v₀ = 8.9 m/s
R = 0.116 m
a= μg=0.34•9.8=3.33 m/s²
The torque M=Iε=(2mR²/5)•ε
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
ω= ω ₀+εt = εt
=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=
E(tr) =mv²/2=4.1•6.37²/2=83.18 J
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