Rotational Motion

The correct answer is Option C. Here's why :

The moment of inertia of a solid sphere about an axis passing through its center is given by :

$I_0 = \frac{2}{5}MR^2$

Where :

• $I_0$ is the moment of inertia about the center
• $M$ is the mass of the sphere
• $R$ is the radius of the sphere

Now, using the parallel axis theorem, we can find the moment of inertia about an axis parallel to the diameter and at a distance of 'x' from it :

$I(x) = I_0 + Mx^2$

Substituting the value of $I_0$ :

$I(x) = \frac{2}{5}MR^2 + Mx^2$

This equation represents a parabola, where :

• The coefficient of $x^2$ is positive (M), indicating an upward-opening parabola.
• The y-intercept (the value of $I(x)$ when $x = 0$) is $\frac{2}{5}MR^2$, which is a positive constant.

Therefore, the graph of $I(x)$ vs. $x$ should be an upward-opening parabola, and Option C accurately represents this.

To determine the moment of the forces about point O, we need to calculate the moments of each force and then combine them vectorially.

First, consider force $\overrightarrow{F_1}$. This force acts along the z-axis at the point $(2\overrightarrow{i} + 3\overrightarrow{j})$.

The position vector of the point of application of $\overrightarrow{F_1}$ relative to the origin O is:

Since $\overrightarrow{F_1}$ acts along the z-axis, we can express it as:

The moment of force $\overrightarrow{F_1}$ about point O is given by the cross product:

Substitute the values:

Compute the cross product:

So, the moment due to force $\overrightarrow{F_1}$ is:

Using parallel axis theorem, moment of inertia about the given axis in the figure below will be

$\begin{aligned} & I_1=\frac{2}{5} M R^2+M(2 R)^2 \\\\ & I_1=\frac{22}{5} M R^2\end{aligned}$

Considering both spheres at equal distance from the axis, moment of inertia due to both spheres about this axis will be

$ 2 I_1=2 \times 22 / 5 M R^2 $ = ${{44} \over 5}$MR²

Moment of inertia of rod

Here, given that L = 2R

$\therefore I_2=\frac{1}{12} M(2 R)^2=\frac{1}{3} M R^2$

So, total moment of inertia of the system is

$ \begin{aligned} I & =2 I_1+I_2=2 \times \frac{22}{5} M R^2+\frac{1}{3} M R^2 \\\\ \Rightarrow I & =\left(\frac{44}{5}+\frac{1}{3}\right) M R^2=\frac{137}{15} M R^2 \end{aligned} $

Centre of mass after the collision will be at O. Therefore,

$2m{L \over 6} = m{L \over 3}$

Now, using conservation of angular momentum, we get

$2m{L \over 6} \times v + m{L \over 3} \times 2v = I\omega $

Therefore,

$2m{L \over 6} \times v + m{L \over 3} \times 2v = \left[ {8m{{{L^2}} \over {12}} + 2m{{\left( {{L \over 6}} \right)}^2} + m{{\left( {{L \over 3}} \right)}^2}} \right]\omega $

$ \Rightarrow \left( {{L \over 6} + {L \over 3}} \right)2mv = \left[ {8m{{{L^2}} \over {12}} + {{2m{L^2}} \over {36}} + {{m{L^2}} \over 9}} \right]\omega $

$ \Rightarrow {{3L} \over 6} \times 2mV = {5 \over 6}m{L^2}\omega $

$ \Rightarrow Lmv = {5 \over 6}m{L^2}\omega \Rightarrow \omega = {{6Lmv} \over {5m{L^2}}} \Rightarrow \omega = {6 \over 5}{v \over L}$

When the end M is released and rod makes angle $\alpha$ with horizontal, the displacement of centre of mass is ${L \over 2}\sin \alpha $.

Now, we know $mg{L \over 2}\sin \alpha = {1 \over 2}I{\omega ^2}$

Here, $I = {{m{L^2}} \over 3}$; therefore,

$mg{L \over 2}\sin \alpha = {1 \over 2}{{m{L^2}} \over 3}{\omega ^2}$

$ \Rightarrow {\omega ^2} = {{3g\sin \alpha } \over L} \Rightarrow \omega = \sqrt {{{3g\sin \alpha } \over L}} \Rightarrow \omega \propto \sqrt {\sin \alpha } $

For proper mixing, the concrete mixture should not stick to the top wall of the cylinder and it should fall down.

The maximum velocity at the top is v, then we have

${{m{v^2}} \over r} = mg$

$ \Rightarrow v = \sqrt {rg} $

Therefore, the maximum rotational speed of the drum is

$\omega = {v \over r} = \sqrt {{g \over r}} = \sqrt {{{9.8} \over {1.25}}} $ rad/s

$ = \sqrt {{{9.8} \over {1.25}}} \times {{60} \over {2\pi }}$ rpm = 26.74 $\approx$ 27 rpm

1. Determine the side length of the cube:

The cube with the maximum possible volume that can be cut from the sphere will have its diagonal equal to the diameter of the sphere. Let the side length of the cube be 'a'. Using Pythagoras in 3D, we have:

$a^2 + a^2 + a^2 = (2R)^2$

$3a^2 = 4R^2$

$a = \sqrt{\frac{4R^2}{3}} = \frac{2R}{\sqrt{3}}$

2. Calculate the mass of the cube:

The volume of the cube is $V = a^3 = \left(\frac{2R}{\sqrt{3}}\right)^3 = \frac{8R^3}{3\sqrt{3}}$

The density of the sphere (and hence the cube) is $\rho = \frac{M}{\frac{4}{3}\pi R^3}$

The mass of the cube is $m = \rho V = \frac{M}{\frac{4}{3}\pi R^3} \cdot \frac{8R^3}{3\sqrt{3}} = \frac{2M}{\sqrt{3}\pi}$

3. Find the moment of inertia of the cube:

The moment of inertia of a cube about an axis passing through its center and perpendicular to one of its faces is given by:

$I = \frac{1}{6}ma^2$

Substituting the values we found:

$I = \frac{1}{6} \cdot \frac{2M}{\sqrt{3}\pi} \cdot \left(\frac{2R}{\sqrt{3}}\right)^2$

$I = \frac{4MR^2}{9\sqrt{3}\pi}$

Therefore, the correct answer is Option A: ${{4M{R^2}} \over {9\sqrt {3\pi } }}$

As $\tau $ is perpendicular to the angular momentum $\overrightarrow L$ of the bob, so the direction of $L$ changes but magnitude remains same.

${I_{A}} = 2 \times m{\left( {{\textstyle{\ell \over {\sqrt 2 }}}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2}$

$ = m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}$