Q.A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

Concept and intuition

Torque $\tau$ is the rotational analogue of force: $\tau = I \alpha$, where $I$ is the moment of inertia about the rotation axis and $\alpha$ is the angular acceleration. Here the axis passes through the centre O and is perpendicular to the square frame. The frame itself is massless, so only the five point masses contribute to $I$. For a point mass $m$ at distance $r$ from the axis, $I = m r^2$. The four corner masses are all at the same distance from O (half the diagonal of the square), and the central mass is exactly on the axis, so its distance is zero. Summing these contributions gives the total $I$; multiplying by $\alpha$ yields the required torque.

Step-by-step solution

1. Find the distance from centre O to a corner. The square has side $a = 0.2\ \mathrm{m}$. The distance from the centre to any corner is half the diagonal:

2. Moment of inertia of the four corner masses. Each corner mass is $m = 2\ \mathrm{kg}$. For one corner:

Four such masses:

3. Moment of inertia of the central mass. The central mass (40 kg) lies exactly on the axis, so its distance from the axis is $0$. Hence:

4. Total moment of inertia.

5. Given angular acceleration.

6. Compute the torque.

Express as a fraction: $0.08 = \frac{8}{100} = \frac{2}{25}$. Thus:

Data. $I_1 = 4\ \mathrm{kg\,m^2},\ r_1 = 0.20\ \mathrm{m},\ I_2 = 20\ \mathrm{kg\,m^2},\ r_2 = 0.30\ \mathrm{m},\ \tau_1 = 10\ \mathrm{N\,m}$.

Step 1 — (a) Angular acceleration of the smaller wheel.

Rotational form of Newton's second law:

So (a) → (iii).

Step 2 — Belt constraint.

A non-slipping belt has the same linear (tangential) speed, hence the same tangential acceleration, at both rims:

Step 3 — (c) Angular acceleration of the larger wheel.

So (c) → (i).

Step 4 — (b) Torque on the larger wheel.

So (b) → (ii).

Step 5 — Assemble the match. a - iii, b - ii, c - i.

ANSWER: D

Concept

Torque about a chosen point is $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ points from that point to the point of application of the force. Here the force acts at the origin, so $\vec{r}$ runs from the reference point $P$ to the origin.

Solution

1. Position vector:

2. Cross product with $\vec{F} = (-F,0,0)$:

3. Components:

4. Assemble:

The string wound on the rim links the block's linear acceleration to the wheel's angular acceleration by $a = \alpha R$; the tension supplies the only torque.

1. Linear acceleration from the fall ($s = \tfrac12 a t^2$):

2. Newton's law for the hanging block (taking $g = 10\text{ m/s}^2$):

3. Rotational law for the wheel, with $\alpha = a/R$ and torque $TR$:

4. Solve for $R$:

Energy check: $mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$ with $v = at = 4$ m/s gives $40 = 8 + \tfrac12\,(16/R^2)$, so $R = 0.5$ m.

The key idea is that work done by a constant torque equals the change in rotational kinetic energy. Since the body starts from rest, the work done is simply the final rotational kinetic energy:

We know the moment of inertia $I = 5\ \mathrm{kg\,m^2}$ and the torque $\tau = 10\ \mathrm{N\,m}$ applied for $t = 25\ \mathrm{s}$. To find $\omega$, we use the rotational analogue of Newton’s second law:

where $\alpha$ is the angular acceleration. Once we have $\alpha$, we can find $\omega$ from $\omega = \alpha t$ (starting from rest). Then plug into the work formula.

1. Find angular acceleration From $\tau = I \alpha$:

2. Find final angular velocity Since the body starts from rest ($\omega_0 = 0$) and accelerates uniformly:

3. Compute work done Work = change in rotational kinetic energy:

Step 1 — Central disc's own moment of inertia.

The central disc's axis passes through its own center, so it simply spins about itself: $I_{central} = \dfrac{1}{2}mR^2$.

Step 2 — Each outer disc, by the parallel-axis theorem.

Every outer disc touches the central disc, so its center sits $2R$ from the axis:

$I_{outer} = \dfrac{1}{2}mR^2 + m(2R)^2 = \dfrac{1}{2}mR^2 + 4mR^2 = \dfrac{9}{2}mR^2$

Step 3 — Sum all six outer discs plus the central disc.

$I_{total} = \dfrac{1}{2}mR^2 + 6 \times \dfrac{9}{2}mR^2 = \dfrac{1}{2}mR^2 + 27mR^2 = \dfrac{55}{2}mR^2$

The alternate reading (option D, $27mR^2$): if the question means only the SIX OUTER discs (excluding the central disc's own spin), the answer is just $6 \times \dfrac{9}{2}mR^2 = 27mR^2$ — this is exactly the genuine ambiguity KEA's own final key resolved by crediting both.

Torque about O is $\tau = rF\sin\theta$, where $\theta$ is the angle between $\vec{OP}$ and $\vec F$.

• $\vec{OP}$ is at $55^\circ$ to the x-axis.
• $\vec F$ is at $25^\circ$ to the horizontal (x-axis).
• Angle between them: $\theta = 55^\circ - 25^\circ = 30^\circ$.

With $r = 2\ \text{cm} = 0.02\ \text{m}$ and $F = 200\ \text{N}$:

Concept: rotational analogue of Newton's second law for a system of particles.

For a system, the internal torques cancel in pairs (Newton's third law, forces along the line joining the particles), so only the EXTERNAL torque changes the total angular momentum:

dL/dt = tau_ext.

(dp/dt = F_ext is the translational law, not the rotational one.)

ANSWER: C

Angular acceleration:

Moment of inertia of a ring about its axis:

Torque:

The problem gives you a wheel starting from rest, reaching 10 rad/s in 5 seconds under uniform angular acceleration. The key is to recognise that this is the rotational analogue of linear motion with constant acceleration. The same kinematic equations apply, just with angular variables: $\theta$ for displacement, $\omega$ for velocity, $\alpha$ for acceleration, and $t$ for time.

Since the acceleration is uniform, the average angular velocity is simply the arithmetic mean of the initial and final velocities. For motion from rest, that average is half the final velocity. The total angle turned is then average angular velocity multiplied by time — a clean, intuitive shortcut.

Let’s work through it step by step.

1. Identify the known quantities.

   Initial angular velocity: $\omega_0 = 0$ (starts from rest).

   Final angular velocity: $\omega = 10\ \text{rad/s}$.

   Time interval: $t = 5\ \text{s}$.

   Acceleration is uniform (constant $\alpha$).

2. Find the angular acceleration.

   Using the definition $\alpha = \frac{\omega - \omega_0}{t}$:

3. Use the angular displacement equation for constant acceleration. The standard kinematic equation is:

Substitute $\omega_0 = 0$, $\alpha = 2$, $t = 5$:

4. Check the options. The result is 25 rad, which matches option (A). Options (C) and (D) involve $\pi$, which would appear only if the motion were circular with a given radius or if the answer were in revolutions — but here the angular velocity is given directly in rad/s, so the angle is in radians, no $\pi$ factor.

Step 1 — Identify the applicable theorem.

The body is a lamina (a plane body), the two given axes $X$ and $Y$ lie in its plane and are mutually perpendicular, and the required axis is perpendicular to the plane through their point of intersection. These are exactly the conditions of the perpendicular axis theorem:

Step 2 — Why the theorem is true.

For a particle of mass $m_i$ at $(x_i, y_i)$ in the plane, its distance from the $Z$-axis is $r_i$ with $r_i^2 = x_i^2 + y_i^2$. Hence

(The distance of a point from the $X$-axis is $|y_i|$ and from the $Y$-axis is $|x_i|$.) Note it works only because $z_i = 0$ for every particle — i.e. only for a lamina.

Step 3 — Substitute.

Common mistakes: subtracting ($25-20 = 5$, option A) — that is not what the theorem says; averaging ($12.5$, option C); or multiplying ($500$, option D). The theorem is a plain sum, and a sum is also physically sensible since moments of inertia are positive additive quantities.

ANSWER: B