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 `9 Hz` over `6 s`?

Answer 1

`L=(25*pi)/2`

Explanation:

`L=alpha*I`
`"(L:torque ;alpha :angular acceleration ;I:the moment of inertia)"`
`alpha=(Delta omega)/(Delta t)`
`alpha=(2*pi*Delta f)/(Delta t)`
`alpha=(2*pi*9)/6`
`alpha=3*pi`
`I=1/12*m*l^2 " for a rod spinning around its center"`
`I=1/12*2*5^2=25/6`
`L=3*pi*25/6`
`L=(25*pi)/2`