A solid cylinder rolls down a fixed inclined plane of height 'h' and inclination θ. Calculate the time taken to reach bottom?
{Assume the mass be M and radius be R}
The Free body drawing of the sphere is as shown
Assume the direction of 'a' and 'α' as shown,
Mgsinθ−fr=Ma ....(1)
Taking torque about centre O,
τ=fr×h=Iα
As we know ,I=MR22 {about centre 0}
⇒R.fr=MR22×α
Also as the cyclinder is rolling purely,
a=Rα and v=Rω.
⇒R.fr=MR22×aR
fr=Ma2 ............(ii)
from (i)and (ii),
Mg sinθ−Ma2=Ma
⇒a=23gsinθ .......(iii)
Now, assume the length of wedge be 'L'.
By simple trigonometry,
L sinθ=h
or L=h cosecθ . ....(iv)
Applying the equation of motion,
s=ut+12at2
From (i),(ii)and(iv)
h cosecθ=0+1/2∝/23gt2sinθ
3h cosec2θg=t2
t=√3h cosec2θg
or {As taking only postive value}
t=cosecθ.√3hg.
Hence (D)