Question

What torque would have to be applied to a rod with a length of `4"m"` and a mass of `2"kg"` to change its horizontal spin by a frequency `9"Hz"` over `8"s"`?

Answer 1

Use following formula and calculate

Explanation:

Torque (`tau`) = moment of inertia (I)X angular acceleration (`alpha` )
I = `1/12ML^2` , where mass M =2kg, Length L= 4m
`alpha=(domega)/dt` = `2pi`xrate of change of frequency
= 2x3.14x9/8 rads^-2
pl proceed and get result