Question

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the intial angular acceleration of the rod is

A. `(3g)/(2L)`
B. `(g)/(L)`
C. `(2g)/(L)`
D. `(2g)/(3L)`

Answer 1

Correct Answer - A

Torque on the rod = Moment of weight of the rod about P,
`tau = "Mg" (L)/(2)` ….(i)
`because` Moment of inertia of rod about P,
`I = (ML^(2))/(3)` ….(ii)
As `tau = I alpha`
From Eqs. (i) and (ii), we get
`"Mg"(L)/(2)=(ML^(2))/(3)alpha`
`:.` The initial angular acceleration of the rod
`alpha = (3g)/(2L)`