Name: Angular Impulse
Uploaded: 2019-07-19
Duration: 4 min 54 s
Description: What is angular impulse? How is it different from angular momentum? Let's watch this video for a better understanding.

Question

A uniform solid sphere of radius $R$ is rolling without sliding on a horizontal surface with a velocity $v_{0}$ . It collides with an obstacle of height $h (< R)$ inelastically. Find the angular speed of the sphere just after the collision.

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Accepted Answer

At moment of collision two impulsive force will act on the sphere. In FBD as shown, friction $f$ and normal contact force $v_{0}$ acting at the point of impact $P$ . The gravitational force is non-impulsive.

For reference point $P$ , the forces $f$ and $N$ are passing through this point, thus, torques produced about $P$ by them is zero. Then, the angular impulses of both forces $f$ and $N$ are zero about $P$ .

Hence, we can apply conserve the angular momentum of the sphere about $P$ just before and after the impact.

Let anticlockwise is positive direction and clockwise and negative direction.

Then, initial angular momentum is,

$L_{i} = - m v_{0} (R - h) - I ω_{0} = - m v_{0} (R - h) - \frac{2}{5} m R^{2} ω_{0} . . . (i)$

As the sphere rotates perfectly about $P$ just after the collision and it does not slide and bounce.

Let angular velocity is $ω$ about point $P$ , then, final angular momentum with direction about $P$ is,

$L_{f} = - I_{P} ω . . . (i i)$

Using parallel axis theorem, $I_{P} = I_{CM} + m R^{2} = \frac{2}{5} m R^{2} + m R^{2} = \frac{7}{5} m R^{2}$

From equations $(i)$ and $(i i)$ , we get,

$v_{0} (R - h) + \frac{2}{5} R^{2} ω_{0} = \frac{7}{5} R^{2} ω$

The sphere was rolling before the impact, so, $v_{0} = R ω_{0}$ , then, evaluating above expression, we get,

$v_{0} R - v_{0} h + \frac{2}{5} v_{0} R = \frac{7}{5} R^{2} ω$ or $\frac{7}{5} R^{2} ω = (\frac{7}{5} R - h) v_{0}$

$ω = (1 - \frac{5 h}{7 R}) (\frac{v_{0}}{R})$

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