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

1 Answer
Rory H.
Feb 10, 2016

The angular acceleration is (3π)radians⋅s−2 and the moment of inertia is 2 kg m2. The required torque is the product of these, which is 6π Nm, or approximately 18.8 Nm

Explanation:

First find the angular acceleration, α.

α=change in angular velocitytime

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

α = 6 (2 π *radians) s−1) / 4 s = (3 π * radians) s−2

Torque = moment of inertia * α

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 * rod length2
so:
moment of inertia = 2kg * 1m2 = 2 kg m2

Now we have:

Torque = (2 kg m2) * ((3π)radians(s−2))

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

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