Question

Ques no. 3 &4 on
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Answer 1

Answer on Question #56596-Physics-Mechanics-Relativity

5-3 A body of radius $r$ and mass $m$ is rolling horizontally without slipping with speed $v$. It then rolls up a hill to a maximum height $h$. If $h = 3v^2 / 4g$, what might the body be? What is the body's moment of inertia?

Ans. $[R / \sqrt{2}]$

5-4 A conical pendulum is formed by a thin rod of length $l$ and mass $m$, hinged at the upper end, rotates uniformly about a vertical axis passing through its upper end, with angular velocity $\omega$. Find the angle $\theta$ between the rod and the vertical.

Ans. $[\theta = \cos^{-1}\left(\frac{3g}{2\omega^2l}\right)]$

3.

Solution

The conservation of energy law:

$$\frac{l \omega^2}{2} + \frac{m v^2}{2} = m g h.$$

No slipping:

$$\omega = \frac{v}{r}.$$

The moment of inertia is

$$I = \frac{2 m g h - m v^2}{\frac{v^2}{r^2}} = \frac{2 m g h r^2 - m v^2 r^2}{v^2} = \frac{2 m g \left(\frac{3 v^2}{4 g}\right) r^2 - m v^2 r^2}{v^2} = \frac{1}{2} m r^2.$$

It is the ring of radius $r$.

4.

Solution

Looking at the rotating reference frame of the rod, there is a centrifugal force and gravity. Their torque must sum to zero because the rod is in equilibrium in that reference frame. For each bit of mass $dm$, we have

$$d \tau = (x \cos \theta) (\omega^2 x \sin \theta) d m.$$

plugging in $dm = \left(\frac{m}{l}\right) dx$ and simplify, I get

$$d \tau = \frac{m \omega^2 x^2}{l} \sin \theta \cos \theta d x.$$ $$\tau = \int_0^l \frac{m \omega^2 x^2}{l} \sin \theta \cos \theta d x = \frac{1}{3} m \omega^2 l^2 \sin \theta \cos \theta.$$

The torque provided by gravity is simply $mg\left(\frac{l}{2}\right)\sin \theta$. Equating with the centrifugal torque, I get

$$\frac {1}{3} m \omega^ {2} l ^ {2} \sin \theta \cos \theta = m g \left(\frac {l}{2}\right) \sin \theta$$ $$\cos \theta = \frac {3 g}{2 \omega^ {2} l}$$ $$\theta = \cos^ {- 1} \left[ \frac {3 g}{2 \omega^ {2} l} \right]$$

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