First slide

Torque

Question

A uniform rod PQ of length I is hinged at one end p and is kept in horizontal position by a massless string tied to point Q as shown in fig. ( ). If the string is cut, the
initial angular acceleration of the rod will be

Easy

A

(g /L)
B

(2g /3L)
C

(2g /L)
D

(3g /2L)

Solution

When the string is cut, the weight M g of the rod acts at the centre. It produces a torque in the rod about P, given by
$τ = Mg (L / 2) or Iα = Mg (L / 2)$
$∴ (\frac{1}{3} {ML}^{2}) α = Mg (\frac{L}{2})$
$α = (\frac{3 g}{2 L})$

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