Question

a uniform rod of mass 10 kg and length 3 m is pivoted at it's end . angular acceleration of rod about pivot, just after it is released from rest in the horizontal position as shown in figure​

Answer 1

Explanation:

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Answer 2

The angular acceleration of rod about pivot is 5 rad/s².

A uniform rod of mass 10 kg and length 3 m is pivoted at its one end.

We have to find the angular acceleration of rod about pivot, just after it is released from rest in the horizontal position as shown in figure.

Concepts :

• The rate of change of angular momentum is known as torque.
• But we know, angular momentum, L = Iω , where I is moment of inertia and ω is angular velocity.
• Torque is a measure of a force of an object than can cause to rotate about an axis. and equation of torque is given by, τ = rFsinθ, where r is distance of centre of mass to its axis(or position) , F is the force acting on it and θ is the angle between position and force.

Here,

Rate of change of angular momentum = torque

⇒ d(Iω)/dt = rFsinθ

⇒ I × dω/dt = rFsinθ

⇒ I × α = rFsinθ

Here just after the rod is released,

• moment of inertia of rod about an end, I = ml²/3
• position, r = l/2 [ from center of mass to the axis ]
• force, F = W = mg
• angle between position to force ( weight) , θ = 90°

Now,

ml²/3 × α = l/2 × mg × sin90°

⇒ α = 3g/2l

Here, l = 3 m and g = 10 m/s²

⇒ α = 5 rad/s²

Therefore the angular acceleration of rod about pivot is 5 rad/s².

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