Question

A uniform rod of length 1meter and mass100g is at one end and is hanging vertically it is displaced throw 60degress from the increase in its potential energy is g=10

Answer 1

Answer:

0.25J

Explanation:

Let the displacement of centre of mass be denoted by 'x' m

x=(l/2)(1-cosø)

Where 'l' is length of rod and ø is angle through which rod is displaced

On plugging in values, we get

Delta (PE) = 0.25J

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