Name: Collision Between a Point Mass and a Rigid Body
Uploaded: 2019-07-19
Duration: 7 min 57 s
Description: Can we treat the collision of a point mass with a rigid body in the same way we treated the collision of point masses? Watch this video to know about it.

Question

A thin rod is passing through the centre of a sphere. The rod is fixed to a vertical axis and the sphere is made to roll on a surface with friction. The radius of the sphere is $r,$ the mass is $m$ and the length of the rod is $l .$ The rod is rotating with an angular velocity $ω_{0} .$ Find the energy of the sphere in terms of $ω_{0}, m, l$ and $r .$ Assume the rod to be of negligible mass.

Accepted Answer

Given,
Radius of sphere is, $r$
Mass of sphere is, $m$
Length of rod is, $l$
Mass of rod is negligible.

Energy possessed by the sphere is due to rotational kinetic energy of sphere about vertical axis and about its own axis passing through its centre.

By using parallel axis theorem, the MI of the sphere about the rod (vertical axis) is
$I_{1} = \frac{2}{5} m r^{2} + m l^{2}$
Also moment of inertia about its own central axis is given by,
$I_{2} = \frac{2}{5} m r^{2}$

Total energy possessed by the sphere is given by,
$E = E_{1} + E_{2}$
$\Rightarrow E = \frac{1}{2} I_{1} {ω_{0}}^{2} + \frac{1}{2} I_{2} {(\frac{v_{s}}{r})}^{2}$ , here $v_{s}$ is the velocity of centre of sphere.
$\Rightarrow E = \frac{1}{2} [\frac{2}{5} m r^{2} + m l^{2}] ω_{0}^{2} + \frac{1}{2} (\frac{2}{5} m r^{2}) {(\frac{ω_{0} l}{r})}^{2}, (∵ v_{s} = ω_{0} l)$
$\Rightarrow E = \frac{1}{5} m r^{2} ω_{0}^{2} + \frac{1}{2} m l^{2} ω_{0}^{2} + \frac{1}{5} m {ω_{0}}^{2} l^{2}$
$\Rightarrow E = \frac{1}{5} m r^{2} ω_{0}^{2} + \frac{7}{10} m l^{2} ω_{0}^{2}$
$\Rightarrow E = \frac{1}{5} m ω_{0}^{2} (r^{2} + \frac{7}{2} l^{2})$

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