A vertical frictionless semicircular track of radius `1m` fixed on the edge og a movable trolley (figure). Initially, the system is rest and a mass `m` is kept at the top of the track. The trolly starts moving to the right with a uniform horizontal acceleration `a=2g//9` . The mass slides down the track, eventually losing contact with it and dropping to the floor `1.3m` below the trolly. This `1.3m` is from the point where mass loses contact. `(g=10m//s^(2))`

(a) Calculate the angle `theta` at which it loses contact with the trolly and
(b) the time taken by the mass to drop on the floor, after losing contact.

Answer

B, C

Explanation

(a) Let `v_(r)` be the velocity of mass relative to track at angular position `theta` .
From work energy theorem, `KE` of particle relative to track
=Work done by force of gravity+work done by pseudo force
:. `(1)/(2)mv_(r)^(2)=mg(1-costheta)+m((2g)/(9))sintheta`
opr `v_(r)^(2=2g(1-costheta)+(4g)/(9)sintheta`
Particle leaves contact with the track where `N=0`

or `mgcostheta-m((2g)/(9))sintheta=mv_(r)^(2)`
or `gcostheta-(2g)/(9)sintheta=2g(1-costheta)+(4g)/(9)sintheta`
or `3costheta-(6)/(9)sintheta2`
Solving this, we get `theta~~37^(@)`
(b) From Eq. (i) we have,
`v_(r)=sqrt(2g(1-cos theta)+(4g)/9 sin theta)`
or `v_(r)=2.58 m//s at theta=37^(@)`
Vertical component of its velocity is
`v_(y)=v_(r) sin theta`
`=2.58xx3/5`
`=1.55 m//s`
Now, `1.3=1.55t+5t^(2)( :' s=ut+1/2 gt^(2))`
or `5t^(2)+1.55t-1.3=0`
or `t=0.38 s`