AS chain of length l and mass m lies o the surface of a smooth sphere of radius `Rgtl` with one end tied to the top of the sphere. a.Find the gravitational potential energy of the chain with reference level at the centre of the sphere. B. suppose the chin is released and slides down the sphere. Find the kinetic eneergy of the chain, when it has slid through an angle `theta` c. find the tangential acceleration `(dv)/(dt)` of the chain when the chain starts sliding down.

Answer

A, B, C

Explanation

(a) Mass per unit length length`=m/l`

`dm =m/(l)R dalpha`
`h=R cos alpha`
`dU =(dm)gh=(mgR^(2))/(l)cos alpha. D alpha`
`:. U=int_(0)^(l//R)dU =(mgR^(2))/(l) sin (l/R)`
(b) `KE=U_(i)-U_(f)`
Here, `U_(i) =(mgR^(2))/(l) sin(l/R)`
and
`U_(f)=int_(0)^(l//R+theta)dU =(mgR^2)/l [sin (1/R+theta)-sin theta] `
`:. KE=(mgR^(2))/l [sin (l/R)+sin theta-sin (thet+a_l/R)]`
(c ) `1/2mv^(2) =(mgR^(2))/l =[sin (l/R)+sin theta-sin(theta_l/R)]`
or `v=sqrt((2gR^(2))/l[ sin (l/R)+sin theta-sin (theta+l/R)])`
`v^(2) =(2gR^(2))/l[sin (l/R) +sin theta -sin(theta+l/R)]`
`:. 2v.(dv)/(dt) =((2gR^(2))/l[cos theta-cos (theta+l/R)])/(2v)(d theta)/(dt)`....(i)
Here `((d theta)/dt)/(v)=omega/v =l/R`
Substituting in Eq. (i), we get
`(dv)/(dt)=(gR)/l[1-cos(l/R)]`.