Question

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

Answer 1

Answer:

Rotational acceleration would be 3g/2L

Explanation:

Let us first see the attached figure to understand the scenario;

Torque on the rod is equal to the moment of weight of the rod about the point P

In equation form it will be;

T = mg x L/2  ……….. eq 1

Moment of interia of rod about

P = (M L^2)/3  …………. Eq 2

As T = Iα

From  eq 1 and eq 2 we will get;

Mg x L/2 = (M L^2)/3 α

α = 3g/2L

So the rotational acceleration would be 3g/2L

Answer 2

Answer:

3g/2L

Explanation:

Torque =I *angular acceleration

MgL/2 = (ML^2 )/3 × alpha

alpha =3g/2L