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 of the edge without slipping. The cylinder rotates through angle before it leaves contact with the edge. Find the value of cos of angle

Answer 1

$\displaystyle \frac{4}{7}$

Answer 2

Step 1: Energy conservation
The cylinder pivots about the edge without slipping. Using the pivot theorem, its moment of inertia about the edge is

$$I_{\rm pivot}=I_{\rm cm}+mR^2=\tfrac12mR^2+mR^2=\tfrac32mR^2.$$

A drop in height of the CM after rotating by angle $\theta$ is $\Delta h=R(1-\cos\theta)$.
Thus,

$$mgR(1-\cos\theta)=\tfrac12I_{\rm pivot}\,\omega^2 =\tfrac12\bigl(\tfrac32mR^2\bigr)\omega^2 =\tfrac34\,mR^2\omega^2,$$

giving

$$\omega^2=\frac{4g(1-\cos\theta)}{3R}.$$

Step 2: Condition for loss of contact
The normal reaction $N$ goes to zero when the component of weight toward the pivot provides exactly the required centripetal force:

$$mg\cos\theta = mR\,\omega^2 = mR\cdot\frac{4g(1-\cos\theta)}{3R} =\frac{4}{3}m g\,(1-\cos\theta).$$

Rearranging,

$$\cos\theta = \frac{4}{3}\,(1-\cos\theta) \quad\Longrightarrow\quad \cos\theta + \frac{4}{3}\cos\theta = \frac{4}{3} \quad\Longrightarrow\quad \frac{7}{3}\cos\theta = \frac{4}{3} \quad\Longrightarrow\quad \cos\theta = \frac{4}{7}.$$