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UNIVERSITY PHYSICS （20 MARKS）
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1. Use equation: s = (1/2).(u+v)t with s = 170 rev., u = 450 rpm, t = 34 s hence, 170 = (1/2)(450 + v).(34/60) v = 150 rpm 2. Deceleration of flywheel = -(170-450)/(34/60) rev/min^2 = -494.1 rev/min^2 Use equation: v = u + at with v = 0 rpm, u = 450 rpm, a = -494.1 rev/min^2, t =? 0 = 450 + (-494.1)t t = 0.9107 min = 54.64 s 3. Use equation: s = (1/2).(u+v)t s = (1/2).(450).(0.9107) rev = 205 rev
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UNIVERSITY PHYSICS （20 MARKS）
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A high-speed flywheel in a motor is spinning at 450rpm rpm when a power failure suddenly occurs. The flywheel has mass 45.0kg kg and diameter 74.0cm cm . The power is off for 34.0s s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes... 顯示更多 A high-speed flywheel in a motor is spinning at 450rpm rpm when a power failure suddenly occurs. The flywheel has mass 45.0kg kg and diameter 74.0cm cm . The power is off for 34.0s s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions. 1 At what rate is the flywheel spinning when the power comes back on? （in angular velocity) 2 How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on? 3 How many revolutions would the wheel have made during this time? 更新: ANS incorrect..... and the first one should be in angular velocity.. 更新 2: SOR,, the first one is correct but the other two incorrect,,,最佳解答:
1. Use equation: s = (1/2).(u+v)t with s = 170 rev., u = 450 rpm, t = 34 s hence, 170 = (1/2)(450 + v).(34/60) v = 150 rpm 2. Deceleration of flywheel = -(170-450)/(34/60) rev/min^2 = -494.1 rev/min^2 Use equation: v = u + at with v = 0 rpm, u = 450 rpm, a = -494.1 rev/min^2, t =? 0 = 450 + (-494.1)t t = 0.9107 min = 54.64 s 3. Use equation: s = (1/2).(u+v)t s = (1/2).(450).(0.9107) rev = 205 rev
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