First slide

Rotational motion

Question

A uniform rod of length L (in between the supports) and mass m is placed on two supports A and B. The rod breaks suddenly at length L/10 from the support B. Find the reaction at support A immediately after the rod breaks.

Difficult

A

$\frac{9}{40} m g$
B

$\frac{19}{40} m g$
C

$\frac{m g}{2}$
D

$\frac{9}{20} m g$

Solution

$\begin{matrix} Torque = τ = \frac{9}{10} m g (\frac{9}{20} L) = I α = \frac{m}{3} {(\frac{9}{10} L)}^{2} α \\ α = \frac{3 g}{2 L} \end{matrix}$
$Acceleration, a_{CM} = α (A C)$
$\begin{array}{l} a_{CM} = \frac{3 g}{2 L} (\frac{9 L}{20}) = \frac{27 g}{40} \\ Now, \frac{9}{10} m g - N_{A} = m a_{CM} = m \cdot \frac{27 g}{40} \\ or N_{A} = \frac{9}{40} m g \end{array}$

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