A man of mass `m_1` stands on the edge of a horizontal uniform disc of mass `m_2` and radius R which is capable of rotating freely about a stationary vertical axis passing through its centre. At a certain moment the man starts moving along the edge of the disc, he shifts over an angle `varphi^'` relative to the disc and then stops. In the process of motion the velocity of the man varies with time as `v^'(t)`.
Assuming the dimensions of the man to be negligible, find:
(a) the angle through which the disc had turned by the moment the man stopped,
(b) the force moment (relative to the rotation axis) with which the man acted on the disc in the process of motion.

Explanation

(a) Let z be the rotation axis of disc and `varphi` be its rotation angle in accordance with right-hand screw rule(figure). (`varphi` and `varphi^'` are to be measured in the same sense algebraically.)
As `M_z` of the system (disc+man) is conserved and `M_(z(i nitial)=0`, we have at any instant,
`0=(m_2R^2)/(2)(dvarphi)/(dt)+m_1[((dvarphi^')/(dt))R+((dvarphi)/(dt))R]R`
or, `dvarphi=[-(m_1)/(m_1+(m_2//2))]dvarphi^'`
On integrating `underset(0)overset(varphi)intdvarphi=-underset(0)overset(varphi^')int((m_1)/(m_1+(m_2//2)))dvarphi^'`
or, `varphi=-((m_1)/(m_1+m_2/2))varphi'` (1)
This gives the total angle of rotation of the disc.
(b) From Eq. (1)
`(dvarphi)/(dt)=-((m_1)/(m_1+(m_2)/(2)))(dvarphi')/(dt)=-((m_1)/(m_1+m_2/2))(v'(t))/(R)`
Differentiating with respect to time
`(d^2vaphi)/(dt^2)=-((m_1)/(m_1+m_2/2))1/R(dv^'(t))/(dt)`
Thus the sought force moment from the Eq. `N_z=Ibeta_z`
`N_z=(m_2R^2)/(2)(d^2varphi)/(dt^2)=-(m_2R^2)/(2)((m_1)/(m_1+m_2/2))1/R(dv^'(t))/(dt)`
Hence `N_z=-(m_1m_2R)/(2m_1+m_2)(dv^'(t))/(dt)`