Posted by **JOJO** on Monday, March 18, 2013 at 5:15am.

A spherical bowling ball with mass m = 4.1 kg and radius R = 0.116 m is thrown down the lane with an initial speed of v = 8.9 m/s. The coefficient of kinetic friction between the sliding ball and the ground is ¦Ì = 0.34. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.

How long does it take the bowling ball to begin rolling without slipping?

4) How far does the bowling ball slide before it begins to roll without slipping?

5) What is the magnitude of the final velocity?

6) After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:

- physics -
**Elena**, Monday, March 18, 2013 at 11:05am
v₀ = 8.9 m/s

m=4.1 kg

R = 0.116 m

μ=0.34

===

F=ma

F(fr)=μmg

F=F(fr)

ma= μmg

a= μg=0.34•9.8=3.33 m/s²

The torque M=Iε=(2mR²/5)•ε

M=F(fr)•R= μmgR

(2mR²/5)•ε= μmgR

ε= 5μmgR/2mR²= =5μg/2R=5•0.34•9.8/2•0.116=71.8 rad/s²

When speed v has decreased enough and angular speed ω has increased enough, the ball stops sliding and then rolls smoothly

v=v₀-at

ω= ω ₀+εt = εt

v= ωR

v₀-at= εtR

v₀ =t(a+εR)

t= v₀/(a+εR)=

=8.9/(3.33 +71.8•0.116)=0.76 s.

(4) s= at²/2=3.33•0.76²/2=0.962 m

(5) v=v₀-at=8.9 -3.33•0.76=6.37 m/s

(6) ω= εt = 71.8•0.76=54.57 rad/s

E(rot) = Iω²/2= 2mR²ω²/ 10=

=2•4.1•0.116²•54.57²/10=32.86 J.

E(tr) =mv²/2=4.1•6.37²/2=83.18 J

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