A torque on a body produces angular acceleration. Consider a rigid body rotatin about an axis passing through the point O and perpendicular to the plane of the figure. Suppose that a torque ` vec(tau)` on the body produces uniform angular acceleration `vec(alpha)` along the axis of rotation.
The body can be considered as made of N particles with masses `m_(1),m_(2),...., r_(N)` respectively from the axis of rotaion. `vec(alpha)` is the same for all the particles as the body is rigid. Let `vec(F_(1)), vec(F_(2)), ...., vecF_(N)` be the external forces on the particles.
the torque `vec(tau_(1)),` on the particle of mass `m_(1),` is `vec(tau_(1)) = vec(r_(1))xxvec(F_(1))`.
`therefore tau_(1) = r_(1)F_(1) sin theta` where `theta` is the smaller of the two angles between `vec(r_(1))`
and `vec(F_(1)). " " therefore tau_(1) = r_(1)F_(1)` (since, in this case, `theta = 90^(@))`
Now, `F_(1) = m_(1)a_(1) = m_(1)r_(1)alpha " "(because a_(1) = r_(1)alpha) " "...(1)`
`therefore tau_(1) = r_(1) (m_(1) r_(1) alpha) = m_(1)r_(1)^(2)alpha" "...(2)`
Similarly , `tau_(2) = m_(2)r_(2)^(2) alpha, ....tau_(N) = m_(N) r_(N)^(2) alpha`
The total torque on the body is
`tau = tau_(1) + tau_(2) + ... + tau_(N)`
` = m_(1)r_(1)^(2)alpha + m_(2)r_(2)^(2)alpha + ... m_(N)r_(N)^(2)alpha`
` = (m_(1)r_(1)^(2) + m_(2)r_(2)^(2) + .... + M_(N)r_(N)^(2))alpha " "....(3)`
` = (sum_(i=1)^(N) m_(i)r_(i)^(2))alpha`
`therefore tau = Ialpha " "...(4)`
where `I = sum_(i=1)^(N)m_(i)r_(i)^(2)` is hte moment of inertia of the body about the axis of roataion.
In vector form `tau = vec(Ialpha)`
This gives the requuired relation. Angular acceleration `vecalpha` has the same direction s the torque `vecatau` adn both of them are along the rotation axis.
