Question

A 150 g uniform rod of length 3.0 m is pivoted about one end. What would be the angular acceleration due to gravity if the rod were held horizontally and released from rest?

Answer 1

Given the length of the rod L= 3m

Mass of rod =150g = 150*10-³ Kg

The moment of inertia of the rod with respect to its end I = 1/3. mL²

I= 1/3 *0.150*3² = 0.45 Kg m²

The angular acceleration can be found by the equation,

Torque T = I* a

Where T is the torque on the rod due to gravity and a is the angular acceleration.

T = F * r, F is the weight of the rod and r is the perpandicular distance form the pivot to centre of mass

Here F =mg and r = 3/2 m

Do torque T = mg * r= 0.150*9.8*3/2

T = 2.205 Nm

We have torque T = I a Or

Angular acceleration a = T/I

Subtituting the values ,

a =2.205/0.45=

Angular acceleration of the rod just after releasing = 4.9 rad /s²

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