Question

A long, thin rod is cut into two pieces, one being twice as long as the other. To the midpoint of piece A (the longer piece), piece B isattached perpendicularly, in order to form the inverted "T" shown in the drawing. The application of a net external torque causes thisobject to rotate about axis 1 with an angular acceleration of 1.8 rad/s2. When the same net external torque is used to cause the objectto rotate about axis 2, what is the angular acceleration?

Answer 1

Let x represent the original length of the rod.

From the information given,

angular acceleration, α1 = 1.8 rad/s^2

Since the rod is divided into A and B, then

A + B = x

A is twice the length of B. It means that

A = 2B

Thus,

2B + B = x

2B = x

B = x/3

A = x - x/3

A = 2x/3

Let M be the initial mass of the rod

Mass of A = 2M/3

Mass of B = M/3

The formula for calculating the moment of inertia of a thin rod with length x whose axis passes through the center and it is perpendicular to its length is

I = Mx^2/12

If the axis passes through one end,

I = Mx^2/3

Calculating moment about axis 1,

I1 = 0 + 1/3 * M/3 * (x/3)^2

I1 = Mx^2/81

Calculating moment about axis 2,

I2 = 1/12 * 2M/3 * (2x/3)^2 + 0

I2 = 8Mx^2/324

In both cases, the external applied torques are equal. This,

I1α1 = I2α2

α2 = I1α1/I2

By substituting the given values,

α2 = (Mx^2/81 * 1.8)/(8Mx^2/324)

α2 = Mx^2/81 * 1.8 * 324/8Mx^2

α2 = 0.9 rad/s^2

the angular acceleration = 0.9 rad/s^2