First slide

Dynamics of rotational motion about fixed axis

Question

A rod of mass 5 kg is connected to the string at point B. The span of rod is along horizontal. The other end
of the rod is hinged at point A If the string is massless, then the reaction of hinge at the instant when string
is cut, is (Take, 9 = 10 ms^-2) [UP CPMT 2015]

Moderate

A

10.1N
B

12.5N
C

5N
D

15N

Solution

When string is cut, the weight of the rod constitutes a torque about the hinge point A.
So, $τ_{A} = m g \frac{l}{2}$ …(i)
Also, from Newton's law, $τ_{A} = I α$ …(ii)
where, $α$ = angular acceleration of the rod
and I = moment of inertia of rod.
From Eqs. (i) and (ii), we get
$I α = m g \frac{l}{2} \Rightarrow α = m g \frac{l}{2} / I$
As, $I = m \frac{l^{2}}{3}$
So, $α = \frac{m g (l / 2)}{\frac{m l^{2}}{3}} = \frac{3 g}{2 l}$
Now, acceleration of centre of mass of rod, $a_{CM} = α \cdot r$
Here, r = distance between hinge point A and centre of the rod.
$\Rightarrow a_{CM} = \frac{3}{2} \frac{g}{l} \frac{l}{2} = \frac{3 g}{4}$
$\Rightarrow m g - R_{A} = m a_{CM}$
where, R_A = reaction at A.
$\Rightarrow m g - R_{A} = m \times \frac{3 g}{4}$
$\Rightarrow R_{A} = \frac{m g}{4} = \frac{5 \times 10}{4} = 12.5 N$

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