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Two different balls are rolled (without slipping) toward a common finish line. Ball 1 has an angular speed of ????1=23.2 rad/s, and ball 2 h
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Two different balls are rolled (without slipping) toward a common finish line. Ball 1 has an angular speed of ????1=23.2 rad/s, and ball 2 has an angular speed of ????2=14.8 rad/s. The first ball, which has a radius of 0.0663 m, is rolling along a conveyor belt that is moving at 1.19 m/s and starts out 8.57 m from the finish line. The second ball has a radius of 0.0488 m and is rolling along the stationary floor. Two balls are vertically offset and are rolling with clockwise angular speed towards a vertical finish line to the right. Ball 1 rolls with an angular speed of omega subscript 1 on a conveyor belt above ball 2. Arrows indicate that the top surface of the conveyor belt that ball 1 is on moves to the right, while the bottom surface of the conveyor belt moves to the left. Ball 2 rolls with an angular speed of omega subscript 2 on a stationary floor. If the second ball starts out 5.84 m from the finish line, how long does each ball take to reach the finish li
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Physics
3 years
2021-07-28T01:03:12+00:00
2021-07-28T01:03:12+00:00 1 Answers
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Answer:
Ball 1 = 3.14s , ball 2 = 8.11s
Explanation:
ω₁ = 23.2rad/s
ω₂ = 14.2 rad/s
r₁ = 0.0663m
r₂ = 0.0488m
S = 8.57m
u = 1.19m/s
Assuming conveyor belt is moving towards the finish line,
W = v + u
v = ωr
ωr + u
(ω₁ * r₁ ) + v = (23.9 * 0.0663) + 1.19 = 2.728m/s
Angular velocity (ω) = distance moved (s) / time taken to complete 1 revolution
ω = s / t
t = s / ω
t = 8.57 / 2.728
t = 3.14s
For ball 2, the ball starts 5.54m from the finish line,
W = v + o (since there’s no longer initial velocity)
ωr = v
V = 14.2 * 0.0488
V = 0.683 m/s
Velocity = distance / time
V = s / t
t = s / v
t = 5.54 / 0.683
t = 8.11s