Question

What torque would have to be applied to a rod with a length of `5 m` and a mass of `2 kg` to change its horizontal spin by a frequency of `7 Hz` over `3 s`?

Answer 1

`244.4 N.m`

Explanation:

Torque is defined as the rate of change in angular momentum,

i.e `tau = (dL)/dt = (d(Iomega))/dt = I (domega)/dt`

Now,moment of inertia(`I`) of a rod w.r.t its one end is `(Ml^2)/3`,(as here its length is the radius of the circle in which it moves)

so,its `I =16.67 Kg.m^2` (given `l`=5 m and `M`=`2Kg`)

Now,its rate of change in angular velocity is `(domega)/dt = 7/3*2pi` (as, `omega = 2pi*nu` where `nu` is frequency)

So,torque acting is `16.67*(7/3)*2pi` 0r, `244.4 N.m`