Question

A sphere of mass `M` and radius `r` shown in figure slips on a rough horizontal plane. At some instant it has translational velocity `V_(0)` and rotational velocity about the centre `(v_(0))/(2r)`. Find the translational velocity after the sphere starts pure rolling.
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Answer 1

Velociyt of the centre `=v_0` and the angular velocity about the centre `=(v_0)/(2r)`. Thus `v_0gtomega_0r.` The sphre slips forward and thsu the friction by the plane on the sphere will act backward. As the friction is kinetic its value is `muN=muMg` and the sphere will be decelerated by `a_(cm)=f/M.` Hence
`v(t)=v_0-f/Mt`............i
This friction will also have a torque `Gamma=fr` about the centre. This torque is clockwise and in the directionof `omega_0`. Hence the angular acceleration about the centre will be
`alpa=f r/((2/5)(Mr^2)=(5f)/(2Mr)`
and the clockwise angular velocity at time t will be
`omega(t)=(5f)/(2Mr)t=v_0/(2r)+(5f)/(2Mr)t`
pure rolling starts when v(t)=romega(t)`
i.e. `v(t)=v_0/2+(5f)/(2M)t` ....ii
elimiN/Atin t from i and ii
`5/2v(t)+v(t) =5/2 v_0+v_0/2`
or `v(t)=2/7xx3v_=6/7v_0` ,brgt Thus the sphere rolls with translation velocity `6v_0/7` in the forward direction.
AlterN/Ative: let us consider the torque bout teh initil point of contact A. The force of firctioni passes through this point and hecne its torque is zero. The norjal force and the weight balance each other. The net torque about A is zero. Hence the angular momentum about A is conserved.
Initial angular omentum is
`L=L_(cm)+Mrv_0+I_(cm)omega+Mrv_)`
`=(2/5Mr^2)(v_0)/(2r)+Mrv_0=6/5Mrv_0`
Thus angular velocity is `v/r`. the angular momentum abut A is
`L=L_(cm)+Mrv`
`=(2/5Mr^2)(v/r)+mrv+7/5Mrv`
Thsu `6/5Mrv_0=7/5Mrv`
`or v=6/7v_0`