Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

A uniform solid cylinder of radius $R = 15 cm$ , rolls over a horizontal plane, passing into an inclined plane, forming an angle $α = 30 °$ , with the horizontal. Find the maximum value of the velocity $v_{0}$ , which still permits the cylinder to roll onto the inclined plane section, without a jump. The sliding is assumed to be absent.

$(Gravitational acceleration, g = 9.8 m s^{- 2})$

Accepted Answer

Let the centre of the cylinder be $C$ . Since the cylinder moves in pure rolling motion, the centre of the cylinder rotates about the point $O$ . We can say that while the cylinder is moving from horizontal to inclined plane, axis through $O$ will act as an instantaneous axis of rotation. Let $v_{0}$ be the velocity of centre at point $C$ .
Let the centre move from point $C$ to point $C_{1}$ , as shown in the figure, after covering an angle $β$ , and at any instant, during this motion, the velocity of the centre is $v_{1}$ . We have, the centripetal force,

The centripetal force acting on the sphere is
$\frac{m v_{1}^{2}}{R} = m g \cos β - N$ ,

where, $N$ is the normal contact force by the edge,

$v_{1}^{2} = g R \cos β - \frac{N R}{m} . . . (1)$

Now, conserving energy between positions $C and C_{1}$ ,
$∆ K_{r o t a t i o n} = - ∆ U$

$\frac{1}{2} I_{0} \frac{v_{1}^{2}}{R^{2}} - \frac{1}{2} I_{0} \frac{v_{0}^{2}}{R^{2}} = m g R (1 - \cos β)$

Where, moment of inertia of the cylinder, about the instantaneous axis, is $I_{0} = \frac{m R^{2}}{2} + m R^{2} = \frac{3}{2} m R^{2}$ (Using the parallel axes theorem)

So, we have, $v_{1}^{2} = v_{0}^{2} + \frac{4}{3} g R (1 - \cos β) . . . (2)$

From equations $(1)$ and $(2)$ , weget

$v_{0}^{2} = \frac{g R}{3} (7 \cos β - 4) - \frac{N R}{m}$

From the figure, it is clear that, $β \leq α$ and that normal reaction $N$ , is decreasing as $β \to α$ . Putting $β = α and N = N_{0}$ in this equation, we get,

$v_{0}^{2} = \frac{g R}{3} (7 \cos α - 4) - \frac{N_{0} R}{M}$

Since for $N \geq N_{0} \geq 0$ there is no jumping, we come to the conclusion that, maximum velocity $v_{\max}$ of the centre can be,

$v_{\max} = \sqrt{\frac{g R}{3} (7 \cos α - 4)}$

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