Question

A sphere of mass `M` is held at rest on a horizontal floor. One end of a light string is fixed at a point that is vertically above the centre of the sphere. The other end of the string is connected to a small particle of mass `m` that rest one the sphere. The string makes an angle `alpha = 30^(@)` with the vertical. Then the acceleration of the spherer immediately after it is released is : (There is no frication anywhere and string is tangene to the sphere) :

A. `(sqrt(3)mg)/(3m + 2M)`
B. `(sqrt(3)mg)/(2m + 4M)`
C. `(sqrt(2)mg)/(3m + 4M)`
D. `(sqrt(3)mg)/(3m + 4M)`

Answer 1

Correct Answer - D
Since the particle is small, the string will be trangential to the sphere.` theta = 60^(@)` (see figure).
Let acceleration of the sphere be `a_(0)` immediately after release
The particle will have its initial acceleration (a) along normal to the string towards `PC`. And, component of `a_(0)` in the direction `overset(vec)(PC)` must be equal to a

`:. a = a_(0)cos(90-theta) = a_(0)sintheta = (sqrt(3))/(2)a_(0)`
Force on the particle has been shown in figure. The equation of motion along `PC` will be
`mg costheta - N = ma rArr (1)/(2)mg - N = (sqrt(3))/(2)ma_(0)` .......(i)
Sphere experiences a forces `N` along `PC`. It has a horizontal component `= N sin theta`.......(ii)
From `(i)` and `(ii) (1)/(2)mg = ((sqrt(3m))/(2)+(2M)/(sqrt(3)))a_(0)`
`a_(0) = (sqrt(3)mg)/(3m + 4M)`