Question

A hollow sphere is rolling along a horizontal floor at 7.00 m/s when it comes to a 31.0 ∘ incline.

How far up the incline does it roll before reversing direction?

Express your answer with the appropriate units.

Answer 1

Let mass of hollow sphere be m

let the radius be r

Moment of inertia of hollow sphere, I = 2/3*m*r²

Angular velocity be

Given,

velocity, v = 7 m/s

let the distance sphere move up the incline be L

Thus, vertical displacement, h = L*sin = L*sin31

                                                   = 0.52*L

Now, Sphere will change the direction when velocity becomes zero

Applying conservation of energy

=> Initial energy = final energy

=> 1/2*m*v² + 1/2*I*² = m*g*h

Since, I = 2/3*m*r²   and   = v/r

=> 1/2*m*v² + 1/2*(2/3*m*r²)*(v/r)² = m*g*h

=> 1/2*m*v² + 1/3*m*v² = m*g*h

=> 5/6*m*v² = m*g*h = m*g*0.52*L

=> 5/6*v² = g*0.52*L

=> 5/6*7*7 = 9.8*0.52*L

=> L = (5/6*49)/(0.52*9.8)

         = 7.98 m