A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius R

Q: Why does the cylinder rotate about the edge instead of just falling straight down?

Because there is sufficient friction at the edge to prevent slipping. This friction acts as a pivot, forcing the cylinder's center of mass to move in a circular arc of radius $R$ around the edge until the normal force drops to zero.

Q: Why is the work done by friction zero if there is friction present?

In pure rolling, the instantaneous point of contact between the cylinder and the edge is momentarily at rest. Since the displacement of the point where the friction force is applied is zero, the work done by friction is exactly zero.

Q: Why does the cylinder eventually leave the edge?

As the cylinder rolls down, its speed increases, which means it needs a larger centripetal force to stay in its circular path. This force is provided by $mg \cos\theta - N$. As the required centripetal force grows and the inward component of gravity $mg \cos\theta$ shrinks, $N$ must decrease. When $N$ reaches zero, gravity alone isn't enough to keep it in the circle, and it flies off.

Q: What happens to the angular velocity after it leaves the edge?

Once it leaves the edge, both the normal reaction and friction become zero. The only force acting on the cylinder is gravity, which acts directly through the center of mass. Since gravity produces no torque about the center of mass, the angular velocity remains perfectly constant.

Q: Why doesn't the rotational kinetic energy increase after separation?

Rotational kinetic energy is given by $\frac{1}{2}I\omega^2$. Since the angular velocity $\omega$ becomes constant due to zero net torque, the rotational kinetic energy also becomes locked and remains constant, even as the cylinder continues to fall and gain translational kinetic energy.

Question

A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius $R$ is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane as shown in figure. There is sufficient friction present at the edge, so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine
(a) the angle $	heta_c$ through which the cylinder rotates before it leaves contact with the edge,
(b) the speed of the centre of mass of the cylinder before leaving contact with the edge and
(c) the ratio of the translational to rotational kinetic energies of the cylinder when its centre of mass is in horizontal line with the edge.

Sigma Percentile · Accepted Answer

Imagine you are watching a heavy solid cylinder perfectly balanced on the edge of a rigid block. A tiny nudge, and it starts rolling. This is a classic physics scenario that beautifully combines rotational dynamics, energy conservation, and circular motion constraints. Let's break down the journey of this cylinder from its initial precarious balance to its eventual free fall.

### Analyzing the Setup

As long as there is no slipping, the edge of the block acts as a fixed pivot point. The center of mass (CM) of the cylinder is forced to move in a circular arc of radius $R$ around this edge.

Because the cylinder rolls without slipping, the instantaneous point of contact is momentarily at rest. This is a crucial detail: it means the work done by friction is exactly zero! With no non-conservative forces doing work, we can safely apply the principle of **conservation of mechanical energy**.

### The Master Equation: Conservation of Energy

Let's set up our energy equation. Initially, the center of mass is at a height $R$ above the edge, so the energy is purely potential, $mgR$. As it rolls down by an angle $	heta$, the height becomes $R \cos	heta$. The lost potential energy converts into both translational and rotational kinetic energy.

$$mgR = mgR \cos	heta + \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$$

To solve this, we need to link the rotational and translational variables. For pure rolling, the angular velocity $\omega$ equals $v/R$. We also know that the moment of inertia of a solid cylinder about its central axis is $I = \frac{1}{2}mR^2$. Substituting these raw values into our equation gives:

$$mgR(1 - \cos	heta) = \frac{1}{2}\left(\frac{1}{2}mR^2ight)\left(\frac{v^2}{R^2}ight) + \frac{1}{2}mv^2$$

The kinetic energy terms simplify beautifully. One-fourth plus one-half gives us three-fourths $mv^2$. Rearranging this, we get an expression for the centripetal acceleration, $\frac{v^2}{R}$:

$$\frac{v^2}{R} = \frac{4}{3}g(1 - \cos	heta)$$

### The Separation Condition: When Does It Fly?

There is a catch here. The cylinder will not stay on the edge forever. It leaves contact exactly when the normal reaction $N$ drops to zero. To find when this happens, we must look at the forces in the radial direction (towards the edge $O$).

Gravity pulls the cylinder inward with a component $mg \cos	heta$, and the normal force $N$ pushes outward. The net force provides the required centripetal force to keep the CM in its circular path:

$$mg \cos	heta - N = \frac{mv^2}{R}$$

By setting $N = 0$ at the critical angle $	heta_c$, we find that the entire radial component of gravity is used just to keep it in the circular path. This gives us our second equation for $\frac{v^2}{R}$:

$$\frac{v^2}{R} = g \cos	heta_c$$

### Solving for the Critical Angle and Velocity

We now have two different expressions for the same physical quantity. Let's equate them. The $g$ cancels out, and a simple linear equation in $\cos	heta$ remains:

$$\frac{4}{3}g(1 - \cos	heta_c) = g \cos	heta_c$$

$$4 - 4\cos	heta_c = 3\cos	heta_c$$

$$7\cos	heta_c = 4 \implies \cos	heta_c = \frac{4}{7}$$

We have our first answer! The cylinder leaves the edge at an angle of $	heta_c = \cos^{-1}\left(\frac{4}{7}ight)$. Substituting this back into our velocity equation, we get the speed at the exact moment of separation:

$$v = \sqrt{gR \left(\frac{4}{7}ight)} = \sqrt{\frac{4gR}{7}}$$

### The Aftermath: Free Fall and Locked Energy

Once the cylinder loses contact, the normal force and friction both vanish. With no friction acting at the edge, there is absolutely no torque about the center of mass.

This leads to a fascinating consequence: the angular velocity $\omega$ becomes constant. This means the rotational kinetic energy gets **locked in** and remains perfectly constant for the rest of the fall. Let's calculate this locked-in rotational kinetic energy, $K_R$:

$$K_R = \frac{1}{2}I\omega^2 = \frac{1}{4}mv^2$$

Substituting our separation velocity squared, the fours cancel out beautifully:

$$K_R = \frac{1}{4}m\left(\frac{4}{7}gRight) = \frac{mgR}{7}$$

### The Final Ratio

Finally, the problem asks for the energies when the center of mass is exactly horizontal with the edge. At this point, it has fallen a total vertical distance of $R$. The total loss in potential energy from the very beginning is $mgR$, which must equal the sum of its final translational ($K_T$) and rotational ($K_R$) kinetic energies.

$$mgR = K_T + K_R$$

The translational kinetic energy is simply the total energy minus the constant rotational part:

$$K_T = mgR - \frac{mgR}{7} = \frac{6}{7}mgR$$

Taking the ratio of translational to rotational kinetic energy, the $\frac{mgR}{7}$ term cancels out completely, leaving us with a clean, elegant integer:

$$\frac{K_T}{K_R} = \frac{\frac{6}{7}mgR}{\frac{1}{7}mgR} = 6$$

What a beautiful problem that perfectly ties together energy, forces, and the subtle nuances of rigid body dynamics!

Visualized Solution

$Initial Setup$

$Δ E = 0$

$E_{i} = E_{f}$

$ω = \frac{v}{R}, I = \frac{1}{2} m R^{2}$

$\frac{v ^{2}}{R} = \frac{4}{3} g (1 - cos θ)$

$N = 0$

$\frac{v ^{2}}{R} = g cos θ_{c}$

$cos θ_{c} = \frac{4}{7}$

$v = \frac{4 g R}{7}$

$τ_{n e t} = 0 ⟹ ω = constant$

$K_{R} = \frac{m g R}{7}$

$m g R = K_{T} + K_{R}$

$\frac{K _{T}}{K _{R}} = 6$

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Master Equation

The Separation Condition

Solving for the Critical Angle and Velocity

The Aftermath

The Final Ratio

Similar Questions

A block of base $10 cm \times 10 cm$ and height $15 cm$ is kept on an inclined plane. The coefficient of friction between them is $3$ . The inclination $θ$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$ . Then,

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ei​=Ef​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=Rv​,I=21​mR2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rv2​=34​g(1−cosθ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rv2​=gcosθc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθc​=74​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=74gR​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τnet​=0⟹ω=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } KR​=7mgR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mgR=KT​+KR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } KR​KT​​=6