27. A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radi

Q: Why does the cylinder rotate about the edge instead of its own center?

Because of the sufficient friction at the edge, the point of contact is momentarily at rest. This makes the edge the instantaneous axis of rotation, causing the center of mass to move in a circular arc around it.

Q: Why do we set the normal reaction $N$ to zero to find the point of losing contact?

The normal reaction is a contact force that pushes the cylinder away from the edge. When the cylinder is just about to fly off, it is no longer pressing against the edge, so $N$ drops to exactly zero.

Q: Why does the rotational kinetic energy become constant after losing contact?

Once the cylinder is airborne, the only force acting on it is gravity, which acts directly through the center of mass. Since there is no friction to provide a torque, the angular velocity $\omega$ cannot change.

Q: Can we use kinematics equations like $v^2 = u^2 + 2as$ here?

No, because the acceleration is not constant. The component of gravity driving the motion ($mg \sin\theta$) changes continuously as the angle $\theta$ increases. Energy conservation is the best tool here.

Q: Why is the total decrease in potential energy $mgR$ in the final part?

Initially, the center of mass is at a height $R$ above the edge. When the center of mass is in a horizontal line with the edge, its height is $0$. Thus, the vertical drop is exactly $R$, leading to a potential energy decrease of $mgR$.

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

### Analyzing the Setup

Imagine a solid cylinder resting peacefully on the edge of a rectangular block. The moment we give it a microscopic nudge, it begins to roll. But here is the fascinating part: because there is sufficient friction, the cylinder doesn't just slide off. Instead, it pivots exactly around the edge of the block.

This means the center of mass (CM) of the cylinder is forced to travel in a perfect circular arc of radius $R$ around the edge. Let's define the angle $	heta$ as the angle the CM makes with the vertical as it falls.

### The Master Equation: Energy Conservation

As the cylinder rolls without slipping, the work done by friction is exactly zero. This is a green light to use the principle of conservation of mechanical energy.

Initially, the CM is at a height $R$ above the edge, giving it a potential energy of $mgR$. As it rotates by an angle $	heta$, its height drops to $R \cos	heta$. The lost potential energy is converted into kinetic energy—both translational ($\frac{1}{2} mv^2$) and rotational ($\frac{1}{2} I \omega^2$).

Equating the initial and final energies:
$mgR = mgR \cos	heta + \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$

For pure rolling, the velocity $v$ and angular velocity $\omega$ are locked in a beautiful dance: $v = \omega R$. Substituting this and the moment of inertia of a solid cylinder ($I = \frac{1}{2} mR^2$), we can simplify the energy equation:
$mgR (1 - \cos	heta) = \frac{1}{4} mv^2 + \frac{1}{2} mv^2 = \frac{3}{4} mv^2$

Rearranging for $v^2$, we get our master velocity equation:
$v^2 = \frac{4}{3} gR (1 - \cos	heta)$

### The Critical Moment: Losing Contact

The cylinder won't hug the edge forever. As it speeds up, it requires more and more centripetal force to stay on its circular path. This force is provided by the radial component of gravity ($mg \cos	heta$) minus the outward normal reaction ($N$).

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

The exact moment the cylinder loses contact and becomes airborne is when the normal reaction $N$ drops to zero. Setting $N = 0$, we find the condition for the critical angle $	heta_c$:
$mg \cos	heta_c = \frac{mv^2}{R}$

Substituting our master velocity equation into this force condition:
$g \cos	heta_c = \frac{4}{3} g (1 - \cos	heta_c)$

Solving this elegant linear equation yields:
$3 \cos	heta_c = 4 - 4 \cos	heta_c \implies 7 \cos	heta_c = 4 \implies \cos	heta_c = \frac{4}{7}$

Thus, the cylinder leaves the edge at $	heta_c = \cos^{-1}\left(\frac{4}{7}ight)$.

To find the speed at this exact moment, we plug $\cos	heta_c$ back into the velocity equation:
$v^2 = \frac{4}{3} gR \left(1 - \frac{4}{7}ight) = \frac{4}{7} gR \implies v = \sqrt{\frac{4gR}{7}}$

### The Aftermath: Airborne Dynamics

What happens the millisecond after contact is lost? The normal reaction is gone, which means the frictional force at the edge also vanishes. Without friction, there is absolutely no force to provide a torque about the center of mass.

By Newton's laws of rotational motion, zero torque means the angular velocity $\omega$ becomes a constant. Consequently, the rotational kinetic energy ($K_R$) is locked in at the value it had at the moment of separation:
$K_R = \frac{1}{2} I \omega^2 = \frac{1}{4} mv^2 = \frac{1}{4} m \left(\frac{4}{7} gRight) = \frac{1}{7} mgR$

### The Final Calculation

The problem asks for the ratio of kinetic energies when the CM is in a horizontal line with the edge. At this point, the CM has dropped a total vertical distance of $R$, meaning the total loss in potential energy is exactly $mgR$.

By conservation of energy, this entire $mgR$ must now be the total kinetic energy of the cylinder. Since the rotational kinetic energy is locked at $\frac{1}{7} mgR$, the remaining energy must be translational ($K_T$):
$K_T = mgR - K_R = mgR - \frac{1}{7} mgR = \frac{6}{7} mgR$

Finally, the ratio of translational to rotational kinetic energy is simply:
$	ext{Ratio} = \frac{K_T}{K_R} = \frac{\frac{6}{7} mgR}{\frac{1}{7} mgR} = 6$

A stunningly clean integer answer to a complex dynamical journey!

Visualized Solution

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$V e l oc i t y a s a F u n c t i o n o f A n g l e$

$C o n d i t i o n f or L os in g C o n t a c t$

$F in d in g t h e C r i t i c a l A n g l e$

$S p ee d a t t h e P o in t o f L os in g C o n t a c t$

$M o t i o n A f t er L os in g C o n t a c t$

$E n er g y W h e n C M i s H or i z o n t a l$

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The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Master Equation

The Critical Moment

The Aftermath

The Final Calculation

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinematics of Pure Rolling

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity as a Function of Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Losing Contact

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Critical Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Speed at the Point of Losing Contact

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motion After Losing Contact

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy When CM is Horizontal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Kinetic Energies