A solid cylinder rolls up an inclined plane of angle of inclination 30^(@). At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5ms^(-1) How far will the cylinder go up theplane ?

A solid cylinder rolls up an inclined plane of angle of inclination 30

1.	A solid cylinder rolls up an inclined plane of angle of inclination 30^(@). At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5ms^(-1) How far will the cylinder go up theplane ?
Answer» Solution :Given, `v_c=5ms^(-1),theta=30^(@)` We know that, `v^2=(2gh)/(1+(K^2)/(R^2))` Here K`=(R^2)/2` for a solid cylinder `:.v^2=(2gh)/(1+1/2)=4/3gh" or "H=(3v^2)/(4g)` Taking `g=10ms^(-2),h=(3xx5xx5)/(40)=1.875` `SIN30^(@)=h/l"":.l=h/(sin30^(@))=1.875xx2` l=3.75m