A small mass m starts from rest and slides down the smooth spherical surface of F. Assume zero potential energy at the top. Find
(a) the change in potential energy,
(b) the kintic energy,
(c) the speed ot the mass as a function of the angle `theta` made by the radius through the mass with the vertical.

Options

A. `- mgR(1 - costheta)`, `mgR(1-cos theta)`, `v=sqrt(2gR(1-cos theta))`
B. `- mgR(1+costheta)`, `mgR(1-cos theta)`, `v=sqrt(2gR(1-cos theta))`
C. `- mgR(1 - costheta)`, `mgR(1+cos theta)`, `v=sqrt(2gR(1-cos theta))`
D. `- mgR(1 - costheta)`, `mgR(1-cos theta)`, `v=sqrt(2gR(1+cos theta))`

Answer

A

Explanation

In the figure, `h=R (1-cos theta)`
(a) As the mass comes down, potential energy will decrease. Hence,
`DeltaU = - mgh = - mgR(1 - costheta)`
(b) Magnitude of decrease in potential energy = increase in kinectic energy
`:.` Kinetic energy`=mgh`
`=mgR(1-cos theta)`
(c) `1/2mv^(2)=mgR(1-costheta)`
`:. v=sqrt(2gR(1-cos theta))`.
.