Question

The sphere shown in figue lies on a rough plane when a particle of mass m travbelling at a speed `v_0` collides and sticks with it. If the line of motion of the particle is at a distance h above the plane, find a. the linear speed o the combine dsystem just after teh collision b. the angular speed of the system about the centre of the sphere just the collistion c. the value of h for which the sphere starts pure rolling on the plane Assume that the mass M of the sphesre is large compared to teh mass of the partcle so that is large compared toteh mass of the particle so that teh centre of mass of the combined system is not appreciably shifted from the centre of the sphere.

Answer 1

Take the particle plus the sphere as the system
a. Using conservation of linear momentum, the inear speed o the combined system v is given by
`mv_0=(M+m)v or, v=(mv_0)/(M+m)` ……..i
b. Next we shall use conservation angular momentum about the centre of mass, which is to be taken at the centre of the sphere `(Mgtgtm)` Angular momentum of the particle before collsion is `mv_0(h-R).` If the system rotates with angular speed `omega` after collision, the angular momentum of the system becomes
`(2/5MR^2+mR^2)omega`
Hence, `mv_0(h-R)=(2/5 M+m)R^2omega`
or, ` omega=(mv_0(h-R))/(2/5 M+m))R^2`
c. The sphere will start roling just after the colision if
`v=omegaR i.e. (mv_0)/(M+m)=(mv_0(h-R))/(2/5M+m)R)`
giving`h=((7/5M+2m)/(M+m))R=7/5R`.