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 the figure. When string is cut, the initial angular acceleration of the rod is

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 the figure. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-gvnmvhbsl9heotpq.jpg" /> When string is cut, the initial angular acceleration of the rod is (A) (3 g/2 L) (B) (g/L) (C) (2 g/L) (D) (2 g/2 L)

Tardigrade Admin · Accepted Answer

Torque on the rod = Moment of weight of the rod about

P

.

τ = M g \frac{L}{2}

..... (i)
Moment of inertia of the rod about P

I = \frac{M L ^{3}}{3}

..... (ii)
As,

τ = I α

From Eqs. (i) and (ii), we get

M g \frac{L}{2} = \frac{M L ^{2}}{3} α

\Rightarrow α = \frac{3 g}{2 L}

Q. 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 the figure.

When string is cut, the initial angular acceleration of the rod is

$\frac{3 g}{2 L}$

$\frac{g}{L}$

$\frac{2 g}{L}$

$\frac{2 g}{2 L}$

Torque on the rod = Moment of weight of the rod about $P$ .

$τ = M g \frac{L}{2}$ ..... (i)
Moment of inertia of the rod about P
$I = \frac{M L ^{3}}{3}$ ..... (ii)
As, $τ = I α$
From Eqs. (i) and (ii), we get
$M g \frac{L}{2} = \frac{M L ^{2}}{3} α$
$\Rightarrow α = \frac{3 g}{2 L}$

Q. 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 the figure. When string is cut, the initial angular acceleration of the rod is

2L3g​

Lg​

L2g​

2L2g​