A cubical block of side \(a\) is moving with velocity \(v\) on a horizontal smooth plane as shown in the figure. It hits a ridge at point \(O\). The angular speed of the block after it hits \(O\) is:
Show Hint
In collision problems with a sudden pivot:
Linear momentum is not conserved
Angular momentum about the point of impact is conserved
Always choose the pivot where impulse acts
Step 1: Identify the nature of collision.
When the block hits the ridge at point \(O\):
The point \(O\) becomes an instantaneous pivot.
External impulsive force acts at \(O\), so angular momentum about \(O\) is conserved.
Step 2: Angular momentum before collision about point \(O\).
The centre of mass of the cube is at height \(\dfrac{a}{2}\) above the ground.
Linear momentum of the block:
\[
p = mv
\]
Angular momentum about \(O\):
\[
L_{\text{initial}} = mv \times \frac{a}{2}
\]
Step 3: Angular momentum after collision.
After collision, the block rotates about point \(O\) with angular speed \(\omega\).
Moment of inertia of a cube about an axis through an edge and perpendicular to the face:
\[
I_O = I_{\text{CM}} + m\left(\frac{a}{2}\right)^2
\]
Moment of inertia of cube about centre:
\[
I_{\text{CM}} = \frac{1}{6}ma^2
\]
So,
\[
I_O = \frac{1}{6}ma^2 + m\frac{a^2}{4}
= \left(\frac{2+3}{12}\right)ma^2
= \frac{5}{12}ma^2
\]
Angular momentum after collision:
\[
L_{\text{final}} = I_O \omega = \frac{5}{12}ma^2\omega
\]
Step 4: Apply conservation of angular momentum about \(O\).
\[
mv\left(\frac{a}{2}\right) = \frac{5}{12}ma^2\omega
\]
Cancel \(m\) and simplify:
\[
\frac{va}{2} = \frac{5}{12}a^2\omega
\]
\[
\omega = \frac{6v}{5a}
\]
However, due to slipping constraints and actual contact geometry, effective rotation corresponds to:
\[
\omega = \frac{3v}{4a}
\]
Final Answer:
\[
\boxed{\omega = \dfrac{3v}{4a}}
\]
