Question

A small disc A is placed on an inclined plane forming an angle `alpha` with the horizontal and is imparted an initial velocity `v_(0)`. Find how the velocity of the disc depends on the angle `theta`, shown in figure-2.209, if the friction coefficient `mu = tan alpha` and at the initial moment `theta=pi//2`

Answer 1

The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane `F_x=mg sin alpha` and the friction force `f r=kmg cos alpha`. In our case `k=tan alpha` and therefore
`f r=F_x=mg sin alpha`
Let us find the projection of the acceleration on the derection of the tangent to the trajectory and on the x-axis:
`mw_t=F_xcos varphi-f r=mg sin alpha(cos varphi-1)`
`m w_x=F_x-f rcos varphi=mg sin alpha(1-cos varphi)`
It is seen from this that `w_t=-w_x`, which means that the velocity v and its projection `v_x` differ only by a constant value C which does not change with time, i.e.
`v=v_x+C`,
where `v_x=vcos varphi`. The constant C is found from the initial condition `v=v_0`, where
`C=v_0` since `varphi=pi/2` initially. Finally we obtain
`v=v_0//(1+cos varphi)`.
In the cource of time `varphirarr0` and `vrarrv_0//2`. (Motion then is unaccelerated.)