Q.A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s$^{-1}$. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Concept & Intuition

The radius of gyration $K$ tells us how far from the axis the object’s mass would need to be concentrated to have the same moment of inertia. For a given shape, $I = M K^2$. So if we know $I$ for the sphere, we can find $K$ for the disc by comparing their moments of inertia — but careful: the masses and radii are different, so we must express everything in terms of the given $K$.

Step-by-step solution

1. Moment of inertia of the solid sphere about its diameter For a solid sphere of mass $M$ and radius $R$, the moment of inertia about any diameter is

Its radius of gyration $K$ satisfies $I_{\text{sphere}} = M K^2$, so

Hence $R^2 = \frac{5}{2} K^2$.

2. Moment of inertia of the uniform circular disc about its diameter For a disc of mass $m$ and radius $r$, the moment of inertia about a diameter is

Here $m = 2M$ and $r = \frac{R}{2}$. Substitute:

3. Express $I_{\text{disc}}$ in terms of $K$ From step 1, $R^2 = \frac{5}{2} K^2$. Plug this in: $$$$

The key idea is that when the two particles are attached to the rotating ring, the moment of inertia of the system increases. Since no external torque is applied, angular momentum must stay the same. A larger moment of inertia means a smaller angular speed — the ring slows down.

Let’s work through it.

1. Moment of inertia of the ring alone For a thin circular ring of mass $M$ and radius $R$, rotating about its central axis, the moment of inertia is

Here $M = 0.2\ \mathrm{kg}$, so $I_\text{ring} = 0.2\,R^2$.

2. Moment of inertia of the two particles Each particle has mass $m = 2\ \mathrm{g} = 0.002\ \mathrm{kg}$. They are attached at diametrically opposite points, so each is at a distance $R$ from the axis. The moment of inertia of a point mass is $m R^2$. For two such particles,

3. Total moment of inertia after attachment

4. Apply conservation of angular momentum Initial angular momentum: $L_i = I_\text{ring}\,\omega_i$, with $\omega_i = 51\ \mathrm{rad\,s^{-1}}$. Final angular momentum: $L_f = I_\text{total}\,\omega_f$. Since $L_i = L_f$,

The $R^2$ cancels (the radius doesn’t matter — nice, isn’t it?).

5. Solve for $\omega_f$ …

The key is to remember that moment of inertia for a circular disc about its geometrical axis (the axis through its centre, perpendicular to the plane) is $I = \frac{1}{2} M R^2$. But here the discs have different radii and different thicknesses, so their masses are not the same. We need to express mass in terms of radius and thickness using density.

Since both discs are made of iron, they have the same density $\rho$. The volume of a disc is area × thickness = $\pi R^2 t$, so mass $M = \rho \pi R^2 t$.

Let’s work through the ratio step by step.

1. Write the moment of inertia formula for each disc.

   For disc X: $I_X = \frac{1}{2} M_X R_X^2$

   For disc Y: $I_Y = \frac{1}{2} M_Y R_Y^2$

2. Express each mass in terms of its radius and thickness.

   $M_X = \rho \pi R_X^2 t_X$

   $M_Y = \rho \pi R_Y^2 t_Y$

   Given $R_X = R$, $t_X = t$, $R_Y = 4R$, $t_Y = t/4$.

3. Substitute into the inertia expressions.

   $I_X = \frac{1}{2} (\rho \pi R^2 t) \cdot R^2 = \frac{1}{2} \rho \pi R^4 t$

   $I_Y = \frac{1}{2} (\rho \pi (4R)^2 \cdot \frac{t}{4}) \cdot (4R)^2$

   Simplify step by step:

   $(4R)^2 = 16R^2$, so $M_Y = \rho \pi (16R^2) \cdot \frac{t}{4} = \rho \pi \cdot 4 R^2 t$

   Then $I_Y = \frac{1}{2} (\rho \pi \cdot 4 R^2 t) \cdot (16R^2) = \frac{1}{2} \rho \pi \cdot 64 R^4 t$

4. Take the ratio. …

The key insight lies in understanding the relationship between orbital motion and escape velocity. A satellite in a stable circular orbit has just enough speed to balance gravitational pull with centripetal acceleration. To escape Earth's gravity entirely, it needs a higher speed—the escape velocity.

For any circular orbit, there's a beautiful connection between the kinetic energy, potential energy, and the energy needed to escape. Let me show you why the answer emerges naturally from energy considerations.

The energy structure of circular orbits

When a satellite orbits at radius $r$ with speed $v_{\text{orb}}$, two conditions hold:

1. Orbital condition: The gravitational force provides exactly the centripetal force needed for circular motion:

This gives $v_{\text{orb}}^2 = \frac{GM}{r}$, so the kinetic energy is:

2. Potential energy: The gravitational potential energy at radius $r$ is:

Notice that $U = -2E$. The total mechanical energy of the orbit is:

The orbit is bound (negative total energy), which makes physical sense—the satellite is trapped in Earth's gravitational well.

What escape requires

To escape means reaching infinite distance with zero (or positive) final energy. The minimum escape condition is:

At the surface (radius $R$), if we give the satellite speed $v_{\text{esc}}$, its total energy must be zero: …

The key concept is the moment of inertia of a thin circular ring about an axis through its center and perpendicular to its plane. For any thin ring, all the mass lies at the same distance from the axis — the radius of the ring. That means the moment of inertia is simply $I = M_{\text{total}} R_{\text{ring}}^2$. No integration is needed; it’s a direct formula.

But here the problem gives the diameter as $R$, not the radius. That’s the classic trap: students often plug in $R$ as if it were the radius. Always check whether a length is a radius or a diameter.

Let’s work through it step by step.

1. Identify the total mass.

   The ring’s mass is given as $2M$. So $M_{\text{total}} = 2M$.

2. Find the radius from the given diameter.

   The diameter is $R$, so the radius is

3. Apply the moment of inertia formula for a thin ring. For a thin circular ring of mass $m$ and radius $r$, about an axis through its center and perpendicular to its plane:

Substituting $m = 2M$ and $r = R/2$:

$I = (2M) \left( \frac{R}{2} \right)^2 = 2M \cdot \frac{R^2}{4} = \frac{MR^2}{2}.$ …

Kinetic energy vs radius. For a circular orbit gravity supplies the centripetal force:

Period vs radius (Kepler's third law). …

When a circular ring rolls up an inclined plane, its initial kinetic energy is transformed into gravitational potential energy. Since the ring is rolling without slipping, the static friction force at the point of contact does no work, meaning there is no energy dissipation. Therefore, the total mechanical energy of the ring is conserved.

The ring will continue to move up the incline until all its initial kinetic energy is converted into potential energy, at which point its speed momentarily becomes zero, and it reaches its maximum height.

Here's how we can determine the maximum distance:

1. Identify Initial and Final Energy States:

   • Initial State (bottom of the incline): The ring has a linear speed $v$ and is rolling. We set the initial gravitational potential energy $PE_i = 0$. Its total initial energy $E_i$ is purely kinetic.
   • Final State (maximum distance up the incline): The ring momentarily stops, so its final linear speed $v_f = 0$ and its final angular speed $\omega_f = 0$. This means its final kinetic energy $KE_f = 0$. Its total final energy $E_f$ is purely gravitational potential energy.

2. Apply Conservation of Mechanical Energy:

   Since mechanical energy is conserved, the total energy at the bottom must equal the total energy at the maximum height:

   $E_i = E_f$

   $KE_i + PE_i = KE_f + PE_f$

3. Calculate Initial Kinetic Energy ($KE_i$):

   The initial kinetic energy of a rolling object is the sum of its translational kinetic energy and its rotational kinetic energy.

   $KE_i = KE_{translational} + KE_{rotational}$

   • Translational kinetic energy: $KE_{translational} = \frac{1}{2}mv^2$, where $m$ is the mass of the ring and $v$ is the speed of its center of mass.
   • Rotational kinetic energy: $KE_{rotational} = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the ring about its center of mass and $\omega$ is its angular speed.

   For a circular ring, the moment of inertia is $I = mr^2$, where $r$ is its radius.

   For pure rolling (without slipping), the linear speed $v$ and angular speed $\omega$ are related by $v = r\omega$, which implies $\omega = \frac{v}{r}$.

   Substitute these into the rotational kinetic energy expression:

   $KE_{rotational} = \frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mr^2\frac{v^2}{r^2} = \frac{1}{2}mv^2$.

   Now, sum the translational and rotational kinetic energies to get the total initial kinetic energy:

   $KE_i = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.

   Watch out

   A common mistake is to forget the rotational kinetic energy component, or to use the incorrect moment of inertia for a ring. For a ring, $I=mr^2$, leading to $KE_{rotational} = \frac{1}{2}mv^2$. If you only considered translational kinetic energy, your answer would be half of the correct value.

4. Calculate Final Potential Energy ($PE_f$): …

The key relationship here is between momentum $p$ and kinetic energy $K$. Momentum is $p = mv$, while kinetic energy is $K = \frac{1}{2}mv^2$. If you double the momentum, you’re doubling $mv$. But kinetic energy depends on $v^2$, not just $v$. So the effect isn’t linear — it’s quadratic. The cleanest way to see this is to eliminate velocity and express $K$ directly in terms of $p$.

1. Express kinetic energy in terms of momentum. Start with $p = mv$, so $v = \frac{p}{m}$. Substitute into $K = \frac{1}{2}mv^2$:

This shows that for a fixed mass $m$, kinetic energy is proportional to the square of momentum: $K \propto p^2$.

2. Apply the change. If momentum is doubled, the new momentum is $p' = 2p$. The new kinetic energy is:

So the kinetic energy becomes four times its original value.

3. Check the intuition. …

When a force is applied to stop a rotating object, it creates a torque that opposes the rotation, causing angular deceleration. This problem involves relating the initial rotational motion and the time taken to stop, to the frictional force and the coefficient of kinetic friction.

The key idea is to connect the kinematics of rotational motion (angular speed, time, angular deceleration) with the dynamics of rotational motion (torque, moment of inertia) and then link this torque to the friction force.

Here's how we approach the problem:

1. Convert initial angular speed to standard units: The initial angular speed is given in revolutions per minute ($\text{rev/min}$). For calculations in physics, we typically use radians per second ($\text{rad/s}$). We know that $1\,\text{revolution} = 2\pi\,\text{radians}$ and $1\,\text{minute} = 60\,\text{seconds}$.

Numerically, $\omega_0 \approx 7.33\,\text{rad/s}$.

2. Calculate the angular deceleration:

The wheel stops in $5.0\,\text{s}$, meaning its final angular speed $\omega_f$ is $0\,\text{rad/s}$. We can use the first equation of rotational kinematics to find the angular acceleration (deceleration, in this case).

> [!IMPORTANT]

> The kinematic equation relating initial angular speed ($\omega_0$), final angular speed ($\omega_f$), angular acceleration ($\alpha$), and time ($t$) is:

> $\omega_f = \omega_0 + \alpha t$

Substituting the known values:

The negative sign indicates that this is an angular deceleration, opposing the initial rotation. For calculating the magnitude of torque, we will use the magnitude of $\alpha$, which is $\frac{7\pi}{15}\,\text{rad/s}^2$.

3. Calculate the torque required to stop the wheel:

The torque ($\tau$) required to produce an angular acceleration ($\alpha$) in an object with moment of inertia ($I$) is given by Newton's second law for rotation.

> [!FORMULA]

> $\tau = I\alpha$

Given $I = 10\,\text{kg}.\text{m}^2$ and $|\alpha| = \frac{7\pi}{15}\,\text{rad/s}^2$:

4. Relate the torque to the friction force: …

The rod is pivoted at one end and falls from vertical to horizontal. As it falls, gravity does work on its centre of mass, converting potential energy into rotational kinetic energy about the hinge. The key is to treat the rod as a rigid body rotating about a fixed axis — not as a point mass falling freely.

Why energy conservation works here: The hinge is fixed and frictionless (assumed), so no external torque does work except gravity, which is conservative. Therefore, mechanical energy is conserved. We don’t need to integrate torque or solve an equation of motion — just compare the initial and final energies.

1. Identify the initial and final configurations. Initially, the rod stands vertical, hinged at the bottom. Its centre of mass is at height $L/2$ from the ground (since the rod is uniform). Finally, the rod lies flat on the ground, so the centre of mass is at height $0$. The loss in gravitational potential energy is:

2. Write the rotational kinetic energy. The rod rotates about the hinge with angular speed $\omega$. Its moment of inertia about one end is:

The rotational kinetic energy is:

The problem asks for a constant tangential force that stops a rotating wheel in a given number of revolutions. The direct route is to think about energy: the force does work over the distance it acts, and that work must exactly remove the wheel’s rotational kinetic energy. No need to involve torque and angular acceleration separately — the work–energy theorem ties it all together in one clean equation.

1. Convert the given data to consistent SI units.

Mass $m = 20\ \mathrm{kg}$, radius $R = 30\ \mathrm{cm} = 0.30\ \mathrm{m}$.

Initial angular speed $\omega_0 = 80\ \mathrm{rev/min}$. Convert to rad/s:

The wheel stops after $5$ revolutions, so the angular displacement $\theta = 5 \times 2\pi = 10\pi\ \mathrm{rad}$.

2. Find the moment of inertia.

The wheel is a solid disc (or a ring? The problem says “wheel” — for a typical solid disc, $I = \frac12 mR^2$).

3. Compute the initial rotational kinetic energy.

First square the angular speed:

Then

Simplify: $0.45 = \frac{9}{20}$, so

4. Relate work done by the tangential force to the change in kinetic energy. …

The key idea here is that moment of inertia depends on the square of a linear dimension. For a uniform rod rotating about its perpendicular bisector, $I = \frac{1}{12} M L^2$. When temperature rises, the rod expands uniformly, so its length changes — but mass stays the same. The fractional change in $I$ then follows directly from the fractional change in $L^2$.

Let’s walk through it.

1. Write the expression for moment of inertia. For a uniform rod of mass $M$ and length $L$, about an axis through its centre and perpendicular to its length:

2. Understand the effect of heating. The coefficient of linear expansion $\alpha$ tells us how length changes with temperature:

Since $\Delta T$ is small, we can treat this as a differential change.

3. Find the change in $I$. Differentiate $I$ with respect to $L$:

But $dL = \alpha L \Delta T$, so:

4. Express the fractional change. Divide $dI$ by $I$: $\frac{dI}{I} = \frac{\frac{1}{6} M L^2 \alpha \Delta T}{\frac{1}{12} M L^2} = \frac{1/6}{1/12} \alpha \Delta T = 2 \alpha \Delta T$ …