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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