Question

Chapter 09, Problem 57 Chalkboard Video A bowling ball encounters a 0.760-m vertical rise 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 5.57 m/s at the bottom of the rise. Find the translational speed at the top. Units Number

Answer 1

Since there is no frictional losses ,the total energy is conserved

In initial condition the ball has Linear kinetic energy and rotational kinetic energy

In final condition the ball has Linear kinetic energy ,rotational kinetic energy and potential energy

$1/2mv_{initial}^{2}+1/2I\omega _{initial}^{2}=1/2mv_{final}^{2}+1/2I\omega _{final}^{2}+mgh$

$I=2/5mr^{2}$ and $\omega =v/r$

Initial condition:

$TE_{initial}=7/10mv_{initial}^{2}$

$TE_{initial}=7/10*m*5.57^{2}$

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Final Condition

$TE_{final}=7/10mv_{final}^{2}+mgh$

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$TE_{initial}=TE_{Final}$

$7/10*m*5.57^{2}=7/10mv_{final}^{2}+mgh$

Cancel mass m

$7/10*5.57^{2}=7/10v_{final}^{2}+gh$

$21.72=0.7v_{final}^{2}+0.76*9.81$

$14.2644=0.7v_{final}^{2}$

$v_{final}=\sqrt{14.2644/0.7}$

ANSWER: $v_{final}=4.514m/s$

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