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 uniform solid sphere is placed on a smooth horizontal surface. An impulse $I$ is given horizontally to the sphere at a height $h = 4 R / 5$ above the centre line. $m$ and $R$ are mass and radius of the sphere, respectively.

$(a)$ Find the angular velocity of the sphere and linear velocity of centre of mass of the sphere after impulse.

$(b)$ Find the minimum time after which the highest point $B$ will touch the ground,

$(c)$ Find the displacement of the centre of mass during this interval.

Accepted Answer

$(a)$ Angular impulse about a point is equal to the product of impulse and the perpendicular distance of application of impulse from the point, mathematically

$J = I r_{⊥} = I h \dots (i)$

Now, angular impulse $=$ change in angular momentum

$B$ Moment of inertia $\times$ change in angular velocity

$\Rightarrow J = (\frac{2}{5} m R^{2}) (∆ ω) \Rightarrow I h = \frac{2 m R^{2} ∆ ω}{5} \Rightarrow I \frac{4 R}{5} = \frac{2 m R^{2} ∆ ω}{5}$

$\Rightarrow ∆ ω = \frac{2 I}{m R}$ .

As initial angular velocity is zero $\Rightarrow ω = \frac{2 I}{m R}$ .

Again impulse $=$ change in linear momentum $=$ Mass $\times$ change in velocity

$\Rightarrow I = m (∆ v) \Rightarrow ∆ v = \frac{I}{m}$ .

As initial velocity is zero $\Rightarrow v = \frac{I}{m}$ .

$(b)$ The highest point $(b)$ will touch the ground when the sphere rotates by an angle $π$

$\Rightarrow ω t = π \Rightarrow t = \frac{π}{ω} = \frac{π}{\frac{2 I}{m R}} = \frac{m R π}{2 I}$ .

$(c)$ As COM is moving horizontally with constant velocity, displacement at the same time $t$ will be

$S = v t = (\frac{I}{m}) (\frac{m R π}{2 I}) = \frac{π R}{2}$ .

Important Questions on Rotational Mechanics

Learn and Prepare for any exam you want !