Rotational Motion

Given, ${I_1} = 2.5{I_2}$ and ${I_3} = n{I_2}$

we know, ${I_1} = {{MR_1^2} \over 4},{I_2} = {{MR_2^2} \over 2},{I_3} = {2 \over 5}MR_3^2$

$ \Rightarrow {{MR_1^2} \over 4} = 2.5{{MR_2^2} \over 2} \Rightarrow R_1^2 = 5R_2^2$

Now, ${I_3} = n{I_2}$

$ \Rightarrow {2 \over 5}MR_1^2 = n{{MR_2^2} \over 2}$ (As ${R_3} = {R_1}$)

$ \Rightarrow {2 \over 5} \times 5R_2^2 = {{nR_2^2} \over 2} \Rightarrow n = 4$

Given, ${m_A} = 1\,kg = {m_B}$

$\overrightarrow {{r_A}} = ({\alpha _1}{t^2}\widehat i + {\alpha _2}t\widehat j + {\alpha _3}t\widehat k)\,m$

$\overrightarrow {{r_B}} = ({\beta _1}t\widehat i + {\beta _2}{t^2}\widehat j + {\beta _3}t\widehat k)\,m$

$({\alpha _1} = 1\,m/{s^2},{\alpha _2} = 3n\,m/s,{\alpha _3} = 2\,m/s$

${\beta _1} = 2\,m/s,{\beta _2} = - 1\,m/{s^2},{\beta _3} = 4p\,m/s)$

$\overrightarrow {{V_A}} = {{d\overrightarrow {{r_A}} } \over {dt}} = 2t\widehat i + 3n\widehat j + 2\widehat k$

$\overrightarrow {{V_B}} = {{d\overrightarrow {{r_B}} } \over {dt}} = 2\widehat i - 2t\widehat j + 4p\widehat k$

$\overrightarrow {{V_B}} \,.\,\overrightarrow {{V_B}} = 0$ (As $\overrightarrow {{V_A}} \bot \overrightarrow {{V_B}} $ given)

$ \Rightarrow 4t - 6nt + 8p = 0$

At $t = 1$, $4 - 6n + 8p = 0$

$ \Rightarrow 2 - 3n + 4p = 0$

$ \Rightarrow 3n = 2 + 4p$ ..... (1)

given, $\left| {\overrightarrow {{V_A}} } \right| = \left| {\overrightarrow {{V_B}} } \right|$

$ \Rightarrow 4 + 9{n^2} + 4 = 4 + 4 + 16{p^2}$

$ \Rightarrow {(2 + 4p)^2} = 16{p^2}$ (from (1)

$ \Rightarrow 4 + 16{p^2} + 16p = 16{p^2}$

$ \Rightarrow p = - {1 \over 4}$

$ \Rightarrow 3n = 2 + 4\left( { - {1 \over 4}} \right) = 1 \Rightarrow n = {1 \over 3}$

Now, $\overrightarrow L = {m_A}\left( {{{\overrightarrow r }_{A/B}} \times {{\overrightarrow v }_A}} \right)$

at $t = 1\,\sec ,$

$\overrightarrow {{r_{{A \over B}}}} = \left( {{\alpha _1} - {\beta _1}} \right)\widehat i + \left( {{\alpha _2} - {\beta _2}} \right)\widehat j + \left( {{\alpha _3} - {\beta _3}} \right)\widehat k$

$ \Rightarrow \overrightarrow {{r_{{A \over B}}}} = (1 - 2)\widehat i + (3n + 1)\widehat j + (2 - 4p)\widehat k$

$ = - \widehat i + 2\widehat j + 3\widehat k$

$\overrightarrow L = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & 2 & 3 \cr 2 & 1 & 2 \cr } } \right|$

$ = \widehat i(4 - 3) - \widehat j( - 2 - 6) + \widehat k( - 1 - 4)$

$\overrightarrow L = \widehat i + 8\widehat j - 5\widehat k$

$ \Rightarrow \left| {\overrightarrow L } \right| = \sqrt {{1^2} + {8^2} + {{( - 5)}^2}} = \sqrt {1 + 64 + 25} $

$ \Rightarrow \left| {\overrightarrow L } \right| = \sqrt {90} $ kg m$^2$ s$^{-1}$ = $\sqrt L$ (given)

So, L = 90

To find the value of $ \alpha $ from the given problem, we need to analyze the motion of a circular disc moving down an inclined plane in two different modes: slipping and rolling.

Slipping:

When the disc slips without rolling, it is primarily subjected to kinetic friction and gravity, without any rolling friction or torque affecting rotational motion. The motion can be considered as purely translational.

1. Equation for Time in Slipping Mode:

The acceleration $ a $ of the disc while slipping is given by:

$a = g \sin \theta$

where $ g $ is the acceleration due to gravity and $ \theta $ is the angle of the inclined plane.

The time $ t $ to travel down the incline of length $ l $ with this acceleration from rest is:

$l = \frac{1}{2} a t^2 \Rightarrow t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l}{g \sin \theta}}$

Rolling:

When the disc rolls, both translational and rotational motions are involved, and the rolling motion means that there is a rotational inertia factor that affects the acceleration.

1. Equation for Time in Rolling Mode:

For a solid disc, the moment of inertia $ I $ is $ \frac{1}{2} MR^2 $, where $ M $ is the mass and $ R $ is the radius of the disc. The acceleration $ a $ when rolling down without slipping is reduced due to the rotational inertia:

$a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2g \sin \theta}{3}$

The time to travel the same distance $ l $ is:

$t_{\text{roll}} = \sqrt{\frac{2l}{a_{\text{roll}}}} = \sqrt{\frac{2l}{\frac{2g \sin \theta}{3}}} = \sqrt{\frac{3l}{g \sin \theta}}$

Compare Times:

Given in the problem is the relation:

$t_{\text{roll}} = \left(\frac{\alpha}{2}\right)^{1/2} t$

From the derived formulas:

$\sqrt{\frac{3l}{g \sin \theta}} = \left(\frac{\alpha}{2}\right)^{1/2} \sqrt{\frac{2l}{g \sin \theta}}$

Solving for $ \alpha $:

$\sqrt{3} = \left(\frac{\alpha}{2}\right)^{1/2} \sqrt{2}$

$3 = \frac{\alpha}{2} \times 2$

$3 = \alpha$

Conclusion:

Thus, $ \alpha $ is 3.

To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$).

The torque ($\tau$) produced by the force ($F$) is given by the product of the force and the radius ($r$) of the wheel through which the force is applied:

$\tau = F \cdot r$

Given that $F = 40 \, \mathrm{N}$ and $r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}$, the torque can be calculated as:

$\tau = 40 \cdot 0.1 = 4 \, \mathrm{Nm}$

The torque is related to the angular acceleration ($\alpha$) and the moment of inertia ($I$) of the wheel by the equation:

$\tau = I \cdot \alpha$

Given that $I = 0.40 \, \mathrm{kg \cdot m}^2$, we can rearrange the above formula to solve for $\alpha$:

$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10 \, \mathrm{rad/s}^2$

With the angular acceleration ($\alpha$), we can calculate the angular velocity ($\omega$) after a given time ($t$) using the formula:

$\omega = \omega_0 + \alpha \cdot t$

Where $\omega_0$ is the initial angular velocity. Since the wheel starts from rest, $\omega_0 = 0$. Thus, for $t = 10 \, \mathrm{s}$:

$\omega = 0 + 10 \cdot 10 = 100 \, \mathrm{rad/s}$

Therefore, the angular velocity ($\omega$) of the wheel after $10$ seconds is $100 \, \mathrm{rad/s}$, so $x = 100$.

$\begin{aligned} & m v_r \frac{d v_r}{d r}=m \omega^2 r \\ & \frac{v_v^2}{2}=\frac{\omega^2\left(r^2-1\right)}{2} \\ & v_r=\omega \sqrt{8}=2 \omega \sqrt{2} \end{aligned}$

Moment of inertia about c and perpendicular to the plane is :

$\begin{aligned} & I=2 \times\left(\frac{a}{2 \sqrt{3}}\right)^2+4 \times\left(\frac{a}{2 \sqrt{3}}\right)^2+6\left(\frac{a}{2 \sqrt{3}}\right)^2 \\\\ & I=4 \mathrm{~kg} \mathrm{~m}^2 \end{aligned}$

Distance between centroid and midpoint of sides

$=\frac{a}{2 \sqrt{3}}$

For a hollow sphere rolling on a plane surface without slipping, its total kinetic energy (K.E.) is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy results from the motion of the center of mass of the sphere, and the rotational kinetic energy is due to its rotation about an axis through its center of mass (in this case, the axis of symmetry).

The translational kinetic energy (TKE) can be expressed as:

$\text{TKE} = \frac{1}{2}mv^2$

Where:

• m is the mass of the sphere,


• v is the linear velocity of the center of mass.

The rotational kinetic energy (RKE) for a rolling object can be given by:

$\text{RKE} = \frac{1}{2}I\omega^2$

For a hollow sphere, the moment of inertia (I) about its axis of symmetry is:

$I = \frac{2}{3}mr^2$

where r is the radius of the sphere. The angular velocity, $\omega$, can be related to the linear velocity, v, by the relation $v = r\omega$, for an object rolling without slipping. We substitute $\omega = \frac{v}{r}$ into the expression for RKE:

$\text{RKE} = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \left(\frac{v}{r}\right)^2$

This simplifies to:

$\text{RKE} = \frac{1}{3}mv^2$

The total kinetic energy (Total K.E.) of the rolling hollow sphere is the sum of its translational and rotational kinetic energies:

$\text{Total K.E.} = \text{TKE} + \text{RKE} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$

Now, we want to find the ratio of the rotational kinetic energy to the total kinetic energy:

$\frac{\text{RKE}}{\text{Total K.E.}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2}$

Since the mass and velocity are common in both the numerator and the denominator, they will cancel out, leaving:

$\frac{\frac{1}{3}}{\frac{5}{6}} = \frac{1}{3} \cdot \frac{6}{5} = \frac{2}{5}$

Therefore, the value of $x$, representing the ratio of the rotational kinetic energy to the total kinetic energy for a hollow sphere rolling on a plane surface about its axis of symmetry, is $2$. Thus, $x = 2$.

To solve this problem, we first note that for both the solid sphere and the hollow cylinder, the total mechanical energy is conserved as they roll up the inclined plane without slipping. The initial kinetic energy (comprised of both translational and rotational kinetic energy) is converted into potential energy at the maximum height.

Kinetic Energy for Each Body at the Start:

For the solid sphere, the moment of inertia $I$ is given by $I = \frac{2}{5}mr^2$, where $m$ is mass and $r$ is the radius of the sphere. The kinetic energy is the sum of translational kinetic energy $\left(\frac{1}{2}mv^2\right)$ and rotational kinetic energy $\left(\frac{1}{2}I\omega^2\right)$, where $\omega$ is the angular velocity. Since the sphere rolls without slipping, $v = r\omega$.

The total initial kinetic energy for the solid sphere is:

$KE_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

For the hollow cylinder, the moment of inertia $I$ is $mr^2$. Thus, its total kinetic energy is:

$KE_{\text{cylinder}} = \frac{1}{2}mv^2 + \frac{1}{2}mr^2\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$

Potential Energy at Maximum Height:

For both bodies, the potential energy at the maximum height is given by $PE = mgh$, where $h$ is the height reached.

Applying Conservation of Energy:

For the solid sphere, the energy conservation equation is:

$\frac{7}{10}mv^2 = mgh_1$

Solving for $h_1$ gives:

$h_1 = \frac{7v^2}{10g}$

For the hollow cylinder, the conservation of energy gives:

$mv^2 = mgh_2$

Thus, $h_2 = \frac{v^2}{g}$.

Finding the Ratio $h_1:h_2$:

The ratio of $h_1$ to $h_2$ is:

$\frac{h_1}{h_2} = \frac{\frac{7v^2}{10g}}{\frac{v^2}{g}} = \frac{7}{10}$

Therefore, the value of $n$, which represents the numerator in the ratio $\frac{n}{10}$, is $7$. Thus, the ratio of maximum heights $h_1:h_2$ reached by the solid sphere and the hollow cylinder, respectively, is $\frac{7}{10}$.

Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$

$ \mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s} $

Angular impuls $(\vec{M})$

$ \begin{aligned} & \vec{M}_c=\int \tau d t \\\\ & =\int F \frac{L}{2} d t \\\\ & =\frac{L}{2} \int F d t=\frac{L}{2} \times J \\\\ & =\frac{0.3}{2} \times 0.2 \\\\ & =0.03 \end{aligned} $

$\begin{aligned} & I_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}=\frac{2 \times(0.3)^2}{12}=\frac{0.09}{6} \\\\ & \mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right) \\\\ & 0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right) \\\\ & \omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}\end{aligned}$

$\theta=\omega \mathrm{t}$

$\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi}{2 \times 2}=\frac{\pi}{4} \mathrm{sec}$

$X=4$

$\begin{aligned} & \mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ & =\mathrm{mu}^3 \frac{1}{4 \sqrt{2} \mathrm{~g}} \Rightarrow \mathrm{x}=8 \end{aligned}$

$\begin{aligned} & \mathrm{I}=\left(\frac{2}{5} \mathrm{mR}^2+\mathrm{md}^2\right) \times 2 \\ & \mathrm{I}=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^2+2 \times\left(\frac{3}{4}\right)^2\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^2 \\ & \mathrm{X}=53 \end{aligned}$

To find the loss in kinetic energy when two spinning discs are brought together, we use the principle of conservation of angular momentum and the formula for kinetic energy. Here's how:

First, because angular momentum before and after they touch must be the same, we have:

$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$

where:

• $I_1$ and $I_2$ are the moments of inertia for the two discs.
• $\omega_1$ and $\omega_2$ are their angular speeds before contact.
• $\omega_0$ is their common angular speed after contact.

Plugging in the given values, we find that:

$\omega_0 = 8 \mathrm{rad/s}$

Next, to calculate the loss in kinetic energy, we first find the total kinetic energy before and after they touch:

Before: $E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = 216 \,\mathrm{J}$

After: $E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = 192 \,\mathrm{J}$

The loss in kinetic energy ($\Delta E$) is the difference:

$\Delta E = E_1 - E_2 = 24 \,\mathrm{J}$

So, when the two discs are brought together, the system loses 24 J of kinetic energy.

$\begin{aligned} & \vec{L}_i=I \omega_i=\frac{M R^2}{2} \cdot \omega=100 \mathrm{~kgm}^2 / \mathrm{s} \\ & E_i=\frac{1}{2} \cdot \frac{M R^2}{2} \cdot \omega^2=500 \mathrm{~J} \\ & \vec{L}_i=\vec{L}_f \Rightarrow 100=2 I \omega_f \\ & \omega_f=5 \mathrm{~rad} / \mathrm{sec} \\ & E_f=2 \times \frac{1}{2} \cdot \frac{5(2)^2}{2} \cdot(5)^2=250 \mathrm{~J} \\ & \Delta E=250 \mathrm{~J} \end{aligned}$

$y-x-4=0$

$d_1$ is perpendicular distance of given line from origin.

$\mathrm{d}_1=\left|\frac{-4}{\sqrt{1^2+1^2}}\right| \Rightarrow 2 \sqrt{2} \mathrm{~m}$

So

$\begin{aligned} |\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \\ & =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \end{aligned}$

To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $ \theta $, the component of gravitational acceleration along the plane is $ g \sin \theta $. However, because the cylinder is rolling and not sliding, not all of this component accelerates the center of mass; some of it goes into causing rotational acceleration about the center of mass.

For a rolling cylinder, the moment of inertia $ I $ is $ I = \frac{1}{2} m r^2 $, where $ m $ is the mass of the cylinder and $ r $ is the radius. The condition for rolling without slipping is that the linear acceleration $ a $ of the center of mass is equal to the radius $ r $ times the angular acceleration $ \alpha $, i.e., $ a = r \alpha $.

To find the linear acceleration $ a $, we use the torque $ \tau $ about the center of mass caused by the gravitational force down the incline. The torque due to gravity is $ \tau = mg \sin \theta \cdot r $, and from Newton's second law for rotation, the angular acceleration is given by

$ \alpha = \frac{\tau}{I} = \frac{mg \sin \theta \cdot r}{\frac{1}{2} m r^2} = \frac{2g \sin \theta}{r} $

Now using $ a = r \alpha $:

$ a = r \left( \frac{2g \sin \theta}{r} \right) = 2g \sin \theta $

But we must account for the fact that only a portion of the gravitational acceleration goes into translating the cylinder down the plane due to the rolling condition. This is where we apply the concept of the ``rolling factor'' for a cylinder, which is $ \frac{2}{3} $ for a solid cylinder, meaning $ \frac{2}{3} $ of the gravitational component is used for translation.

The acceleration of the center of mass for the cylinder is therefore:

$ a = \frac{2}{3} g \sin \theta $

Now we plug in the values of $ \theta = 60^{\circ} $ (which has $ \sin 60^{\circ} = \frac{\sqrt{3}}{2} $) and $ g = 10 \; m/s^2 $:

$ a = \frac{2}{3} \cdot 10 \cdot \frac{\sqrt{3}}{2} = \frac{10 \sqrt{3}}{3} $

To match the given expression $ \frac{x}{\sqrt{3}} m / s^2 $, let's manipulate our expression for $ a $:

$ a = \frac{10 \sqrt{3}}{3} = \frac{10 \sqrt{3}}{3} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{10 \cdot 3}{3 \cdot \sqrt{3}} = \frac{10}{\sqrt{3}} m/s^2 $

Hence, the value of $ x $ is $ 10 $.

In pure rolling work done by friction is zero. Hence potential energy is converted into kinetic energy. Since initially the ring and the sphere have same potential energy, finally they will have same kinetic energy too.

$\therefore$ Ratio of kinetic energies $=1$

$ \Rightarrow \frac{7}{x}=1 \Rightarrow x=7 $

$\begin{aligned} & \mathrm{I}=\mathrm{ma}^2+\mathrm{ma}^2+\mathrm{m}(\sqrt{2 \mathrm{a}})^2 \\ & =4 \mathrm{ma}^2 \\ & =4 \times 1 \times(2)^2=16 \end{aligned}$

Let the mass of both the solid sphere and the solid cylinder be $M$, and let their common radius be $R$. The moment of inertia $I$ of a solid sphere and a solid cylinder are given by:

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

$I_{cyl} = \frac{1}{2}MR^2$

The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I = Mk^2$. Therefore, we can find the radius of gyration for both the solid sphere and the solid cylinder using their respective moments of inertia:

$k_{sph}^2 = \frac{I_{sph}}{M} = \frac{2}{5}R^2$

$k_{cyl}^2 = \frac{I_{cyl}}{M} = \frac{1}{2}R^2$

Now, let's find the ratio of their radius of gyrations:

$\frac{k_{sph}}{k_{cyl}} = \frac{2}{\sqrt{x}}$

Squaring both sides:

$\frac{k_{sph}^2}{k_{cyl}^2} = \frac{4}{x}$

Substituting the expressions for $k_{sph}^2$ and $k_{cyl}^2$:

$\frac{\frac{2}{5}R^2}{\frac{1}{2}R^2} = \frac{4}{x}$

Simplifying and solving for $x$:

$\frac{2}{5} \cdot \frac{2}{1} = \frac{4}{x}$

$\frac{4}{5} = \frac{4}{x}$

Thus, $x = 5$.

In this problem, the net force on the cylinder is the tension $T$ in the rope, which is equal to the force applied to the rope:

$F=T=52.5~\mathrm{N}$

The force causes the cylinder to accelerate with an angular acceleration $\alpha$, which is related to its linear acceleration $a$ and the radius of the cylinder $R$ by the equation:

$\alpha=\frac{a}{R}$

The linear acceleration $a$ of the cylinder can be found using the formula $F=ma$:

$ma=F=52.5~\mathrm{N}$

where $m=5~\mathrm{kg}$ is the mass of the cylinder. Solving for $a$, we get:

$a=\frac{F}{m}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}}=10.5~\mathrm{m/s^2}$

Substituting this value of $a$ into the equation for $\alpha$, we get:

$\alpha=\frac{a}{R}=\frac{10.5~\mathrm{m/s^2}}{0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$

Therefore, the angular acceleration of the cylinder is $15~\mathrm{rad/s^2}$.

Alternate Method:

Let's first draw a free body diagram of the cylinder. The force $F$ applied to the rope creates a tension in the rope, which in turn exerts a force on the cylinder in the opposite direction. This force is given by:

$T=F$

where $T$ is the tension in the rope. The cylinder also experiences a torque due to the tension in the rope, which causes it to rotate. The torque is given by:

$\tau=TR$

where $R$ is the radius of the cylinder.

The net torque on the cylinder is equal to the product of the moment of inertia $I$ of the cylinder and its angular acceleration $\alpha$:

$\tau=I\alpha$

The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:

$I=MR^2$

where $M$ is the mass of the cylinder.

Substituting the given values, we get:

$\tau=TR=I\alpha=MR^2\alpha$

Solving for $\alpha$, we get:

$\alpha=\frac{T}{MR}=\frac{F}{MR}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}\cdot0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$

Therefore, the angular acceleration of the cylinder is $15~\mathrm{rad/s^2}$.

Given that the solid sphere is rolling without slipping, we have:

Angular momentum $L = \left(I_{\text{com}}\right)(\omega)$

Kinetic energy $K = \frac{1}{2}(I_{\text{com}})(\omega^2) + \frac{1}{2}MV_{\text{com}}^2$

For a solid sphere, the moment of inertia is $I_{\text{com}} = \frac{2}{5}MR^2$, and the relationship between linear and angular velocity is $V_{\text{com}} = R\omega$.

Substituting these values into the expressions for $L$ and $K$:

$L = \frac{2}{5}MR^2 \frac{V_{\text{com}}}{R} = \frac{2MRV_{\text{com}}}{5}$

$K = \frac{1}{2}\left(\frac{2}{5}MR^2\right) \frac{V_{\text{com}}^2}{R^2} + \frac{1}{2}MV_{\text{com}}^2 = \frac{7}{10}MV_{\text{com}}^2$

Now, the given ratio of $\frac{L}{K}$ is $\frac{\pi}{22}$:

$\frac{L}{K} = \frac{4}{7} \frac{R}{V_{\text{com}}} = \frac{\pi}{22}$

Since $V_{\text{com}} = R\omega$, we can substitute this relationship into the equation and solve for $\omega$:

$\frac{4}{7} \frac{R}{R\omega} = \frac{\pi}{22}$

$\frac{4}{7\omega} = \frac{\pi}{22}$

$\omega = \frac{4}{7} \times \frac{22}{\pi} \times 7 = 4$

Thus, the value of the angular speed is $4 \ \text{rad/s}$.

For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy ($K_{total}$) can be expressed as the sum of the translational kinetic energy ($K_{trans}$) and the rotational kinetic energy ($K_{rot}$):

$K_{total} = K_{trans} + K_{rot}$

The translational kinetic energy of an object with mass (m) and linear velocity (v) is given by:

$K_{trans} = \frac{1}{2}mv^2$

The rotational kinetic energy of a rolling spherical shell with moment of inertia (I) and angular velocity (ω) is given by:

$K_{rot} = \frac{1}{2}Iω^2$

For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:

$v = Rω$

Where R is the radius of the spherical shell.

The moment of inertia for a spherical shell is given by:

$I = \frac{2}{3}mR^2$

Now, we can substitute the moment of inertia and the relationship between linear and angular velocity into the equation for rotational kinetic energy:

$K_{rot} = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2$

Simplifying the equation:

$K_{rot} = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\frac{v^2}{R^2}$

$K_{rot} = \frac{1}{3}mv^2$

Now, we can find the ratio of rotational kinetic energy to total kinetic energy:

$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2 + \frac{1}{3}mv^2}$

Simplifying the equation:

$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}}{\frac{1}{2} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{5}{6}}$

Multiplying both the numerator and the denominator by 6:

$\frac{K_{rot}}{K_{total}} = \frac{2}{5}$

Comparing this to the given ratio of $\frac{x}{5}$, we can determine that the value of $x$ is 2.

The conservation of angular momentum states that the angular momentum (L) remains constant. The relation between kinetic energy (KE), angular momentum (L), and moment of inertia (I) is given by:

$ \mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}} $

Using this relation, we can find the ratio of the final kinetic energy ($\mathrm{KE}_{\text{final}}$) to the initial kinetic energy ($\mathrm{KE}_{\text{initial}}$ or E):

$ \frac{\mathrm{KE}_{\text{final}}}{\mathrm{KE}_{\text{initial}}}=\frac{\mathrm{I}_{\text{initial}}}{\mathrm{I}_{\text{final}}} $

Since the moment of inertia triples, we have $\mathrm{I}_{\text{final}} = 3\mathrm{I}_{\text{initial}}$. Therefore,

$ \frac{\mathrm{KE}_{\text{final}}}{\mathrm{E}}=\frac{\mathrm{I}_{\text{initial}}}{3\mathrm{I}_{\text{initial}}}=\frac{1}{3} $

This means that the final kinetic energy of the system is:

$ \mathrm{KE}_{\text{final}}=\frac{E}{3} $

So, the value of $x$ is 3.

$ \begin{aligned} & L_{\text {diameter }}=\frac{2}{5} M R^2 \omega ; \quad I_{\text {tangent }}=\frac{7}{5} M R^2 \\\\ & \begin{aligned} \frac{I_{\text {tangent }}}{L_{\text {diameter }}} & =\frac{7 / 5}{2 / 5} \times \frac{1}{\omega}=\frac{7}{2 \omega} \\\\ & =\frac{7}{2 \times 10}=\frac{7}{20} \\\\ &= \frac{700}{20} \times 10^{-2}=35 \times 10^{-2} \end{aligned} \end{aligned} $

Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $\vec{F} = -P\hat{k}$.

To find the torque, we first find the position vector of point A with respect to point B:

$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 2)\hat{i} + (0 - (-3))\hat{j} + (0 - 0)\hat{k} = -2\hat{i} + 3\hat{j}$

To calculate the cross product, we can use the determinant method with a 3x3 matrix:

$\vec{\tau} = \vec{r}_{AB} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ 0 & 0 & -P \\ \end{vmatrix}$

Now, we will calculate the cross product components by expanding the determinant along the first row:

1. $\tau_i = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & -P \\ \end{vmatrix} = \hat{i}((3)(-P) - (0)(0)) = -3P\hat{i}$

2. $\tau_j = -\hat{j} \begin{vmatrix} -2 & 0 \\ 0 & -P \\ \end{vmatrix} = -\hat{j}((-2)(-P) - (0)(0)) = -2P\hat{j}$

(Notice the negative sign in front of the $\hat{j}$ term, as it comes from the expansion of the determinant.)

3. $\tau_k = \hat{k} \begin{vmatrix} -2 & 3 \\ 0 & 0 \\ \end{vmatrix} = \hat{k}((-2)(0) - (3)(0)) = 0\hat{k}$

Now, combine the components to get the torque vector:

$\vec{\tau} = -3P\hat{i} - 2P\hat{j} + 0\hat{k} = -3P\hat{i} - 2P\hat{j}$

Comparing this to the given torque vector $\vec{\tau} = P(a\hat{i} + b\hat{j})$, we find that:

$a = -3$

$b = -2$

Thus, the ratio $\frac{a}{b} = \frac{-3}{-2} = \frac{x}{2}$.

Therefore, $x = 3$.

Total initial kinetic energy

$ =\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \mathrm{I} \omega^2 $

$\mathrm{v}=\mathrm{R} \omega$ (for pure rolling)

$ \mathrm{K} . E C=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \times \frac{2}{3} m \mathrm{R}^2 \times \frac{\mathrm{v}^2}{\mathrm{R}^2}=\frac{5}{6} m \mathrm{v}^2 $

Energy remains conserve during whole journey.

$\mathrm{K}_{\cdot} \mathrm{E}_i+\mathrm{P.E}_{\cdot i}=\mathrm{K}_{\cdot} \mathrm{E}_f+\mathrm{P.E}_{\cdot f}$

$ \begin{aligned} & \Rightarrow \frac{5}{2} m \mathrm{v}^2=m g \mathrm{H} ~~~~~~~(\because {K.E.}_f=0)\\\\ & \Rightarrow \mathrm{H}=\frac{5}{6} \times \frac{\mathrm{v}^2}{g} \\\\ & =\frac{5 \times(3)^2}{6 \times 10} \\\\ & =\frac{15}{20} \mathrm{~m}=0.75 \mathrm{~m}=75 \mathrm{~cm} \end{aligned} $

To solve this problem, we need to understand the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass distribution of the object and the axis of rotation.

For a continuous object like a semicircular ring, we can calculate the moment of inertia by integrating over the entire object. Here's how we can approach this problem:

1. Divide the semicircular ring into small mass elements: Imagine the semicircular ring divided into infinitesimally small mass elements, each with mass $dm$.

2. Calculate the moment of inertia of each element: The moment of inertia of each element about the axis passing through the center and perpendicular to the plane of the ring is given by $dI = dmR^2$, where R is the radius of the ring.

3. Integrate to find the total moment of inertia: To find the total moment of inertia, we need to integrate $dI$ over the entire ring. This means integrating from $0$ to $\pi$ (the angle spanned by the semicircle) with respect to the angle $\theta$.

4. Relate $dm$ to the total mass: Since the ring has a uniform mass distribution, we can express the mass of each element $dm$ as a fraction of the total mass $M$: $dm = \frac{M}{πR} Rd\theta = \frac{M}{\pi} d\theta$.

Now, let's perform the integration:

$I = \int_{0}^{\pi} dI = \int_{0}^{\pi} dmR^2 = \int_{0}^{\pi} \frac{M}{\pi} d\theta R^2$

$I = \frac{MR^2}{\pi} \int_{0}^{\pi} d\theta = \frac{MR^2}{\pi} [\theta]_{0}^{\pi}$

$I = \frac{MR^2}{\pi} [\pi - 0] = MR^2$

Therefore, the moment of inertia of the semicircular ring about the given axis is $MR^2$. Comparing this to the given formula, we find that $x = \boxed{1}$.

Given that the radii of gyration for the ring and the solid sphere are equal, we have:

$ K_1 = K_2 $

For the ring, the moment of inertia is:

$ I_{ring} = mR_1^2 = mK_1^2 $

Thus, the radius of gyration for the ring is:

$ K_1 = R_1 $

For the solid sphere, the moment of inertia is:

$ I_{sphere} = \frac{2}{5}m'R_2^2 = m'K_2^2 $

Hence, the radius of gyration for the solid sphere is:

$ K_2 = \sqrt{\frac{2}{5}}R_2 $

Since the radii of gyration are equal:

$ R_1 = \sqrt{\frac{2}{5}}R_2 $

Therefore, the ratio of the radius of the ring to that of the sphere is:

$ \frac{R_1}{R_2} = \sqrt{\frac{2}{5}} $

So, the value of $x$ is:

$ x = 5 $

The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.

First, let’s identify the moment of inertia of each solid sphere about its own center, which is given by the formula:

$I_{\text{sphere}} = \frac{2}{5} m r^2$

Where:

• $m$ is the mass of the sphere = $2 \mathrm{~kg}$
• $r$ is the radius of the sphere = $0.1 \mathrm{~m}$

Substituting the values:

$I_{\text{sphere}} = \frac{2}{5} \times 2 \mathrm{~kg} \times (0.1 \mathrm{~m})^2 = \frac{4}{5} \times 0.01 \mathrm{~kg}~\mathrm{m}^2 = 0.008 \mathrm{~kg}~\mathrm{m}^2$

Now, we need the moment of inertia of the two spheres about the axis passing through the midpoint of the rod. This requires using the parallel axis theorem, which states:

$I_{\text{total}} = I_{\text{sphere}} + m d^2$

Where:

• $d$ is the distance from the center of the sphere to the axis through the rod’s midpoint = $0.2 \mathrm{~m}$

Calculating the additional inertia due to the parallel axis theorem for one sphere:

$I_{\text{parallel}} = m d^2 = 2 \mathrm{~kg} \times (0.2 \mathrm{~m})^2 = 2 \mathrm{~kg} \times 0.04 \mathrm{~m}^2 = 0.08 \mathrm{~kg}~\mathrm{m}^2$

The total moment of inertia for one sphere about the midpoint of the rod is:

$I_{\text{one sphere, total}} = I_{\text{sphere}} + I_{\text{parallel}} = 0.008 \mathrm{~kg}~\mathrm{m}^2 + 0.08 \mathrm{~kg}~\mathrm{m}^2 = 0.088 \mathrm{~kg}~\mathrm{m}^2$

Since there are two identical spheres, the total moment of inertia of the system is:

$I_{\text{system}} = 2 \times 0.088 \mathrm{~kg}~\mathrm{m}^2 = 0.176 \mathrm{~kg}~\mathrm{m}^2$

Converting the result to the given form:

$0.176 \mathrm{~kg}~\mathrm{m}^2 = 176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$

So, the moment of inertia of the system about the given axis is:

$176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$

$\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 \\\\ & =\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2 \\\\ & =\frac{3}{2} \mathrm{MR}^2 \\\\ & \mathrm{x}=3\end{aligned}$

Loss in potential energy $=m g h=m g\left[60 \sin 30^{\circ} \mathrm{cm}\right]$

$\Rightarrow m g\left[\frac{30}{100}\right]=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m v^{2}}{2}$

$\Rightarrow 0.3 \times 10=\frac{3}{4} v^{2}$

$\Rightarrow v^{2}=4$

$\Rightarrow v=2 \mathrm{~m} / \mathrm{s}$

$m=\rho \pi R^{2} t$

$ \begin{aligned} & \text { so } R^{2}=\frac{m}{\rho \pi t} \\\\ & I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t} \end{aligned} $

So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$

So $x=5$

$\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=7 \times 10^{-3}$

$ \Rightarrow $ $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{MR}^{2}\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=7 \times 10^{-3}$

$ \Rightarrow $ $\frac{1}{2} \mathrm{MV}^{2}\left[1+\frac{2}{5}\right]=7 \times 10^{-3}$

$ \Rightarrow $ $\frac{1}{2}(1)\left(\mathrm{V}^{2}\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}$

$ \Rightarrow $ $\mathrm{V}^{2}=10^{-2}$

$ \Rightarrow $ $\mathrm{V}=10^{-1}=0.1 \mathrm{~m} / \mathrm{s}=10 \mathrm{~cm} / \mathrm{s}$

$v = {v_0} - \mu gt$

$ \Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12$ m/s

$\Rightarrow$ Kinetic energy $ = {1 \over 2}m{v^2} + {1 \over 2}{{m{v^2}} \over 2}$

$ = {3 \over 4}m{v^2} = {3 \over 4} \times 0.5 \times 144\,\mathrm{J} = 54\,\mathrm{J}$

Kinetic energy of rod $E = {1 \over 2}{{m{l^2}} \over {12}}{\omega ^2}$

or $\omega = \sqrt {{{24E} \over {m{l^2}}}} = \sqrt {{{24E} \over {d \times A \times {l^3}}}} $

$ \Rightarrow \omega = \sqrt {{{24E} \over {dA{2^3}}}} $

$ = \sqrt {{{3E} \over {Ad}}} $

So, $\alpha = 3$

Horizontal displacement $x = v\cos \theta t$

$ = 10\sqrt 2 t$

So torque of weight about point of projection is

$\tau = mgx\,.\,( - \widehat k)$

${{d\overrightarrow L } \over {dt}} = mgx( - \widehat k)$

$\int_0^{\overrightarrow L } {d\overrightarrow L = 0.1 \times 10 \times 10\sqrt 2 \int_0^2 {t\,dt( - \widehat k)} } $

$\overrightarrow L = - 20\sqrt 2 \widehat k$

$|\overrightarrow L | = 20\sqrt 2 = \sqrt {800} $ kg m$^2$/s

$\frac{1}{2} m v_{\mathrm{cm}}^{2}+\frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v_{\mathrm{cm}}^{2}}{R^{2}}=2240 \mathrm{~J}$

$ \begin{aligned} & \frac{7}{10} m v_{\mathrm{cm}}^{2}=2240 \\\\ & v_{\mathrm{cm}}=\sqrt{\frac{2240 \times 10}{7 \times 2}}=40 \mathrm{~m} / \mathrm{sec} \end{aligned} $

Solid Sphere :

$ \begin{aligned} & I_{\text {tangent }}=I_{\mathrm{cm}}+m R^{2} \\\\ & =\frac{2}{5} m R^{2}+m R^{2}=\frac{7}{5} m R^{2} \\\\ & =7 R^{2} \quad(m=5 \mathrm{~kg}) \end{aligned} $

CIRCULAR DISC :

$ \begin{aligned} I_{\text {disc }} & =I_{\mathrm{cm}}+m R^{2} \\\\ & =\frac{m R^{2}}{4}+m R^{2} \\\\ & =\frac{5}{4} m R^{2} \\\\ & =5 R^{2} \end{aligned} $

$\frac{l_{\text {disc }}}{l_{\text {tangent }}}=\frac{5}{7}$

Concept :

1. For Solid Sphere :

Position of the axis of rotation : About its diametric axis which passes through its centre of mass


Moment of Inertia (I) = ${2 \over 5}M{R^2}$

Radius of gyration (K) = $\sqrt {{2 \over 5}} R$

Position of the axis of rotation : About a tangent to the sphere


Moment of Inertia (I) = ${7 \over 5}M{R^2}$

Radius of gyration (K) = $\sqrt {{7 \over 5}} R$



2. For CIRCULAR DISC :

Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre


Moment of Inertia (I) = ${1 \over 2}M{R^2}$

Radius of gyration (K) = ${R \over {\sqrt 2 }}$

Position of the axis of rotation : About the diametric axis


Moment of Inertia (I) = ${1 \over 4}M{R^2}$

Radius of gyration (K) = ${R \over { 2 }}$

$ I_{A B}=I_{\mathrm{cm}}+M \times\left(\frac{2}{3} R\right)^2 $

$ =\frac{1}{2} M R^{2}+\frac{4}{9} M R^{2} $

$ \begin{aligned} & =\frac{(9+8) M R^{2}}{18}=\left(\frac{17}{18}\right) M R^{2} \end{aligned} $

$\frac{I_{A B}}{I_{\mathrm{cm}}}=\frac{17 / 18}{1 / 2}=\left(\frac{17}{9}\right)$

Value of $x=17$

$I_{1}=\frac{\left(\rho \pi R^{2} L\right) R^{2}}{2} \quad$ ( $\rho$ : density of cylinder)

$ \begin{aligned} & I_{2}=\frac{\left[\rho \pi\left(\frac{R}{2}\right)^{2} \frac{L}{2}\right]\left(\frac{R}{2}\right)^{2}}{2} \\\\ & \therefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1} \end{aligned} $

$ \begin{aligned} & \mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{MR}^2 \\\\ & \mathrm{I}_{\mathrm{PQ}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2 \\\\ & \mathrm{I}_{\mathrm{PQ}}=\frac{2}{5} \mathrm{mR}^2+\mathrm{m}(10 \mathrm{~cm})^2 \end{aligned} $

For radius of gyration $\mathrm{I}_{\mathrm{PQ}}=\mathrm{mk}^2$

$ \begin{aligned} & \mathrm{k}^2=\frac{2}{5} \mathrm{R}^2+(10 \mathrm{~cm})^2 \\\\ & =\frac{2}{5}(5)^2+100 \\\\ & =10+100=110 \\\\ & \mathrm{k}=\sqrt{110} \mathrm{~cm} \\\\ & \mathrm{x}=110 \end{aligned} $