A solid ball encounters a 0.78 m vertical rise (h) on the way back to the ball r
ID: 1263536 • Letter: A
Question
A solid ball encounters a 0.78 m vertical rise (h) on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.60 m/s at the bottom of the rise. Find the translational speed at the top.
A solid ball encounters a 0.78 m vertical rise (h) on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.60 m/s at the bottom of the rise. Find the translational speed at the top.Explanation / Answer
the ball has initial kinetic and rotational energy; the kinetic energy is 1/2 mv^2 and the rotational
is 1/2 Iw^2 where I is the moment of inertia and w is the angular velocity
the moment of inertia of a solid sphere is 2/5mr^2 where r is the radius of the ball, and the angular velocity is related to the linear velocity by w=v/r, so that the rotational energy can be written as
1/2*2/5 mr^2(v/r)^2 = 1/5 mv^2
therefore, the total kinetic energy before the rise is 1/2 mv^2+1/5mv^2 = 7/10 mv^2
the ball loses an amount of energy equal to mg h where h = 0.78m, so we have from energy conservation
7/10 mv^2 = 0.78mg + 7/10 mvf^2 where vf is the final velocity
canceling out m's, we have
7/10(3.6m/s)^2 = 0.78m*9.8m/s/s+7/10 vf^2
vf=1.428 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.