Rotational Motion

According to the law of conservation of angular momentum.

Total initial angular momentum = Total final angular momentum

$ \begin{aligned} \left(I+m R^2\right) \omega & =I \omega^{\prime} \\ \therefore \quad \omega^{\prime} & =\frac{I+m R^2}{I} \omega \Rightarrow \omega^{\prime}=\frac{\frac{M R^2}{2}+m R^2}{\frac{M R^2}{2}} \omega \\ \text { or } \quad \omega^{\prime} & =\frac{M+2 m}{M}=\frac{200+2 \times 80}{200} \times 12 \\ \omega^{\prime} & =\frac{360}{200} \times 12=21.6 \mathrm{rpm} \end{aligned} $

To calculate the moment of inertia of a cube about one of its edges, we can use the theorem of perpendicular axes. Here's how it is derived:

First, we consider the moment of inertia about the center of the cube. According to the theorem of perpendicular axes, the total moment of inertia $ I $ can be found by adding the contribution from each axis. For a cube, with side length $ a $ and mass $ m $, the moment of inertia about one of its center axes is:

$ I_{\text{center}} = \frac{m a^{2}}{12} $

Since the cube is symmetric, this value is the same for any of the three perpendicular axes intersecting at the center. The moment of inertia through the axis parallel to a face through the center can similarly be calculated.

Now, when calculating the moment of inertia about an edge, we can add the parallel axis contribution:

$ I = I_{\text{center}} + m \left(\frac{a}{\sqrt{2}}\right)^{2} $

This accounts for the shift in the axis position. With simplification:

$ \begin{aligned} I & = \left[\frac{m a^{2}}{12} + \frac{m a^{2}}{12}\right] + \frac{m a^{2}}{2} \\ & = \frac{2}{12} m a^{2} + \frac{m a^{2}}{2} \\ & = \frac{m a^{2}}{6} + \frac{m a^{2}}{2} \\ & = \frac{2}{3} m a^{2} \end{aligned} $

Thus, the moment of inertia of a cube of mass $ m $ and side $ a $ about one of its edges is $ \frac{2}{3} m a^{2} $.

The angular velocity of the rigid body is given by $ \omega = \alpha - \beta t $, where $ \alpha $ and $ \beta $ are constants. By comparing this with the equation $ \omega = \omega_0 - \alpha t $, we can identify the initial angular velocity $ \omega_0 $ as $ \alpha $ and the angular retardation (or deceleration) as $ \beta $.

To find the angle through which the body rotates before it comes to a stop, we use the kinematic equation for rotational motion:

$ \omega^2 = \omega_0^2 + 2 \beta \theta $

We know that when the body stops, $ \omega = 0 $. Therefore, by substituting into the equation, we get:

$ 0 = \alpha^2 - 2 \beta \theta $

Solving for $ \theta $, the angle rotated, we find:

$ \theta = \frac{\alpha^2}{2 \beta} $

Thus, the angle through which the body rotates before stopping is $ \frac{\alpha^2}{2 \beta} $.

To determine the value of $n$, let's analyze the problem by considering the moment of inertia for each ring about an axis passing through its center. For a ring, the moment of inertia about an axis through its center can be expressed as:

$I = mR^2$

Where:

• $I$ is the moment of inertia
• $m$ is the mass of the ring
• $R$ is the radius of the ring

Given that both rings are made of the same material, they will have the same density. Let's denote the moment of inertia of the first ring (radius $R$) as $I_1$ and the moment of inertia of the second ring (radius $nR$) as $I_2$. According to the problem:

$\frac{I_1}{I_2} = \frac{1}{64}$

For the first ring (radius $R$):

$I_1 = m_1 R^2$

For the second ring (radius $nR$):

$I_2 = m_2 (nR)^2 = m_2 n^2 R^2$

Assuming both rings are made of the same material and have the same mass per unit length, the mass of the second ring with radius $nR$ will be $n$ times the mass of the first ring (since circumference scales linearly with radius):

$m_2 = n m_1$

Then we can write:

$I_2 = n m_1 (nR)^2 = n m_1 n^2 R^2 = n^3 m_1 R^2$

Using the ratio of the moments of inertia:

$\frac{I_1}{I_2} = \frac{m_1 R^2}{n^3 m_1 R^2} = \frac{1}{n^3} = \frac{1}{64}$

The above equality simplifies to:

$n^3 = 64$

$n = \sqrt[3]{64} = 4$

Thus, the value of $n$ is:

Option C

$\begin{aligned} \frac{x}{\sin \theta} & =\frac{2 r}{\sin 90^{\circ}} \Rightarrow x=2 r \sin \theta \\ \therefore \quad \frac{d x}{d t} & =2 r \cos \theta \frac{d \theta}{d t} \\ \frac{d \theta}{d t} & =\frac{d x / d t}{2 r \cos \theta}=\frac{2 v_0}{2 r \cos 60^{\circ}}=\frac{2 v_0}{r} \end{aligned}$

Area mass density, $\sigma = {M \over {16{R^2}}}$ ($\because$ Area = $4R \times 4R = 16{R^2}$)

Mass of each hole ${m_1} = \sigma \pi {R^2} = {M \over {16{R^2}}}\pi {R^2} = {{\pi M} \over {16}}$

Distance between centre of plate and centre of hole

$x = {{\sqrt {{{(2R)}^2} + {{(2R)}^2}} } \over 2} = {{2\sqrt 2 R} \over 2}$

Moment of inertia of one hole at about Z-axis

${I_1} = {1 \over 2}{m_1}{R^2} + {m_1}{x^2} = {{5\pi } \over {32}}M{R^2}$

Moment of inertia of whole plate about Z-axis

$I = {{M{{(4R)}^2}} \over 6} = {8 \over 3}M{R^2}$

Required moment of inertia

${I_0} = I - 4{I_1},$ $ = \left[ {{8 \over 3} - 4\left( {{{5\pi } \over {32}}} \right)} \right]M{R^2} = \left[ {{8 \over 3} - {{5\pi } \over 8}} \right]M{R^2}$

Given, R = 5 cm and M = 1 kg

So, ${I_0} = \left[ {{8 \over 3} - {{5\pi } \over 8}} \right]1 \times 25 \times {10^{ - 4}} = 0.0017$ kg-m²

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, 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

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