Question

A cubical block of side a moving with velocity v on a smooth horizontal plane. It hits a rigid point. What will be the angular speed of the block after it hits the rigid point.

Answer 1

A cubical block of side a moving with velocity v on a smooth horizontal plane. It hits a rigid point.The angular speed of the block after it hits the rigid point will be $\frac{3v}{4a}$

• As soon as the block hits the stop, the impact force impulse produces a torque about the centre of mass of the cube.

• According to Angular momentum conservation about o(corner point).

                    Mv*a/2 = I(about 0)*w

• Moment of Inertia about an axis through centre of mass = $Ma^{2} b$
• I about an axis through o and ⊥ to the plane of paper =

        $\frac{Ma^{2} }{6} + M(\frac{a}{\sqrt{2}} )^2  = \frac{4Ma^{2} }{6} = \frac{2Ma^{2} }{3}$

$Mv\frac{a}{2} = \frac{2}{3} Ma^{2}$ω

Therefore,

              ω=$\frac{3v}{4a}$