Find the translational speed of the bowling ball after it hits the bottom. Assum

Question

Find the translational speed of the bowling ball after it hits the bottom. Assume that it is a uniform solid sphere. Moment of inertia=2/5mr^2 004 10.0 points A 50 kg bowling ball with a radius of 25 cm starts from rest at the top of an incline 1.6 m ae ton of an ineline 1.eo in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume that the ball is a uniform solid sphere.)The acceleration of gravity is 9.81 m/s2. Answer in units of m/s.

Explanation / Answer

The total energy of the ball on top of the incline is
Etop = mgh = 50 *9.81*1.6 = 784.8 Nm

And the total energy changes into rotational and kinetic energy at the base of the incline:
784.8 Nm = 1/2 mv^2 + 1/2 I^2 and v = r
.................= 1/2 m(r^2^2) + 1/2 (2/5*mr^2)*^2
.................= 1/2 mr^2^2(1+2/5)
.................= 1/2*7/5*mr^2^2
.................= 7/10mv^2
v^2 = 784.8*10/(7m)
v^2 = 784.8*10/(7*50)=
v^2 = 22.42
v = 4.735 m/s = lateral velocity

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