Question

[ (44) A uniform bar of length L hinged at one end is released from horizontal position so as to fall unde ],[ the initial angular acceleration of the rod will be ]​

Answer 1

Answer:

Initial angular acceleration of the rod about its one end is $\alpha = \frac{3g}{2L} rad/s^2$

Explanation:

At Initial position of the rod the torque about hinge point is due to weight of the rod

So we will have

$\tau = mg(\frac{L}{2})$

now we know that the moment of inertia about one end of the rod is given as

$I = \frac{mL^2}{3}$

now we know that

$\frac{mL^2}{3} \alpha = mg(\frac{L}{2})$

so we will have

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

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Topic : Torque on rigid body about fixed axis

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Answer 2

Answer:

Initial angular acceleration of the rod about its one end is  

Explanation:

At Initial position of the rod the torque about hinge point is due to weight of the rod

So we will have

now we know that the moment of inertia about one end of the rod is given as

now we know that

so we will have

Explanation: