Question

A uniform rod of mass 20 kg and length 1.6 m is pivoted at its end and swings in the vertical plane. Angular acceleration of the rod just after the rod is released from rest in the horizontal position

Answer 1

The angular acceleration of the rod is α = 15 g / 16

Explanation:

Mass of rod = 20 Kg

Length of rod = 1.6 m

Torque of the weight mg=Iα

mg×  l  / 2 =  ml^2   / 3α

α = 3 g / 2 l

α = 3 x g / 2 x 1.6

α = 30 g / 2 x 16

α = 30 g / 3

α = 15 g / 16

Thus the angular acceleration of the rod is α = 15 g / 16

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Answer 2

Answer:

15g/12

Explanation:

Given mass:20kg

Length:1.6m

Torque=mg=I alpha

=mg×l/2=ml²/3 alpha

By calculating

alpha=3g/2l

=3g/2×1.6

=30g/2×16

=30g/32

By cancelling with 2 we get

alpha=15g/16