Rotational Motion

Let moment of inertia of body $ =I $

The initial kinetic energy $ =\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2} $

Final potential energy $ =m g h $

Given, $ h=\frac{v^{2}}{g} $

After putting value, potential energy

$ =m \cdot g \cdot \frac{v^{2}}{g}=m v^{2} $

When a body roll up an inclined plane of height $ h $, its kinetic energy converted into potential energy.

So, $ \mathrm{KE}=\mathrm{PE} $

$ \begin{array}{l} \frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}=m v^{2} \\\\ \frac{1}{2} I \omega^{2}=\frac{1}{2} m v^{2} \\\\ I=\frac{m v^{2}}{\omega^{2}}=\frac{m(r \omega)^{2}}{\omega^{2}} \\\\ I=m r^{2} \end{array} $

So, body is a ring.

Time taken by body to roll down from a slope without slipping is given by,

$ t=\sqrt{\frac{2 s}{g \sin \theta \beta}} $

where $ \beta=\frac{1}{1+\frac{K^{2}}{R^{2}}} $

From Eq. (i)

$ t \propto \frac{1}{\beta} $

$ t_{1}= $ time to reach bottom (by hollow cylinder)

$ t_{2}= $ time to reach bottom (by solid cylinder)

$ \frac{t_{1}}{t_{2}}=\sqrt{\frac{\beta_{2}}{\beta_{1}}} $

$ \beta_{1} $, for hollow cylinder

$ \beta_{1}=\frac{1}{1+\frac{K^{2}}{R^{2}}}=\frac{1}{1+1}=\frac{1}{2} \quad\left[\text { as } \frac{K^{2}}{R^{2}}=1\right] $

For solid cylinder

$ \beta_{2}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3} $

Substitute all value in Eq. (iv)

$ \begin{aligned} \frac{2}{t_{2}} & =\sqrt{\frac{\frac{2}{3}}{\frac{1}{2}}} \Rightarrow \frac{2}{t_{2}}=\sqrt{\frac{4}{3}} \\\\ t_{2} & =2 \times \sqrt{\frac{3}{4}} \\\\ t_{2} & =1.732 \mathrm{~s} \end{aligned} $

Disc and sphere has same radius and mass.

Now, moment of inertia of sphere at centre (I)

$ I=\frac{2 M R^{2}}{5} $

Using parallel axis theorem, moment of inertia at point $ X $ of sphere

$ \begin{aligned} I_{X S} & =I+M R^{2} \\\\ & =\frac{2}{5} M R^{2}+M R^{2}=\frac{7 M R^{2}}{5} \end{aligned} $

Moment of inertia of disc at centre perpendicularly

$ I=\frac{M R^{2}}{2} $

Using parallel axis theorem, moment of inertia at point $ X $ of disc.

$ I_{\chi D}=I+M R^{2}=\frac{3 M R^{2}}{2} $

Total moment of inertia at point $ X $

$ \begin{aligned} I_{X S}+I_{X D} & =\frac{3 M R^{2}}{2}+\frac{7 M R^{2}}{5} \\\\ & =\frac{29 M R^{2}}{10} \end{aligned} $

Given, Mass of wire, $=m$

linear mass density $=\rho$

Let length of wire be $l$ and radius ber

$ \rho=\frac{m}{l} $

or

$ m=\rho l \quad \ldots \text { (i) } $

also,

$ \begin{aligned} 2 \pi r & =1 \\\\ \Rightarrow r & =\frac{1}{2 \pi} \quad \ldots \text { (ii) } \end{aligned} $

$\therefore$ The moment of inertia around a circular loop around COM, will be

$ \begin{aligned} I & =\frac{1}{2} m r^2 \\\\ I & \left.=\frac{1}{2}(\rho l)\left(\frac{l}{2 \pi}\right)^2 \text { \{using Eqs. (i) and (ii)\}}\right\} \\\\ & =\frac{1}{2} \rho \frac{l^3}{4 \pi^2} \\\\ & =\frac{1}{8 \pi^2} \rho l^3 \end{aligned} $

$ \begin{aligned} &\text { From parallel axis theorem, }\\\\ &\begin{aligned} I_{\text {diameter }} & =I_{\mathrm{COM}}+M R^2 \\\\ & =\frac{l^3 \rho}{8 \pi^2}+\frac{l^3 \rho}{4 \pi^2} \\\\ & =\frac{3 l^3 \rho}{8 \pi^2} \\\\ \text { or } \quad I_{\text {diameter }} & =\frac{3}{8} \frac{m^3}{\pi^2 \rho^2} \end{aligned} \end{aligned} $

Given, height of plane $=28 \mathrm{~m}$

angle of inclination, $\theta=30^{\circ}$

By conservation of energy,

Potential energy lost by the solid sphere in rolling down the inclined plane

$=$ kinetic energy gain by the sphere

$ \begin{aligned} m g h & =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\\\ & =\frac{1}{2} m v^2+\frac{1}{5} m v^2 \\\\ m g h & =\frac{7}{10} m v^2 \\\\ v^2 & =\frac{10}{7} g h=\sqrt{\frac{10}{7} \times 10 \times 28} \\\\ v & =20 \mathrm{~m} / \mathrm{s} \end{aligned} $

The velocity of the sphere, when it reaches the bottom of the plane is $20 \mathrm{~m} / \mathrm{s}$.

Let $I_{\rm solid}=\tfrac25MR^2$ and $I_{\rm shell}=\tfrac23MR^2$. Since

$I_{\rm solid}=20\;\Rightarrow\;\frac25\,MR^2=20\quad\Longrightarrow\quad MR^2=\frac{5}{2}\times20=50,$

we get

$I_{\rm shell}=\frac23\,MR^2=\frac23\times50= \;33.3\;\mathrm{kg\cdot m^2}.$

Answer: Option C (33.3 kg·m²).

For the ring rolling down the inclined plane, the moment of inertia $(I)$,

$ I_{\text {ring }}=m r^2 $

For solid sphere, $I_{\text {sphere }}=\frac{2}{5} m r^2$ For solid cylinder, $I_{\text {cylinder }}=\frac{1}{2} m r^2$

Since, the solid sphere has the smallest moment of inertia, it will have the greatest translational kinetic energy when it reaches the bottom of the incline, thus solid sphere will reach the bottom of the inclined plane with maximum velocity.

$ \text { Given, } $

Coeffficient of linear expansion,

$ \alpha=20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} $

Temperature change, $\Delta T=50^{\circ} \mathrm{C}$

Initial angular speed $=\omega_0$

Using conservation of angular momentum,

$ \frac{2}{5} m r^2 \omega_0=\frac{2}{5} m[r(1+\alpha \Delta T)]^2 \omega^{\prime} $

$\therefore$ Radius of the sphere after the temperature increase $\left[r^{\prime}=r(1+\alpha \Delta T)]\right.$

$ \begin{aligned} \omega^{\prime} & =\frac{\omega_0}{(1+\alpha \Delta T)^2}=\omega_0[1-2 \alpha \Delta t] \\ & =\omega_0\left[1-2 \times 20 \times 10^{-5} \times 50\right] \\ & =\omega_0[1-0.02]=0.98 \omega_0 \end{aligned} $

The angular speed will be $0.98 \omega_0$.

Let:

$ m $ be the mass

$ r $ be the radius

$ v $ is the linear velocity

$ \omega $ is the angular velocity

The condition for rolling without slipping gives us $ v = r\omega $.

The total kinetic energy (KE) for any rolling object is the sum of its rotational and translational kinetic energies:

$ \mathrm{KE}_{\text{total}} = \mathrm{KE}_{\text{rotational}} + \mathrm{KE}_{\text{translational}} $

The formula for total kinetic energy is:

$ \mathrm{KE}_{\text{total}} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2 $

where $ I $ is the moment of inertia.

For the ring:

$ I = mr^2 $

For the disc:

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

Now, using these values, we can calculate the kinetic energies:

Kinetic Energy of the Disc

$ \mathrm{KE}_{\text{disc}} = \frac{1}{2} \cdot \frac{mr^2}{2} \cdot \left(\frac{v^2}{r^2}\right) + \frac{1}{2} mv^2 = \frac{3}{4} mv^2 \quad \quad \quad \quad \quad \quad \quad \quad \left( \text{i} \right) $

Kinetic Energy of the Ring

$ \mathrm{KE}_{\text{ring}} = \frac{1}{2} \cdot mr^2 \cdot \left(\frac{v^2}{r^2}\right) + \frac{1}{2} mv^2 = mv^2 \quad \quad \quad \quad \left( \text{ii} \right) $

Ratio of Kinetic Energy

From equations (i) and (ii):

$ \frac{\mathrm{KE}_{\text{ring}}}{\mathrm{KE}_{\text{disc}}} = \frac{mv^2}{\frac{3}{4} mv^2} = \frac{4}{3} $

Therefore, the ratio of the kinetic energy of the ring to the disc is 1.33.

The total length of a rod is $L$. So, when the rod is bent at the middle, then each part have same length $\left(\frac{L}{2}\right)$ and mass $\left(\frac{M}{2}\right)$.

Moment of inertia of each part about axis passing through its one end.

$ \begin{aligned} &=\frac{1}{3}\left(\frac{M}{L}\right)\left(\frac{1}{2}\right)^2\\ &\text { Thus, Net moment of inertia }\\ &\begin{aligned} I & =\frac{1}{3}\left[\frac{M}{2}\right]\left[\frac{L}{2}\right]^2+\frac{1}{3}\left[\frac{M}{2}\right]\left[\frac{L}{2}\right]^2 \\ & =\frac{1}{3} \frac{M L^2}{8}+\frac{1}{3} \frac{M L^2}{8} \\ \Rightarrow I & =\frac{M L^2}{12} \end{aligned} \end{aligned} $

By the conservation of energy

Loss in potential energy of rod = Gain of rotational kinetic energy

$ \begin{aligned} & m g \times \frac{2 L}{2} \sin \alpha=\frac{1}{2} I \omega^2 \\ & m g \times \frac{2 L}{2} \sin \alpha=\frac{1}{2} \times \frac{m(2 L)^2}{3} \times \omega^2 \\ & \text { Here, } \omega=\sqrt{\frac{3 g \sin \alpha}{2 L}} \end{aligned} $

$ \text { According to work energy theorem, } $

Total energy at point $A=$ Total energy at point $B$

$ M g h=\frac{1}{2} M v^2+\frac{1}{2} I \omega^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(Rotational energy)

For cylinder, $I=\frac{1}{2} M R^2$ and

Linear velocity, $v=R \omega \Rightarrow \omega=\frac{v}{R}$

Substitute it in Eq. (i)

$ \begin{aligned} & \quad M g h=\frac{1}{2} M v^2+\frac{1}{2} \times \frac{1}{2} \times M R^2 \times \frac{v^2}{R^2} \\ & \Rightarrow \quad g h=\frac{3}{4} v^2 \Rightarrow v=\sqrt{\frac{4}{3} g h} \end{aligned} $

$ \text { Let the axis of rotation be } B C \text {. } $

So, moment of inertia is given by,

$ \begin{aligned} & I=I_A+I_B+I_C \\ & =m_A R^2+m_B(0)+m_C(0) \end{aligned} $

As both masses lie on axis of rotation, separation distance is zero.

$ I=m_A R^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using Pythagoras theorem,

$ \begin{aligned} l^2 & =\frac{l^2}{4}+R^2 \Rightarrow R^2=\frac{3}{4} l^2 \\ I & \left.=m_A \frac{3}{4} l^2 \quad \text { [From Eq. (i) }\right] \\ \Rightarrow \quad I & =\frac{3}{4} m l^2 \quad\left(\text { as } m_A=m\right) \end{aligned} $

Given:

Mass of solid cylinder, $ m_s = 3.2 \, \text{kg} $

Mass of hollow cylinder, $ m_h = 1.6 \, \text{kg} $

Acceleration of the solid cylinder, $ a_s = 4 \, \text{m/s}^2 $

Moment of Inertia

For a solid cylinder:

$ I_s = \frac{1}{2} m R^2 $

For a hollow cylinder:

$ I_h = m R^2 $

Equation of Motion for Rolling Objects

The equation for motion is given by:

$ m g \sin \theta = m a + I \frac{a}{R^2} $

$ g \sin \theta = a \left(1 + \frac{I}{m R^2}\right) $

Calculations

Solid Cylinder

For the solid cylinder:

$ g \sin \theta = 4 \left( 1 + \frac{m R^2}{2m R^2} \right) $

$ g \sin \theta = 6 $

Hollow Cylinder

For the hollow cylinder:

$ g \sin \theta = a_h \left( 1 + \frac{m R^2}{m R^2} \right) $

$ 6 = a_h (1 + 1) $

$ a_h = 3 \, \text{m/s}^2 $

Therefore, the acceleration of the hollow cylinder is $ 3 \, \text{m/s}^2 $.

Given:

Mass of the sphere, $ m = 50 \, \text{kg} $

Radius of the sphere, $ r = 20 \, \text{cm} = 0.2 \, \text{m} $

Angular velocity, $ \omega = 420 \times \frac{2\pi}{60} = 14\pi \, \text{rad/s} $

The moment of inertia $ I $ of a solid sphere rotating about its diameter is given by:

$ I = \frac{2}{5} m r^2 $

Calculating the moment of inertia:

$ \begin{align*} I &= \frac{2}{5} \times 50 \times (0.2)^2 \\ &= \frac{2}{5} \times 50 \times 0.04 \\ &= 20 \times 0.04 \\ &= 0.8 \, \text{kg} \cdot \text{m}^2 \end{align*} $

The angular momentum $ L $ of the rotating sphere is calculated as:

$ L = I \cdot \omega $

Substitute the known values:

$ \begin{align*} L &= 0.8 \times 14\pi \\ &= 0.8 \times 14 \times \frac{22}{7} \\ &= 44 \times 0.8 \\ &= 35.2 \, \text{kg} \cdot \text{m}^2/\text{s} \\ &= 35.2 \, \text{J} \cdot \text{s} \end{align*} $

Therefore, the angular momentum of the sphere is $ 35.2 \, \text{J} \cdot \text{s} $.

Given, $ \boldsymbol{R}_{A} < \boldsymbol{R}_{B} < \boldsymbol{R}_{C} $

Frictional force opposes the body from sliding

Here, $f=m g \quad \ldots \text { (i) } $

Normal reaction provided necessary centripetal force.

$N=m \omega^2 R \quad$ (where $R$ is radius)

$ f=\mu N=\mu m \omega^2 R \quad \ldots \text { (ii) } $

Substitute the value of $f$ in Eq. (i),

$ \begin{aligned} m g & =\mu m \omega^2 R \\ \omega & =\frac{\sqrt{g}}{\mu R} \quad \ldots \text { (iii) } \end{aligned} $

From here, $\omega \propto \frac{1}{R}$

$\therefore$ The correct sequence of angular velocity is

$ \omega_C<\omega_B<\omega_A $

To find the moment of inertia of a solid sphere with a mass of 20 kg and a diameter of 20 cm about a tangent to the sphere, we follow these steps:

First, find the moment of inertia of the sphere about its center of mass. The formula for the moment of inertia ($MI$) of a solid sphere about its center is given by:

$ MI = \frac{2}{5} M R^2 $

where:

$M$ is the mass of the sphere (20 kg),

$R$ is the radius of the sphere.

Given the diameter is 20 cm, the radius $R$ is:

$ R = \frac{20}{2} \, \text{cm} = 10 \, \text{cm} = 0.10 \, \text{m} $

Substituting the values into the formula:

$ MI = \frac{2}{5} \times 20 \times (0.10)^2 = 0.08 \, \text{kg} \, \text{m}^2 $

Next, we use the theorem of parallel axes. This theorem states that the moment of inertia of a body about any axis parallel to an axis through its center of mass is the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance between the two axes.

For a tangent to the sphere, the distance is $R$ (which is the radius), so:

$ \text{MI about tangent} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 $

Substitute the values:

$ =\frac{7}{5} \times 20 \times (0.10)^2 = 0.28 \, \text{kg} \, \text{m}^2 $

Therefore, the moment of inertia of the solid sphere about a tangent is $0.28 \, \text{kg} \, \text{m}^2$.

The problem discusses a solid cylinder rolling down an inclined plane without slipping. Here's a step-by-step breakdown to find the total kinetic energy of the cylinder:

Translation Kinetic Energy

The translation kinetic energy ($ K_{\text{trans}} $) is given by:

$ K_{\text{trans}} = \frac{1}{2} m v^2 $

Given that:

$ K_{\text{trans}} = 140 \, \text{J} $

We know the relationship between linear velocity ($ v $) and angular velocity ($ \omega $) for rolling without slipping is:

$ v = \omega R $

This gives:

$ 140 = \frac{1}{2} m (\omega R)^2 $

Rearranging, we find:

$ 280 = m R^2 \omega^2 \, \ldots \text{(i)} $

Rotational Kinetic Energy

The rotational kinetic energy ($ K_{\text{rot}} $) can be expressed as:

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

For a solid cylinder, the moment of inertia ($ I $) is:

$ I = \frac{m R^2}{2} $

Substitute this into the expression for rotational kinetic energy:

$ K_{\text{rot}} = \frac{1}{2} \times \frac{m R^2}{2} \times \omega^2 $

$ K_{\text{rot}} = \frac{280}{4} \quad \text{[Using Eq. (i)]} $

$ K_{\text{rot}} = 70 \, \text{J} $

Total Kinetic Energy

The total kinetic energy of the cylinder is the sum of its translation and rotational kinetic energies:

$ K = K_{\text{trans}} + K_{\text{rot}} $

Thus:

$ K = 140 + 70 = 210 \, \text{J} $

Therefore, the total kinetic energy of the cylinder is 210 J.

Given:

Mass of both cylinders = $ M $

Radius of both cylinders = $ R $

The moment of inertia for a solid cylinder and a hollow cylinder about their axes are as follows:

For the solid cylinder:

$ I_1 = \frac{M R^2}{2} $

For the hollow cylinder:

$ I_2 = M R^2 $

From these formulas, it is clear that:

$ I_1 < I_2 $

$ \begin{aligned} & \text { Displacement of } C M=R \omega \frac{T}{2}=\pi r \\\\ & \Rightarrow \text { Displacement of } A =\sqrt{(2 R)^2+(\pi R)^2} \\\\ & =R \sqrt{\pi^2+4} \end{aligned} $

The correct answer is Both A and R are correct and R is the correct explanation of A.

Explanation:

Assertion A: An electric fan continues to rotate for some time after the current is switched off. This is a correct statement. When you switch off the fan, it doesn't stop immediately but continues to rotate for some time.

Reason R: Fan continues to rotate due to inertia of motion. This is also a correct statement. Inertia is the resistance of any physical object to any change in its state of motion. This includes changes to the object's speed or direction of motion. An object will stay in its state of motion unless a force acts on it. In the case of the fan, after the current is switched off, the fan blades have inertia and continue to move due to this inertia until the frictional forces (like air resistance and friction in the fan's bearings) cause it to stop. Therefore, inertia of motion is the correct reason for the fan's continued motion after the current is switched off.

Torque balance about ' $O$ '

$\frac{T}{2} \times 0.6=20 \times 0.5+80 \times 1$

$T \times 0.3=10+80=90$

$T=\frac{900}{3}=300 \mathrm{~N}$

A bar, having a mass of $ M = 1 $ kg and a length of $ L = 0.2 $ m, is hinged at a point P. A particle with a mass of $ m = 0.1 $ kg moves in a direction perpendicular to the bar at a velocity of $ u = 5 $ m/s. This particle collides elastically with the bar at a point that is $ \frac{L}{2} $ (0.1 m) away from the pivot point P. As a result of this collision, the bar begins to rotate with an angular velocity $ \omega $, and the particle recoils with a new velocity $ v $.

We can analyze the bar and the particle as a combined system. Initially, before the impact occurs, the system's angular momentum around the pivot point P is in a clockwise direction. The magnitude of this initial angular momentum can be expressed as

$ L_i = m u \left( \frac{L}{2} \right) $

Following the collision, the angular momentum of the system about the pivot point P changes. This final angular momentum can be represented as

$ L_f = I_{\text{bar}} \omega - m v \left( \frac{L}{2} \right) $

Here, $ I_{\text{bar}} $, the moment of inertia of the bar, is given by $ \frac{1}{3} M L^2 $, and the term $ -\frac{1}{2} m v L $ accounts for the particle's contribution post-collision.

During the collision, the angular momentum of the system around the pivot point P is conserved. This means $ L_i = L_f $. Applying this principle, we derive the following relationship :

$ 2 M L \omega = 3 m (u + v) $ ..........(i)

This equation represents the conservation of angular momentum in the system, correlating the initial and final states of the bar and the particle.

The kinetic energy of the system is conserved in an elastic collision. The kinetic energy of the system before the collision is

$ K_i=\frac{1}{2} m u^2 . $

In an elastic collision, the total kinetic energy of the system is preserved. Prior to the collision, the kinetic energy of the system can be quantified as

$ K_i = \frac{1}{2} m u^2 $

This equation denotes the initial kinetic energy, primarily attributed to the particle's motion.

After the collision takes place, the kinetic energy of the system can be expressed as :

$ K_f = \frac{1}{2} m v^2 + \frac{1}{2} I_{\text{rot}} \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{1}{3} M L^2 \omega^2 $

Here, the first term represents the kinetic energy due to the particle's velocity post-collision, and the second term represents the rotational kinetic energy of the bar.

By applying the principle of conservation of kinetic energy, which states $ K_i = K_f $, we obtain the following relation :

$ M L^2 \omega^2 = 3 m(u^2 - v^2) $ ............(ii)

This equation links the initial and final kinetic energies of the system, comprising the particle and the rotating bar.

By resolving equations (i) and (ii), we determine that the velocity $ v $ of the particle post-collision is $ 4.3 $ m/s and the angular velocity $ \omega $ of the bar is $ 6.98 $ rad/s.