Question

A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in figure. It hits a ridge at point O. The angular speed of the block after it hits O is …. 

(A) 3v / 4a 

(B) 3v / 2a 

(C) √(3v) / √(2a) 

(D) zero

Answer 1

Correct option: (A) (3v / 4a)

Explanation:

when block hits ridge at O, it will start rotating about an axis passing through O and perpendicular to the plane of paper. No external torque acts on block hence angular momentum is conserved Angular momentum before hitting, Li = M.V × AC

= MV ∙ (a/2) = [(mva) / 2]

after hitting, L_f = I_oω

I_o is MI of block about axis through O and perpendicular to plane of block.

if I_c is MI about C then I_C = [(Ma²) / 6]

From parallel axes theorem,

I_o = I_c + Mr² 

and r² = (a/2)² + (a/2)² = (a² / 2)

I_o = (1/6)Ma² + (1/2)Ma² = (2/3)Ma² 

∴ L_f = (2/3)Ma² ∙ ω

also L_i = L_f hence

[(mva) / 2] = (2/3)Ma²ω hence ω = (3v / 4a) 

Comments

Here we use conservation of momentum because here no torque is applied on external force.