Question

A hemisphere of radius 2 m and mass 4 kg is kept on smooth floor. A small ball of mass 1 kg is placed on top of hemisphere. Hemisphere is attached with a light pipe along its periphery inside which the ball can move freely and it does not effectively lose contact with hemisphere. Find angular velocity of the ball w. sphere when velocity of sphere is 0.6 m/s and line joining ball and centre of sphere make angle 60° with vertical.

Answer 1

ω = 3 rad/s

Answer 2

We use conservation of horizontal momentum. There being no external horizontal force, the net horizontal momentum remains zero. At the moment of interest:

• Let the hemisphere (mass 4 kg) have a horizontal velocity $u=0.6\,\rm m/s$ to the right.
• Let the ball (mass 1 kg) move relative to the hemisphere along the inner circular pipe (radius $R=2\,\rm m$); denote its angular speed (relative to the hemisphere) as $\omega$.
• The ball’s position is at an angle $\phi=60^\circ$ from the vertical.

A point on the circle (in the hemisphere frame) can be parameterized as:

$$(x,y) = \bigl(R\sin\phi,\; R\cos\phi\bigr).$$

Its unit tangent is

$$(\cos\phi,\;- \sin\phi).$$

Since we require the ball’s horizontal velocity (relative to the ground) to oppose that of the hemisphere (to keep net momentum zero), we choose the tangent direction so that its horizontal component is to the left. Thus, the ball’s velocity relative to the hemisphere is taken as:

$$\mathbf{v}_{\text{rel}} = 2\omega\,(-\cos\phi,\;\sin\phi).$$

At $\phi=60^\circ$, $\cos 60^\circ=\tfrac{1}{2}$ and $\sin 60^\circ=\tfrac{\sqrt3}{2}$. Therefore,

$$\mathbf{v}_{\text{rel}} = 2\omega\,\Bigl(-\frac{1}{2},\;\frac{\sqrt3}{2}\Bigr)= (-\omega,\;\omega\sqrt3).$$

Hence, the absolute velocity (in the ground frame) of the ball is the sum of the hemisphere’s velocity (assumed purely horizontal) and its relative velocity:

$$v_{x,\text{ball}} = u + (-\omega) = 0.6 - \omega.$$

Now, applying conservation of horizontal momentum (initially, taken as zero):

$$\text{Total }p_x = m_{\text{hemisphere}}\,u + m_{\text{ball}}\, (0.6-\omega) = 0.$$

Substitute the masses:

$$4(0.6) + 1(0.6-\omega)= 0 \quad \Longrightarrow \quad 2.4+0.6-\omega=0.$$

Solve for $\omega$:

$$3.0 - \omega = 0\quad \Longrightarrow \quad \omega = 3\,\rm rad/s.$$

Summary of the solution:

1. Write the ball’s relative velocity on the circular path (with radius 2 m) as $\mathbf{v}_{rel} = (-\omega,\;\omega\sqrt3)$ at $60^\circ$.
2. The ball’s absolute horizontal speed is $0.6 - \omega$.
3. Conservation of horizontal momentum gives: $$4(0.6)+ (0.6 - \omega)=0 \quad\Longrightarrow\quad \omega = 3\,\rm rad/s.$$