 m g \sin \theta}{k} \end{aligned}$

Given, mass of girl = M

Radius of marry-go-round = R

Rotational inertia of girl = I

Mass of rock = m

Speed of rock with respect to ground = v

Let linear speed of girl be v.

By using law of conservation of angular momentum,

Initial angular momentum $\left(L_i\right)=$ Final angular momentum $L_f$

$\begin{array}{llrl} \Rightarrow & 0 =m v R-\left(I+M R^2\right) \frac{v^{\prime}}{R} \\\\ \Rightarrow & m v R =\left(I+M R^2\right) \frac{v^{\prime}}{R} \\\\ \Rightarrow & m v R^2 =\left(I+M R^2\right) v^{\prime} \\\\ & \therefore \quad v^{\prime}=\frac{m v R^2}{I+M R^2} \end{array}$

Given, mass $M$ and radius $R$ of wheels are same

As we know that,

Moment of inertia (I) for following bodies are

(i) $\operatorname{Ring}=M R^2$

(ii) Angular disc $=\frac{M}{2}\left(R^2-r^2\right)$

(iii) Solid disc $=M R^2 / 2$

(iv) Cylindrical disc $=M R^2 / 2$

Hence, moment of inertia of ring among all four is maximum.

Given, m = 1 kg, $r = 3t\widehat i + 4\widehat j$

Velocity of body, $v = {{dr} \over {dt}}$

$ = {d \over {dt}}(3t\widehat i + 4\widehat j) \Rightarrow v = 3\widehat i$

Angular momentum, $L = m(r \times v)$

$ = 1 \times [(3t\widehat i + 4\widehat j) \times 3\widehat i]$

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {3t} & 4 & 0 \cr 3 & 0 & 0 \cr } } \right|$

$ = [3t \times 0 - 4 \times 3]\widehat k$

$ \Rightarrow L = - 12\widehat i$ J-s

Let us consider an equilateral $\Delta$ABC of side b.

From figure,

In $\Delta$AOD,

$\cos 30^\circ = {{AD} \over {AO}}$

$ \Rightarrow {{\sqrt 3 } \over 2} = {{{b \over 2}} \over r}$ [$\because$ $\cos 30^\circ = {{\sqrt 3 } \over 2}$]

$ \Rightarrow r = {b \over 2} \times {2 \over {\sqrt 3 }} = {b \over {\sqrt 3 }}$

Moment of inertia about line passing through centroid,

$I = 3 \times m{r^2}$ [due to 3 masses present in system]

$ \Rightarrow 3 \times m \times {\left( {{b \over {\sqrt 3 }}} \right)^2}$

$ \Rightarrow I = m{b^2}$

Let O be the centre of mass of part AB and Q that of CD. Let M be the mass per unit length of the parts AB and CD.

As, no rotation is set up about point P, so

moment of part AB about P = moment of part CD about P

or, (ml) x = (2 ml) (I $-$ x) [$\because$ $m = {M \over I}$]

or, x = 2I $-$ 2x

or, x = 2I / 3

Distance of P from the end C = 2I $-$ 2I/3 = 4I/3