Q.For a point on a rotating rigid body, the graph of its angular position $\theta$ (plotted on the vertical axis) against time $t$ (plotted on the horizontal axis) is a straight line of constant positive slope: $\theta$ increases uniformly with $t$, passing steadily through successive instants $t_1 < t_2 < t_3$. Is the body rotating clockwise or anti-clockwise? Give the reason.

For SHM, $v^2=\omega^2\left(A^2-x^2\right)$. Using the two data points:

Dividing:

$256\left(A^2-64\right)=144\left(A^2-36\right)\;\Rightarrow\;112A^2=11200\;\Rightarrow\;A^2=100$ …

Concept & Intuition

The key idea is the relationship between force, work, and energy. If a force is always perpendicular to the velocity, then the instantaneous power (rate of doing work) is zero:

No work means no change in kinetic energy. This is exactly what happens in uniform circular motion: a centripetal force (constant magnitude, always perpendicular to velocity) keeps the speed constant while changing only the direction. So kinetic energy stays constant, but velocity (a vector) changes direction, and acceleration also changes direction (though its magnitude may be constant). Let’s check each option carefully.

Step-by-step reasoning

1. Force perpendicular to velocity → no work done Work done by a force over a small displacement $d\vec{s} = \vec{v}\,dt$ is

Since $\vec{F} \perp \vec{v}$ at every instant, $\vec{F} \cdot \vec{v} = 0$, so $dW = 0$. No work means no change in kinetic energy:

2. Kinetic energy constant → speed constant

   Kinetic energy $K = \frac{1}{2} m v^2$. If $K$ is constant and mass is constant, then speed $v = |\vec{v}|$ is constant. But velocity $\vec{v}$ is a vector; its direction can change. So option (C) “velocity is constant” is false — only its magnitude is constant.

3. Linear momentum

   Linear momentum $\vec{p} = m\vec{v}$. Since $\vec{v}$ changes direction, $\vec{p}$ changes direction too. Hence momentum is not constant. Option (A) is false.

4. Acceleration …

When an object rotates, its angular speed tells us how quickly its angular position changes. It's a measure of how many radians (or degrees, or rotations) it covers per unit of time. For a merry-go-round rotating at a constant rate, we can find its angular speed by dividing the total angular displacement by the total time taken.

The key idea here is understanding the relationship between rotations and radians. One complete rotation corresponds to an angular displacement of $2\pi$ radians.

1. Identify the given information:

   We are given:

   • Number of rotations = 9
   • Time taken ($t$) = 18 seconds

2. Calculate the total angular displacement:

   Each full rotation corresponds to an angular displacement of $2\pi$ radians. Therefore, for 9 rotations, the total angular displacement ($\theta$) is:

$$ \theta = 18\pi \text{ radians} $$ …

The problem describes the rotational motion of a wheel, which can be broken down into two distinct phases based on its angular acceleration. To solve this, we need to apply the principles of rotational kinematics, which are directly analogous to linear kinematics.

Concept and Intuition

1. Rotational Kinematics with Constant Angular Acceleration:

   When a wheel rotates with a constant angular acceleration ($\alpha$), its angular velocity ($\omega$) and angular displacement ($\Delta\theta$) change over time ($t$) according to specific equations. These are very similar to the equations for linear motion:

   • Angular velocity: $\omega = \omega_0 + \alpha t$
   • Angular displacement: $\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ Here, $\omega_0$ is the initial angular velocity.

2. Motion with Zero Angular Acceleration:

   If the angular acceleration is zero, it means the angular velocity is constant. In this case, the angular displacement is simply the product of the constant angular velocity and the time duration:

   • Angular displacement: $\Delta\theta = \omega \times t$

3. Converting Radians to Revolutions:

   Angular displacement is typically measured in radians. To express it in terms of revolutions, we use the conversion factor: $1 \text{ revolution} = 2\pi \text{ radians}$.

We will first determine the angular acceleration and then calculate the angular displacement for each phase, finally summing them up and converting to revolutions.

Step-by-step Solution

1. Identify the initial conditions and define the phases of motion.

   The problem states that the wheel undergoes constant angular acceleration from $t=0$ to $t=20$ s. After $t=20$ s, the angular acceleration becomes zero. We are given that the angular velocity at $t=2$ s is $5$ rad/s.

   Since the initial angular velocity at $t=0$ is not explicitly given, we make a standard assumption:

   Important

   In problems where the initial state of motion is not specified, it is generally assumed that the object starts from rest. Therefore, we assume the initial angular velocity ($\omega_0$) at $t=0$ s is $0$ rad/s.

   The motion can be divided into two phases:

   • Phase 1: From $t=0$ s to $t=20$ s. During this phase, the angular acceleration ($\alpha$) is constant.
   • Phase 2: From $t=20$ s to $t=50$ s. During this phase, the angular acceleration is zero, meaning the angular velocity is constant.

2. Determine the angular acceleration ($\alpha$) during Phase 1.

   We use the kinematic equation for angular velocity:

   $\omega = \omega_0 + \alpha t$

   Given:

   • $\omega_0 = 0$ rad/s (initial angular velocity at $t=0$ s)
   • $\omega = 5$ rad/s (angular velocity at $t=2$ s)
   • $t = 2$ s

   Substitute these values into the formula:

   $5 = 0 + \alpha (2)$

   $2\alpha = 5$

   $\alpha = \frac{5}{2} = 2.5 \text{ rad/s}^2$

3. Calculate the angular displacement ($\Delta\theta_1$) during Phase 1 ($t=0$ s to $t=20$ s).

   We use the kinematic equation for angular displacement:

   $\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2$

   Given:

   • $\omega_0 = 0$ rad/s
   • $\alpha = 2.5$ rad/s$^2$
   • $t_1 = 20$ s (duration of Phase 1)

   Substitute these values:

   $\Delta\theta_1 = (0)(20) + \frac{1}{2}(2.5)(20)^2$

   $\Delta\theta_1 = 0 + \frac{1}{2}(2.5)(400)$ …

The key idea is that a hoop placed on a rough surface with initial spin but no forward speed will experience kinetic friction. That friction does two things: it reduces the angular speed and increases the translational speed, until the condition for pure rolling (no slipping) is met: $v = R\omega$. Because friction is the only horizontal force, we can use either torque-impulse or conservation of angular momentum about a point where friction produces no torque — the contact point is the clever choice.

1. Set up the initial and final states.

   Initially, the hoop spins at $\omega_0$ and its centre is at rest: $v_0 = 0$. The hoop has radius $R = 0.5\ \text{m}$ and mass $m = 1\ \text{kg}$.

   When slipping stops, the hoop rolls without slipping, so $v = R\omega$, where $\omega$ is the final angular speed.

2. Why friction is the only agent of change.

   The rough surface exerts a kinetic friction force $f_k$ opposite to the slipping direction. Initially the bottom point of the hoop moves backward relative to the ground (since the hoop spins but doesn’t translate), so friction acts forward on the hoop. This force accelerates the centre of mass forward and simultaneously produces a torque that slows the rotation.

3. Use angular momentum conservation about the contact point.

   About the instantaneous point of contact with the ground, friction exerts zero torque (its line of action passes through that point). The only other force is gravity and the normal reaction, both vertical and passing through the centre — they also produce no torque about the contact point. Therefore, angular momentum about the contact point is conserved.

   Tip

   Choosing the contact point eliminates the unknown friction force entirely. This is the fastest route.

4. Compute initial angular momentum about the contact point.

   Initially the centre of mass is at rest, so the hoop’s angular momentum about the contact point is just its spin angular momentum about its own centre (since the centre is directly above the contact point, and the centre’s velocity is zero).

   For a hoop, moment of inertia about its centre is $I_{\text{cm}} = mR^2$.

   Initial angular momentum:

5. Compute final angular momentum about the contact point. …

The key idea here is that "linear acceleration" in uniform circular motion doesn't mean tangential acceleration — that's zero because the speed is constant. Instead, it refers to the centripetal acceleration, which always points toward the center and is responsible for changing the direction of velocity.

Let’s walk through it step by step.

1. Find the speed. The particle completes one full circle of radius $r = 5$ cm in time $T = 5$ s. The distance traveled in one revolution is the circumference, $2\pi r = 2\pi \times 5 = 10\pi$ cm. Since speed is uniform,

2. Recall the formula for centripetal acceleration. For any object moving in a circle of radius $r$ with constant speed $v$, the magnitude of the centripetal acceleration is

This is the only acceleration present here — there is no tangential component.

3. Plug in the numbers. $a_c = \frac{(2\pi)^2}{5} = \frac{4\pi^2}{5} = 0.8\pi^2 \ \text{cm/s}^2.$ …