Question

A torque is applied to a long thin rod making it rotate with an angular acceleration of 5.10 rad/s² about an axis through its center and perpendicular to its length (see the Figure (a) below). If the same torque is now applied to make the rod rotate about an axis about one end perpendicular to its length (see Figure (b) below), determine the new angular acceleration of the rod.
rad/s²

Answer 1

When rod is rotating about an axis through its center and perpendicular to its length, then moment of inertia of rod is given by:

I1 = M*L^2/12

Now when rod is rotating about an axis about one end perpendicular to its length, then moment of inertia of rod is given by:

I2 = M*L^2/3

Now given that same torque is applied on rod in both case, So

1 = 2

I1*1 = I2*2

2 = 1*(I1/I2)

1 = Angular acceleration in 1st case = 5.10 rad/sec^2

So,

2 = 5.10*(ML^2/12/(ML^2/3))

2 = 5.10*(3/12) = 5.10*(1/4)

2 = 1.275 rad/sec^2

In three significant figures if required, 2 = 1.28 rad/sec^2

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