What torque would have to be applied to a rod with a length of 1 m and a mass of 2 kg to change its horizontal spin by a frequency of 6 Hz over 4 s?

1 Answer
Feb 10, 2016

The angular acceleration is (3pi) text{radians} * s^-2 and the moment of inertia is 2 kg m^2. The required torque is the product of these, which is 6pi Nm, or approximately 18.8 Nm

Explanation:

First find the angular acceleration, alpha.

alpha = text{change in angular velocity} / text{time}

The angular velocity is 6 Hz, and 1 Hz means that a circle of 2pi radians is traversed in 1 second. Thus the angular velocity is 6 (2 pi radians) s^-1, which is 12 pi radians s^-1

alpha = 6 (2 pi *radians) s^-1) / 4 s = (3 pi * radians) s^-2

Torque = moment of inertia * alpha

moment of inertia depends on the shape of the spinning object and can be complicated. Fortunately for a mass on a rod it's simple:
moment of inertia = mass * text{rod length}^2
so:
moment of inertia = 2kg * 1m^2 = 2 kg m^2

Now we have:

Torque = (2 kg m^2) * ((3 pi) text{radians} (s^-2))

Torque = 6pi Nm,
or approximately :
Torque ~ 18.8 Nm