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A billiard ball of mass 165 g and radius 2.5 cm starts with a translational spee

ID: 1265824 • Letter: A

Question

A billiard ball of mass 165 g and radius 2.5 cm starts with a translational speed of 2 m/s at point A on the track as shown in the figure below. If point B is at the top of a hill that has a radius of curvature of 89 cm, what is the normal force acting on the ball at point B? Assume the billiard ball rolls without slipping on the track.

A billiard ball of mass 165 g and radius 2.5 cm starts with a translational speed of 2 m/s at point A on the track as shown in the figure below. If point B is at the top of a hill that has a radius of curvature of 89 cm, what is the normal force acting on the ball at point B? Assume the billiard ball rolls without slipping on the track.

Explanation / Answer

The moment of inertia of the solid ball is given as ::

I = 2/5 m r2

r = 0.025 m

Total kinetic energy at A = translational KE + Rotational KE

Total kinetic energy at A = 1/2 m v2         (since W=0 at A)

Total kinetic energy at A = 1/2 m v2       

Total kinetic energy at A = 1/2 m v2

Potential energy at A = mgh                               ( h= height from dotted reference line = 10 cm = 0.1 m)

Total energy at A = Total kinetic energy at A + Potential energy at A

Total energy at A = 1/2 m v2   + mgh                       Eq-1

Total energy at B = KE at B =1/2 m vf2 + 1/2 I W^2                  Eq-2          ( Vf = speed at B )

Total energy at B = 1/2 m vf2 + 1/2 (2/5 m r2) (V/r)2 = 7/10 m vf2

Using conservation of energy at A and B ::

Total energy at A   = Total energy at B

1/2 m v2   + mgh     = 7/10 m vf2      

5 v2   + 10gh     = 7 vf2      

inserting the values   , V = 2 m/s , h = 0.1 m

5 (2)2 + 10 (9.8) (0.1) = 7 vf2

Vf = 2.06 m/s

At Point B ::

Fn = normal force acts up

mg = weight of ball acts down

mv2 /R = centripetal force acts towards the center

force equation for the ball can be given as ::

mg - Fn = mvf2 /R

Fn = mg - mvf2 /R

inserting values m = 165 g = 0.165 kg    , Vf = 2.06 m/s   , R = 89 cm + 2.5 cm = 0.915 m

Fn = 0.165 x 9.8 - 0.165 (2.06)2 /(0.915)

Fn = 1.62 - 0.765

Fn = 0.855 N

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