Step 1: Identifying the Nature of Rolling Motion

In a case of perfect rolling (rolling without slipping) on an inclined plane, the point of contact between the cylinder and the surface is instantaneously at rest relative to the surface [Per the Kinematic Condition for Pure Rolling].

Since the velocity of the point of contact $v_{contact} = 0$, the force of static friction $f$ acts at this point but does not undergo any displacement relative to the incline during the interval of rolling.

Step 2: Defining Work Done by Friction

Work done ($W$) by a force $\vec{F}$ is defined by the line integral of the dot product of the force and the infinitesimal displacement vector $d\vec{r}$:

$W = \int \vec{f} \cdot d\vec{r}$

In pure rolling motion, the displacement of the point of contact $d\vec{r}$ is zero at every instant $[d\vec{r} = 0]$. Consequently, the power developed by the static frictional force $P = \vec{f} \cdot \vec{v}_{contact} = 0$.

Step 3: Geometrical Representation of the Rolling Cylinder

Step 4: Analytical Conclusion

As established by the principles of classical mechanics, for any object undergoing pure rolling, the work done by the static friction force is identically zero because the point of application of the force does not move relative to the surface [Work-Energy Theorem Application].

Parameter	Value/Condition
Displacement of contact point	$0$
Force of friction	$f$ (Static)
Work done ($W = f \cdot \Delta x$)	$f \cdot 0 = 0$

Final Solution: The work done against friction during rolling is 0 J.

Step 1: Theoretical Definition of the Parallel Axes Theorem

The Theorem of Parallel Axes states that the moment of inertia $I$ of a rigid body about any axis is equal to the sum of the moment of inertia $I_{cm}$ of the body about a parallel axis passing through its centre of mass and the product of the mass of the body $M$ and the square of the perpendicular distance $d$ between the two axes. Mathematically, this is expressed as:

$I = I_{cm} + Md^2$

Step 2: Coordinate System Setup

Let the centre of mass (CM) of a system of $n$ particles be at the origin $(0,0,0)$ of the coordinate system. By the definition of the centre of mass, the sum of the moments of mass about the origin is zero: $\sum m_i \vec{r}_i = 0$ [Definition of Centre of Mass].

Consider two parallel axes: one passing through the CM ($z$-axis) and another passing through a point $P$ at distance $d$ from the CM (let this be the $z'$-axis). Let the position vector of a particle $i$ be $\vec{r}_i$ with coordinates $(x_i, y_i, z_i)$.

Step 3: Algebraic Derivation

The moment of inertia about the axis passing through the CM (the $z$-axis) is:

$I_{cm} = \sum m_i (x_i^2 + y_i^2)$

The distance of the $i$-th particle from the parallel axis passing through point $P$ (let $P$ be at $x=d, y=0$) is given by $[(x_i - d)^2 + y_i^2]$. The moment of inertia $I$ about this new axis is:

$I = \sum m_i ((x_i - d)^2 + y_i^2)$

Step 4: Expansion and Simplification

Expand the quadratic term:

$I = \sum m_i (x_i^2 - 2x_id + d^2 + y_i^2)$

Distribute the summation index:

$I = \sum m_i (x_i^2 + y_i^2) + \sum m_i d^2 - 2d \sum m_i x_i$

Step 5: Applying Centre of Mass Constraints

1. The first term $\sum m_i (x_i^2 + y_i^2)$ is exactly $I_{cm}$ [By Definition].

2. The second term $\sum m_i d^2$ becomes $d^2 \sum m_i = Md^2$, where $M$ is the total mass of the system.

3. The third term $2d \sum m_i x_i$: Since the CM is at the origin, $\sum m_i x_i = 0$ [Property of Centre of Mass]. Thus, this term vanishes.

Conclusion

Substituting these values back into the expression for $I$:

$I = I_{cm} + Md^2 + 0$

Final Solution: $I = I_{cm} + Md^2$. This completes the proof of the Parallel Axes Theorem.

Step 1: Establishing Given Parameters

We are provided with the moment of inertia ($I_{diameter}$) of a solid sphere about an axis passing through its center (diameter):

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

Where:

• $M$ = Total mass of the sphere
• $R$ = Radius of the sphere

Step 2: Theorem Selection

To determine the moment of inertia about a tangent, we must utilize the Parallel Axis Theorem. [Per the Parallel Axis Theorem, $I = I_{cm} + Mh^2$, where $I_{cm}$ is the moment of inertia about the center of mass, and $h$ is the perpendicular distance between the two parallel axes.]

Step 3: Calculating the Perpendicular Distance

A tangent to the sphere is parallel to the diameter passing through the center of mass. The distance $h$ between the center of the sphere and its surface (the tangent line) is equal to the radius of the sphere, $R$.

$h = R$

Step 4: Algebraic Derivation

Applying the values to the Parallel Axis Theorem:

$I_{tangent} = I_{cm} + MR^2$

Substitute $I_{cm} = I_{diameter} = \frac{2}{5}MR^2$:

$I_{tangent} = \frac{2}{5}MR^2 + MR^2$

$I_{tangent} = (\frac{2}{5} + 1)MR^2$

$I_{tangent} = (\frac{2 + 5}{5})MR^2$

$I_{tangent} = \frac{7}{5}MR^2$

Step 5: Visual Representation

Final Solution: The moment of inertia of the sphere about a tangent is $\frac{7}{5}MR^2$.

Step 1: Identifying Governing Kinematic Variables

To determine which sphere reaches the bottom first, we must analyze the acceleration of a body undergoing pure rolling motion down an inclined plane of angle $\theta$ and height $h$.

The total energy $E$ at the top of the incline is purely potential: $E = mgh$. At the bottom, the energy is the sum of translational and rotational kinetic energy: $E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.

For a solid sphere, the moment of inertia $I = \frac{2}{5}mR^2$ and the condition for pure rolling is $v = \omega R$. Substituting these into the energy equation [Per the Principle of Conservation of Mechanical Energy]:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$

$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

$v = \sqrt{\frac{10gh}{7}}$

Step 2: Calculating Acceleration and Time

The acceleration $a$ of a body rolling down an incline is given by [Newton’s Second Law for Rotation]:

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

Substituting $I = \frac{2}{5}mR^2$ for the solid sphere:

$a = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{5}{7}g \sin \theta$

The distance $s$ along the incline of height $h$ is $s = \frac{h}{\sin \theta}$. Using the kinematic equation $s = \frac{1}{2}at^2$:

$\frac{h}{\sin \theta} = \frac{1}{2} (\frac{5}{7}g \sin \theta) t^2$

$t^2 = \frac{14h}{5g \sin^2 \theta} \implies t = \frac{1}{\sin \theta} \sqrt{\frac{14h}{5g}}$

Step 3: Comparative Analysis of Inclination Angles

We observe that time $t$ is inversely proportional to $\sin \theta$:

$t \propto \frac{1}{\sin \theta}$

If there are two inclined planes with angles $\theta_1$ and $\theta_2$, where $\theta_2 > \theta_1$:

• $\sin \theta_2 > \sin \theta_1$
• Therefore, $\frac{1}{\sin \theta_2} < \frac{1}{\sin \theta_1}$
• Resulting in $t_2 < t_1$

Step 4: Conclusion

Since the time $t$ taken to reach the bottom is inversely proportional to the sine of the angle of inclination ($\sin \theta$), the sphere on the plane with the larger angle of inclination will reach the bottom first. This is because a steeper incline results in a higher component of gravitational force acting along the plane, leading to a greater linear acceleration.

Final Solution: The sphere on the plane with the larger angle of inclination reaches the bottom first, because the time taken is inversely proportional to $\sin \theta$, meaning a steeper angle results in higher acceleration and less transit time.

Step 1: Definition of Angular Momentum

The angular momentum $\vec{L}$ of a particle with respect to the origin is defined as the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$ [Per the definition of rotational dynamics]:

$\vec{L} = \vec{r} \times \vec{p}$

Given the components of the vectors as:

• $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$
• $\vec{p} = p_x\hat{i} + p_y\hat{j} + p_z\hat{k}$

Step 2: Determinant Expansion for Components

Calculating the cross product using the determinant form [By the properties of vector products]:

$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}$

Expanding the determinant along the first row:

$\vec{L} = \hat{i}(yp_z - zp_y) - \hat{j}(xp_z - zp_x) + \hat{k}(xp_y - yp_x)$

Comparing the result with $\vec{L} = L_x\hat{i} + L_y\hat{j} + L_z\hat{k}$, we extract the components:

• $L_x = yp_z - zp_y$
• $L_y = zp_x - xp_z$
• $L_z = xp_y - yp_x$

Step 3: Constraint Analysis for Motion in the x-y Plane

If the particle is restricted to the $x-y$ plane, its position and momentum along the $z$-axis must be zero:

$z = 0 \quad \text{and} \quad p_z = 0$

Substituting these constraints into the derived components [By substitution principle]:

• $L_x = y(0) - (0)p_y = 0$
• $L_y = (0)p_x - x(0) = 0$
• $L_z = xp_y - yp_x$

Since $L_x = 0$ and $L_y = 0$, the vector $\vec{L}$ simplifies to $\vec{L} = (xp_y - yp_x)\hat{k}$, which is directed entirely along the $z$-axis.

Final Solution: The components are $L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, and $L_z = xp_y - yp_x$. For motion in the $x-y$ plane ($z=0, p_z=0$), $L_x=0$ and $L_y=0$, proving that $\vec{L}$ resides exclusively along the $z$-axis.

Step 1: Definition of Velocity Vectors

Let the position vector of the $i^{th}$ particle in a laboratory frame be $\vec{r}_i$ and the position of the Centre of Mass (CM) be $\vec{R}$. The position of the $i^{th}$ particle relative to the Centre of Mass is defined as:

$\vec{r}'_i = \vec{r}_i - \vec{R}$ [Definition of Relative Position Vector]

To find the corresponding velocities, we differentiate both sides with respect to time $t$:

$\frac{d\vec{r}'_i}{dt} = \frac{d\vec{r}_i}{dt} - \frac{d\vec{R}}{dt}$

$\vec{v}'_i = \vec{v}_i - \vec{V}$ [Where $\vec{v}_i$ is absolute velocity and $\vec{V}$ is the velocity of the CM]

Step 2: Derivation of Momentum Transformation

Given the velocity relationship $\vec{v}_i = \vec{v}'_i + \vec{V}$, we multiply both sides by the mass of the $i^{th}$ particle, $m_i$:

$m_i \vec{v}_i = m_i (\vec{v}'_i + \vec{V})$

$m_i \vec{v}_i = m_i \vec{v}'_i + m_i \vec{V}$

Substituting the definition of momentum $\vec{p}_i = m_i \vec{v}_i$ and $\vec{p}'_i = m_i \vec{v}'_i$:

$\vec{p}_i = \vec{p}'_i + m_i \vec{V}$

[This confirms the transformation of momentum into the CM frame components.]

Step 3: Proving the Summation of Relative Momenta is Zero

We analyze the sum of the relative momenta: $\sum \vec{p}'_i = \sum m_i \vec{v}'_i$.

Substitute $\vec{v}'_i = \vec{v}_i - \vec{V}$ into the expression:

$\sum \vec{p}'_i = \sum m_i (\vec{v}_i - \vec{V})$

$\sum \vec{p}'_i = \sum m_i \vec{v}_i - \sum m_i \vec{V}$

Since $\vec{V}$ is constant for the system at any instant, we factor it out:

$\sum \vec{p}'_i = \sum \vec{p}_i - \vec{V} \sum m_i$

Using the definition of total mass $M = \sum m_i$ and the definition of CM velocity $\vec{V} = \frac{\sum m_i \vec{v}_i}{M}$:

$\sum \vec{p}_i = M \vec{V}$

Substituting these back into the summation equation:

$\sum \vec{p}'_i = M \vec{V} - \vec{V} (M)$

$\sum \vec{p}'_i = 0$

[Per the definition of the Centre of Mass, the net momentum in the CM frame is identically zero.]

Visual Representation of Frames

Final Solution: The relation $\vec{p}_i = \vec{p}'_i + m_i \vec{V}$ holds by definition of relative velocity, and the sum of relative momenta $\sum \vec{p}'_i = 0$ is proven by substituting the definition of the Centre of Mass, confirming the CM frame is the inertial frame where total momentum is zero.

Given Variables & Initial Setup

• Total mass of the car ($M$) = $1800 \, \text{kg}$
• Distance between axles ($L$) = $1.8 \, \text{m}$
• Distance of the centre of gravity (CG) from front axle ($d_f$) = $1.05 \, \text{m}$
• Distance of the centre of gravity (CG) from back axle ($d_b$) = $L - d_f = 1.8 \, \text{m} - 1.05 \, \text{m} = 0.75 \, \text{m}$
• Acceleration due to gravity ($g$) = $9.8 \, \text{m/s}^2$
• Weight of the car ($W$) = $Mg = 1800 \times 9.8 = 17640 \, \text{N}$

Visual Representation of Forces

Step 1: Applying Translational Equilibrium

For the car to be in vertical equilibrium, the sum of all upward normal forces must equal the weight of the car acting downwards [Per Newton's First Law of Motion]:

$F_f + F_b = Mg = 17640 \, \text{N}$ — (Equation 1)

Where $F_f$ is the total force on both front wheels and $F_b$ is the total force on both back wheels.

Step 2: Applying Rotational Equilibrium

To find the individual forces, we take the moment of forces about the front axle. For rotational equilibrium, the sum of torques about any point must be zero [Per the Principle of Moments]:

$\sum \tau_{\text{front}} = 0$

$(F_b \times L) - (Mg \times d_f) = 0$

$F_b \times 1.8 = 17640 \times 1.05$

$F_b = \frac{18522}{1.8} = 10290 \, \text{N}$

Step 3: Calculating Forces on Wheels

Using Equation 1 to find $F_f$:

$F_f = 17640 - 10290 = 7350 \, \text{N}$

Since the force is distributed equally between the two wheels on each axle:

• Force per front wheel = $\frac{F_f}{2} = \frac{7350}{2} = 3675 \, \text{N}$
• Force per back wheel = $\frac{F_b}{2} = \frac{10290}{2} = 5145 \, \text{N}$

Final Solution

The force exerted on each front wheel is 3675 N, and the force exerted on each back wheel is 5145 N.

1. Role of Friction in Rolling Motion

To initiate rolling motion of a disc on a horizontal surface, an external torque must be applied to overcome inertia. Consider a disc of mass $M$ and radius $R$. When a force $F$ is applied at the center, the disc tends to slide forward. [Per Newton's First Law], an object in motion stays in motion; however, to induce rotation, an angular acceleration $\alpha$ is required.

• Translational motion: The force $F$ causes a linear acceleration $a = F/M$.
• Rotational motion: For the disc to roll, there must be a torque $\tau$ about the center of mass, where $\tau = I\alpha$.
• Mechanism: If there were no friction, the disc would merely translate without rotation. Friction $f$ acts at the contact point in the direction opposite to the potential sliding motion. This creates a torque $\tau = f \cdot R$ which provides the necessary angular acceleration $\alpha = \frac{fR}{I}$.

2. Force of Friction in Perfect Rolling

Perfect rolling (rolling without slipping) is defined by the condition $v = \omega R$ and $a = \alpha R$. Let us derive the force of friction $f$ for a disc ($I = \frac{1}{2}MR^2$) being pushed by a force $F$ at its center.

Step 1: Write the equations of motion.

Translational: $F - f = Ma$ [By Newton's Second Law]

Rotational: $fR = I\alpha = \left(\frac{1}{2}MR^2\right) \left(\frac{a}{R}\right)$ [By $\tau = I\alpha$]

Step 2: Simplify the rotational equation.

$fR = \frac{1}{2}MRa \implies f = \frac{1}{2}Ma$

Step 3: Substitute into the translational equation.

$F - \frac{1}{2}Ma = Ma$

$F = \frac{3}{2}Ma \implies a = \frac{2F}{3M}$

Step 4: Solve for friction $f$.

$f = \frac{1}{2}M \left(\frac{2F}{3M}\right) = \frac{1}{3}F$

3. Visual Representation of Forces

Analytical Summary

Parameter	Expression
Linear Acceleration ($a$)	$\frac{2F}{3M}$
Angular Acceleration ($\alpha$)	$\frac{2F}{3MR}$
Frictional Force ($f$)	$\frac{F}{3}$

Once perfect rolling is established at a constant velocity, the acceleration $a$ becomes zero if the external driving force $F$ is removed. In the absence of an external force, $f = 0$. However, as long as the accelerating force $F$ is applied to the center of the disc, the frictional force required to maintain the rolling condition is exactly one-third of the applied force.

Final Solution: The force of friction after perfect rolling begins (assuming the force $F$ is maintained) is $f = \frac{1}{3}F$.

Step 1: Theoretical Definition of Rolling Motion

Rolling motion is defined as the combination of translational motion of the center of mass (CM) and rotational motion about that CM. For a body of radius $R$ and angular velocity $\omega$, the condition for pure rolling without slipping is $v_{CM} = R\omega$.

Step 2: Analysis of Force of Friction in Rolling

When a body rolls on a surface, the point of contact experiences a tendency to slip. The direction of the static frictional force $f$ is determined by the requirement to oppose this relative motion at the contact point.

• If the body is rolling freely on a horizontal surface (without external driving torque), the force of friction is zero.
• If an external force is applied (e.g., a pulling force at the CM), the frictional force acts to maintain the rolling condition.

Step 3: Evaluating the Statement

The statement claims the frictional force acts in the same direction as the motion of the CM. Let us evaluate this under standard conditions:

1. Acceleration case: If a force $F$ is applied at the CM to accelerate the body, the body tends to slip backward at the point of contact. Friction acts forward to oppose this sliding, aligning with $v_{CM}$.
2. Deceleration/Braking case: If the body is rolling and a retarding torque is applied, the contact point tends to move forward. Friction acts backward to oppose this, which is opposite to $v_{CM}$.
3. Pure Rolling (Free): No friction exists ($f=0$).

Step 4: Conclusion

The statement is False. [Per the laws of classical mechanics, specifically Newton's Second Law for rotation and translation], the direction of frictional force is determined by the relative motion of the point of contact. Because it is possible for friction to act in the direction opposite to the motion of the CM (e.g., during braking or when a torque is applied to reduce angular velocity), the assertion that it always acts in the same direction is physically incorrect.

Final Solution: The statement is False. The direction of the force of friction depends on the nature of the motion (accelerating vs. decelerating) and the points of application of external forces/torques. It is not constrained to act in the same direction as the motion of the center of mass.

Step 1: Identifying Given Parameters

The system comprises a hollow cylinder rotating about its central axis. The following parameters are provided:

• Mass of cylinder ($M$) = $3\text{ kg}$
• Radius of cylinder ($R$) = $40\text{ cm} = 0.4\text{ m}$
• Applied Force ($F$) = $30\text{ N}$

Step 2: Moment of Inertia Calculation

For a hollow cylinder, the mass is concentrated at the distance of the radius $R$ from the axis. [By the definition of Moment of Inertia for a thin cylindrical shell/hollow cylinder]:

$I = MR^2$

$I = 3\text{ kg} \times (0.4\text{ m})^2 = 3 \times 0.16 = 0.48\text{ kg}\cdot\text{m}^2$

Step 3: Calculating Torque and Angular Acceleration

Torque ($\tau$) is the product of the applied force and the perpendicular distance from the axis of rotation [Per the definition $\vec{\tau} = \vec{r} \times \vec{F}$]:

$\tau = F \times R = 30\text{ N} \times 0.4\text{ m} = 12\text{ N}\cdot\text{m}$

Using Newton’s Second Law for Rotation, $\tau = I\alpha$, where $\alpha$ is the angular acceleration:

$\alpha = \frac{\tau}{I} = \frac{12\text{ N}\cdot\text{m}}{0.48\text{ kg}\cdot\text{m}^2}$

$\alpha = 25\text{ rad/s}^2$

Step 4: Calculating Linear Acceleration

Given the condition of no slipping, the linear acceleration ($a$) of the rope is related to the angular acceleration of the cylinder by the tangential acceleration formula [Per the kinematic constraint $a = r\alpha$]:

$a = R\alpha = 0.4\text{ m} \times 25\text{ rad/s}^2 = 10\text{ m/s}^2$

Visual Representation of the System

Summary of Calculated Values

Parameter	Value
Angular Acceleration ($\alpha$)	$25\text{ rad/s}^2$
Linear Acceleration ($a$)	$10\text{ m/s}^2$

Final Solution: The angular acceleration of the cylinder is $25\text{ rad/s}^2$ and the linear acceleration of the rope is $10\text{ m/s}^2$.

Step 1: Identifying Given Parameters and Constants

To determine the new angular speed, we define the initial and final states of the system based on the Principle of Conservation of Angular Momentum [Per Newton's Second Law for Rotation, $\tau_{ext} = \frac{dL}{dt}$; since friction is neglected, $\tau_{ext} = 0$, implying $L_i = L_f$].

• Initial angular speed ($\omega_i$): $30 \text{ rpm} = 30 \times \frac{2\pi}{60} \text{ rad/s} = \pi \text{ rad/s} \approx 3.14159 \text{ rad/s}$.
• Mass of each weight ($m$): $5 \text{ kg}$. Total mass of weights $= 10 \text{ kg}$.
• Initial distance ($r_1$): $90 \text{ cm} = 0.9 \text{ m}$.
• Final distance ($r_2$): $20 \text{ cm} = 0.2 \text{ m}$.
• Moment of inertia of man + platform ($I_p$): $7.6 \text{ kg m}^2$.

Step 2: Calculating Moments of Inertia

The total moment of inertia ($I$) is the sum of the moment of inertia of the man/platform and the point-mass weights ($I = I_p + \sum mr^2$).

Initial Moment of Inertia ($I_i$):

$I_i = I_p + 2(mr_1^2)$ $I_i = 7.6 + 2(5 \times 0.9^2)$ $I_i = 7.6 + 2(5 \times 0.81) = 7.6 + 8.1 = 15.7 \text{ kg m}^2$

Final Moment of Inertia ($I_f$):

$I_f = I_p + 2(mr_2^2)$ $I_f = 7.6 + 2(5 \times 0.2^2)$ $I_f = 7.6 + 2(5 \times 0.04) = 7.6 + 0.4 = 8.0 \text{ kg m}^2$

Step 3: Applying Conservation of Angular Momentum

Since the net external torque is zero, the angular momentum remains constant [Law of Conservation of Angular Momentum, $L = I\omega$]:

$I_i \omega_i = I_f \omega_f$ $15.7 \times \pi = 8.0 \times \omega_f$

Step 4: Solving for New Angular Speed ($\omega_f$)

$\omega_f = \frac{15.7 \times \pi}{8.0}$ $\omega_f = 1.9625 \times \pi \text{ rad/s}$

Converting back to revolutions per minute (rpm):

$\text{New speed in rpm} = \frac{\omega_f \times 60}{2\pi}$ $\text{New speed in rpm} = \frac{1.9625 \times \pi \times 60}{2\pi} = 1.9625 \times 30 = 58.875 \text{ rpm}$

Final Solution: The new angular speed is $58.875 \text{ rpm}$ (or approximately $6.16 \text{ rad/s}$).

1. Theoretical Foundation of Rolling Motion

For a disc to undergo rolling motion, it must satisfy two conditions simultaneously: linear motion of its center of mass ($v_{cm}$) and rotational motion about its center of mass ($\omega$). Pure rolling is defined by the constraint $v_{cm} = R\omega$. When a disc is placed on a surface with an initial velocity $v_0$ but without sufficient initial rotation, the contact point $B$ experiences a relative sliding velocity $v_B = v_{cm} - R\omega$ [Per the principle of relative motion].

2. Role of Friction in Rolling

Friction is the external agency that provides the torque necessary to change the angular velocity of the disc. Without friction, the disc would continue to slide indefinitely at constant $v_{cm}$ and $\omega = 0$. Friction at the point of contact $B$ acts to oppose the direction of sliding, thereby decelerating the linear motion and accelerating the angular motion [Per Newton’s Second Law for Rotation: $\vec{\tau} = I\vec{\alpha}$].

3. Direction of Frictional Force and Torque

Given the disc is moving forward with $v_{cm}$ and initially has no rotation or insufficient rotation ($\omega$ is small), the contact point $B$ has a velocity relative to the ground pointing forward. Therefore, the kinetic frictional force $f_k$ acts in the direction opposite to the sliding velocity to oppose it [Per the Law of Kinetic Friction].

• Direction of Frictional Force at B: Backward (opposite to the direction of $v_{cm}$).
• Sense of Frictional Torque: Since the force acts at the bottom point $B$ directed backward, the torque $\vec{\tau} = \vec{r} \times \vec{f}$ relative to the center of mass $C$ is directed into the page. According to the right-hand rule, this produces an angular acceleration $\alpha$ that induces a clockwise rotation, increasing the magnitude of $\omega$ [Per Right-Hand Thumb Rule].

4. Visual Representation of Dynamics

5. Conclusion

The frictional force $f_k$ acts in the direction opposite to the center-of-mass velocity to reduce $v_{cm}$ and simultaneously creates a torque $\tau = f_k R$ about the center of mass. This torque provides the angular acceleration $\alpha = \frac{\tau}{I} = \frac{f_k R}{MR^2/2} = \frac{2f_k}{MR}$, which increases the angular velocity $\omega$ until the condition of pure rolling ($v = R\omega$) is achieved.

Final Solution: The frictional force at point B acts in the backward direction. The frictional torque is directed such that it produces a clockwise angular acceleration to initiate or increase the rotation of the disc.

Step 1: Establishing the Principle of Conservation of Angular Momentum

In a rotating system where no external torques act on the body (since the turntable is frictionless and the system is isolated), the total angular momentum $L$ of the system must remain conserved. This is defined by the law of conservation of angular momentum:

$L_i = L_f$ [Per the Principle of Conservation of Angular Momentum]

Where:

• $L = I\omega$
• $I$ is the moment of inertia of the system.
• $\omega$ is the angular velocity of the system.

Step 2: Identifying Given Variables and Initial Conditions

Let the initial state of the child and turntable be denoted by subscript $i$ and the final state by subscript $f$.

• Initial angular speed, $\omega_i = 40 \text{ rev/min}$
• Initial moment of inertia = $I_i$
• Final moment of inertia, $I_f = \frac{2}{5} I_i$
• Final angular speed = $\omega_f$ (to be determined)

Step 3: Algebraic Derivation

Substituting the expression for angular momentum into the conservation equation:

$I_i \omega_i = I_f \omega_f$

Rearranging to solve for $\omega_f$:

$\omega_f = \frac{I_i \omega_i}{I_f}$

Substituting the relationship $I_f = \frac{2}{5} I_i$ into the equation:

$\omega_f = \frac{I_i \times 40}{\frac{2}{5} I_i}$

Canceling $I_i$ from the numerator and denominator:

$\omega_f = 40 \times \frac{5}{2}$

$\omega_f = 20 \times 5$

$\omega_f = 100 \text{ rev/min}$

Step 4: Geometric Representation of the Rotational System

Step 5: Analysis of Results

Because the angular momentum is conserved, any reduction in the moment of inertia ($I$) must result in a proportional increase in angular velocity ($\omega$). Since $I_f < I_i$, it is physically necessary that $\omega_f > \omega_i$. The factor of decrease in moment of inertia is $\frac{2}{5}$, which necessitates an inverse increase in velocity by a factor of $\frac{5}{2} = 2.5$.

Final Solution: The angular speed of the child is 100 rev/min.

Part 1: Location of the Centre of Mass of a Uniform Ring

A uniform ring is defined as a circular lamina of negligible thickness with a constant linear mass density ($\lambda$). By definition, the centre of mass ($R_{cm}$) of a system of particles is given by the weighted average of the position vectors of all mass elements $dm$:

$\vec{R}_{cm} = \frac{1}{M} \int \vec{r} \, dm$ [Definition of Centre of Mass]

For a uniform ring of radius $R$ centered at the origin $(0,0)$ in the $xy$-plane:

• The mass elements $dm$ are symmetrically distributed around the center.
• For every mass element at coordinates $(x, y)$, there exists an identical mass element at $(-x, -y)$.
• When integrating over the entire circumference ($2\pi R$), the vector contributions cancel out: $\int x \, dm = 0$ and $\int y \, dm = 0$.

Consequently, the centre of mass of a uniform ring is located exactly at its geometric center.

Part 2: Analytical Inquiry into the Position of Centre of Mass

Question: Does the centre of mass of a body necessarily lie inside the body?

Answer: No, the centre of mass of a body does not necessarily lie within the material of the body itself.

Theoretical Justification: The centre of mass is a mathematical point defined by the distribution of mass. It represents the point where the total mass of the system could be considered concentrated for the purpose of translational motion analysis [Newton’s Second Law for a system of particles: $\vec{F}_{ext} = M\vec{a}_{cm}$].

Object	CM Location	Inside Body?
Uniform Sphere	Geometric Center	Yes
Uniform Ring	Geometric Center	No (It is in the empty space)
Uniform "L" shaped rod	Intersection of symmetry lines	No (It is in the empty space)

The position of the centre of mass is dependent solely on the distribution of the mass elements $dm$ and their respective position vectors $\vec{r}$. If the mass distribution is such that there is no matter at the calculated coordinates of the centre of mass, the point will reside in space outside the boundaries of the physical object.

Final Solution: The centre of mass of a uniform ring is at its geometric center. The centre of mass does not necessarily lie inside the body; it is a point that may exist in the empty space surrounding or adjacent to the object, depending on the spatial distribution of its mass.

Step 1: Identifying the Condition for Pure Rolling

For a body of mass $M$ and radius $R$ rolling down an incline of angle $\theta$ without skidding, the force of static friction $f$ required to provide the necessary torque for rotation is given by the equation of motion:

$Mg \sin\theta - f = Ma$ [Newton's Second Law]

The torque $\tau$ about the center of mass providing angular acceleration $\alpha$ is:

$\tau = fR = I\alpha$

Since the cylinder is rolling without slipping, the kinematic constraint is $a = R\alpha$. For a solid cylinder, the moment of inertia is $I = \frac{1}{2}MR^2$.

Step 2: Deriving the Frictional Force

Substituting $\alpha = a/R$ into the torque equation:

$fR = \left(\frac{1}{2}MR^2\right) \left(\frac{a}{R}\right) \implies f = \frac{1}{2}Ma$

Substitute $f = \frac{1}{2}Ma$ into the force equation $Mg \sin\theta - f = Ma$:

$Mg \sin\theta - \frac{1}{2}Ma = Ma$

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

Now, calculate $f$ in terms of $\theta$:

$f = \frac{1}{2}M\left(\frac{2}{3}g \sin\theta\right) = \frac{1}{3}Mg \sin\theta$

Step 3: Defining the Skidding Threshold

Pure rolling continues as long as the required static friction does not exceed the maximum limiting friction:

$f \leq f_{max} = \mu_s N$

The normal force on the incline is $N = Mg \cos\theta$. Therefore, the condition for rolling without skidding is:

$\frac{1}{3}Mg \sin\theta \leq \mu_s Mg \cos\theta$

$\frac{1}{3} \sin\theta \leq \mu_s \cos\theta \implies \tan\theta \leq 3\mu_s$

Step 4: Calculation of the Critical Angle

The cylinder begins to skid when the required friction exceeds the limit, i.e., when $\tan\theta = 3\mu_s$.

Given $\mu_s = 0.25$:

$\tan\theta = 3 \times 0.25 = 0.75$

$\theta = \arctan(0.75)$

$\theta \approx 36.87^\circ$

Final Solution: The cylinder begins to skid when the inclination angle $\theta$ reaches $\arctan(0.75)$, which is approximately $36.9^\circ$.

1. Analysis of the Centre of Mass of a Uniform Cylinder

For a body of uniform mass density (homogeneous body), the centre of mass (CM) coincides with the geometric centre of the body. In a right circular cylinder of uniform density:

• The mass is symmetrically distributed about the central axis of the cylinder.
• The vertical position of the CM lies exactly at the midpoint of the central axis.
• The horizontal position of the CM lies exactly on the central longitudinal axis.

Therefore, for a cylinder of length $L$ and radius $R$, if the cylinder is placed with its base on the $xy$-plane, the CM is located at coordinates $(0, 0, \frac{L}{2})$ [By the principle of symmetry for uniform density distributions].

2. Geometric Visualization

3. Does the Centre of Mass Necessarily Lie Inside the Body?

No, the centre of mass of a body does not necessarily lie within the material of the body. The CM is a mathematical point defined by the mass-weighted average position of all particles in the system, given by the formula:

$$\vec{R}_{cm} = \frac{1}{M} \int \vec{r} \, dm$$

The location of the CM depends on the distribution of mass, not the existence of material at that specific coordinate. Examples include:

• Hollow Objects: A metallic ring or a hollow sphere has its CM at the geometric centre, which is a void space (empty).
• Curved Objects: A boomerang or a horseshoe-shaped object has its CM located in the space between its arms, outside the physical material [By the property of non-convex mass distributions].

Final Solution: The centre of mass of a uniform cylinder is located at the geometric centre (the midpoint of its longitudinal axis). The centre of mass of a body does not necessarily lie inside the body; it can lie in the empty space surrounding or within the body, depending on the object's geometry and mass distribution.

Step 1: Given Variables & Initial Setup

The system under consideration is a solid cylinder rotating about its longitudinal symmetry axis. The parameters provided are as follows:

• Mass of the cylinder ($M$): $20\text{ kg}$
• Radius of the cylinder ($R$): $0.25\text{ m}$
• Angular velocity ($\omega$): $100\text{ rad s}^{-1}$

The moment of inertia ($I$) for a solid cylinder rotating about its central axis is defined by the formula: $I = \frac{1}{2}MR^2$ [Per the definition of the Moment of Inertia for a uniform solid cylinder]

Step 2: Calculation of Moment of Inertia ($I$)

Substitute the given values into the formula:

$I = \frac{1}{2} \times 20\text{ kg} \times (0.25\text{ m})^2$

$I = 10 \times 0.0625\text{ kg m}^2$

$I = 0.625\text{ kg m}^2$

Step 3: Calculating Rotational Kinetic Energy ($K_r$)

The rotational kinetic energy is governed by the relation: $K_r = \frac{1}{2}I\omega^2$ [Per the Principle of Rotational Kinetic Energy]

$K_r = \frac{1}{2} \times 0.625 \times (100)^2$

$K_r = 0.5 \times 0.625 \times 10,000$

$K_r = 0.5 \times 6,250$

$K_r = 3,125\text{ J}$

Step 4: Calculating Angular Momentum ($L$)

The angular momentum is defined as the product of the moment of inertia and angular velocity: $L = I\omega$ [Per the definition of Angular Momentum for a rigid body]

$L = 0.625\text{ kg m}^2 \times 100\text{ rad s}^{-1}$

$L = 62.5\text{ kg m}^2\text{ s}^{-1}$ (or $62.5\text{ J s}$)

Visual Representation of Rotational Parameters

Final Solution: The rotational kinetic energy of the cylinder is $3,125\text{ J}$, and the magnitude of its angular momentum is $62.5\text{ kg m}^2\text{ s}^{-1}$.

Step 1: Identifying Physical Principles and Kinematics

To determine the time taken for a solid cylinder to return to the bottom of an inclined plane, we recognize that the motion is governed by the conservation of mechanical energy and the equations of translational and rotational kinematics. [Per the Law of Conservation of Energy].

The time taken to ascend to the highest point ($t_{up}$) is equal to the time taken to descend from the highest point back to the bottom ($t_{down}$), as the acceleration $a$ is constant throughout both phases. Thus, the total time is $T = 2 \times t_{up}$.

Step 2: Calculating Acceleration along the Incline

For a rolling body on an incline of angle $\theta$, the acceleration $a$ is given by the formula:

$a = \frac{g \sin\theta}{1 + \frac{K^2}{R^2}}$

Where $K$ is the radius of gyration and $R$ is the radius of the object. For a solid cylinder, the moment of inertia $I = \frac{1}{2}MR^2$. Since $I = MK^2$, we have $K^2 = \frac{1}{2}R^2$, implying $\frac{K^2}{R^2} = 0.5$.

Substituting the values: $\theta = 30^\circ$, $g \approx 9.8 \, m/s^2$:

$a = \frac{9.8 \cdot \sin(30^\circ)}{1 + 0.5} = \frac{9.8 \cdot 0.5}{1.5} = \frac{4.9}{1.5} \approx 3.267 \, m/s^2$

Step 3: Calculating Time of Ascent ($t_{up}$)

The cylinder moves against gravity. At the maximum height, the final velocity $v = 0$. Using the first equation of motion [By the Definition of Kinematic Acceleration]:

$v = u - at_{up}$

$0 = 5 - (3.267)t_{up}$

$t_{up} = \frac{5}{3.267} \approx 1.53 \, s$

Step 4: Calculating Total Time of Flight

Since the motion is symmetric in a vacuum/ideal rolling condition [Per the Principle of Reversibility of Motion], the total time $T$ is:

$T = 2 \times t_{up} = 2 \times 1.53 = 3.06 \, s$

Visual Representation of the Mechanical System

Summary of Data

Parameter	Value
Initial Velocity ($u$)	$5 \, m/s$
Acceleration ($a$)	$3.267 \, m/s^2$
Time to ascend ($t_{up}$)	$1.53 \, s$
Total time ($T$)	$3.06 \, s$

Final Solution: The total time taken for the solid cylinder to return to the bottom of the inclined plane is 3.06 seconds.

Step 1: System Definition and Coordinate Geometry

Consider a Cartesian coordinate system where the two particles, $P_1$ and $P_2$, each of mass $m$, move with velocity vectors $\vec{v}_1 = v\hat{i}$ and $\vec{v}_2 = -v\hat{i}$. The particles travel along lines parallel to the x-axis, separated by a distance $d$ along the y-axis.

• Position of $P_1$: $\vec{r}_1 = x_1\hat{i} + y_1\hat{j}$
• Position of $P_2$: $\vec{r}_2 = x_2\hat{i} + (y_1 + d)\hat{j}$

Step 2: Angular Momentum about an Arbitrary Point O

Let an arbitrary point $O$ be the origin $(0,0,0)$. The angular momentum $\vec{L}$ of the system is the sum of the angular momenta of the individual particles [Per the Principle of Superposition of Angular Momentum]:

$\vec{L} = \vec{r}_1 \times \vec{p}_1 + \vec{r}_2 \times \vec{p}_2$

Substituting $\vec{p}_1 = mv\hat{i}$ and $\vec{p}_2 = -mv\hat{i}$:

$\vec{L} = (x_1\hat{i} + y_1\hat{j}) \times (mv\hat{i}) + (x_2\hat{i} + (y_1 + d)\hat{j}) \times (-mv\hat{i})$

Step 3: Algebraic Expansion and Vector Cross Product

Utilizing the properties of cross products ($\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$):

$\vec{L} = [x_1mv(\hat{i} \times \hat{i}) + y_1mv(\hat{j} \times \hat{i})] + [x_2(-mv)(\hat{i} \times \hat{i}) + (y_1 + d)(-mv)(\hat{j} \times \hat{i})]$

$\vec{L} = 0 - y_1mv\hat{k} + 0 + (y_1 + d)mv\hat{k}$

$\vec{L} = (-y_1mv + y_1mv + dmv)\hat{k}$

$\vec{L} = dmv\hat{k}$

Step 4: Analysis of Invariance

The resulting expression $\vec{L} = dmv\hat{k}$ is independent of the coordinates $(x_1, y_1)$ and $(x_2, y_1+d)$. This demonstrates that the choice of the origin (the point about which angular momentum is calculated) does not affect the total angular momentum of the system.

Parameter	Value
Net Angular Momentum	$mv d$
Direction	Perpendicular to the plane of motion

Final Solution: Since the coordinates $(x, y)$ of the particles cancel out during the vector summation, the angular momentum $\vec{L} = dmv\hat{k}$ is invariant and remains constant regardless of the reference point O.