Question

A solid sphere is released from height h from the top of an incline making an angle θ with the horizontal. (a) Calculate the speed of the sphere when it reaches the bottom of the incline in the case that it rolls without slipping. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

Answer 1

Answer:

v =√ (gh/.700)

Explanation:

As per the conservation of mechanical energy,

( K+ U)i = ( K+U)f

where i is the initial energy and f is the final energy.

Kinetic energy, K(i) at top = 0

when call will reach the bottom , U(f) = 0

U(i) = mgh

K(f) = 1/2 (mv²) + 1/2 I ω²

For solid sphere , I = 2/5 m R²

and  ω = v/R

so , K = 1/2 mv² + 1/2*2/5*m*R^2*v^2/R^2

     K = .700 mv²

now .700mv² = mgh

                  v =√ (gh/.700)