Question

A uniform solid sphere is placed on a smooth horizontal surface. An impulse $I$ is given horizontally to the sphere at a height $h=\frac{4R}{5}$ above the centre line. $m$ and $R$ are mass and radius of sphere respectively.
(a) Find angular velocity of sphere & 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.

Answer 1

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

To find:

(a) The angular velocity of sphere & linear velocity of centre of mass of the sphere after impulse.

(b) The minimum time after which the highest point B will touch the ground.

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

Solution:

As per the given criteria,

Mass = m, Radius: R, Sphere is solid.

$h=\frac{4}{5}R$ above the horizontal center line of the sphere

We know

impulse = $I=F\times \Delta t=\Delta p$ as per the conservation of linear momentum.

Linear Momentum of sphere after impulse $=mv=p=I$

As $h>\frac{2}{5}R$ , The ball probably slips along with rolling. So, perhaps $v\ne R\omega$. If the horizontal impulse is given at a height $\frac{2}{5}R$, for a solid sphere, then it rolls without slipping.

Moment of inertia MI about center of mass, $I_{cm}=\frac{2}{5}\times m{R}^2$

Angular momentum, $L=I_{cm}\omega =\frac{2}{5}m{R}^2\omega$

$L=h\times p$= Angular impulse given, as initial angular momentum = 0

$⟹>h\times p=\frac{4}{5}Rmv=\frac{2}{5}m{R}^2\omega ⟹\omega =\frac{2v}{R}...\left(ii\right)$

So $\omega =\frac{2I}{mR}$

We need to find the Time duration for the ball to rotate by $\pi$ radians, 180 deg.:

$=t=\frac{\pi }{\omega }=\frac{\pi \omega R}{2I}$

So the horizontal (translational, linear) displacement

$=s=vt=\frac{I}{m}\times \frac{\pi mR}{2I}=\frac{\pi }{2}\times R$

Also, the Energy K.E. given by the impulse:

$KE=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2⟹KE=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2\times {\omega }^2⟹KE=\frac{1}{2}m{\left(\frac{\omega R}{2}\right)}^2+\frac{1}{5}m{R}^2{\omega }^2⟹KE=\frac{5+8}{40}m{R}^2{\omega }^2⟹KE=\frac{13}{40}m{R}^2{\left(\frac{2v}{R}\right)}^2⟹KE=\frac{13}{10}m{v}^2$