Question

A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an axis which coincides with the parallelepiped’s large diagonal.

a. What is the parallelepiped’s kinetic energy?
b. What torque must be applied to the axis of rotation in order to keep it still? (neglect the gravity.)

Answer 1

(a) k =
`(Mw^(2) )/(6) (a^(2) +b^(2) )`

(b) τ =
`(M)/(3) (a^(2) +b^(2) )` ∝

Step-by-step explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I =
`(M)/(3) (a^(2) +b^(2) )` ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =
`(1)/(2) Iw^(2)` --------------------------------2

Putting equation 1 into equation 2, we have;

k =
`(M)/(6) (a^(2) +b^(2) )w^(2)`

k =
`(Mw^(2) )/(6) (a^(2) +b^(2) )`

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ =
`(Mw)/(3) (a^(2) +b^(2) )`

But

τ = dτ/dt =
`(M)/(3) (a^(2) +b^(2) )(dw)/(dt)` ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ =
`(M)/(3) (a^(2) +b^(2) )` ∝