A high-speed flywheel

Anonymous

2013-07-05 3:20 am

A high-speed flywheel in a motor is spinning at 500rpm when a power
failure suddenly occurs. The flywheel has mass 45.0kg and diameter 80.0cm
. The power is off for 28.0s and during this time the flywheel slows due to
friction in its axle bearings. During the time the power is off, the flywheel
makes 180 complete revolutions.

1. At what rate is the flywheel spinning when the power comes back on?

unit = rad/s

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?

unit = s

3. How many revolutions would the wheel have made during this time?

unit = rev

更新1:

Q1 is correct Q2 is incorrect For Q3, how to convert 43075 rad to 6856 rev?

更新2:

help me la

回答 (1)

天同

2013-07-05 7:53 am

✔ 最佳答案

1. Use equation of motion: s = (1/2)(u+v)t
with s = 180 rev, u = 500 rpm, t = 28/60 minutes, v =?
hence, 180 = (1/2).(500 + v).(28/60)
v = 271.4 rpm = 271.4 x 2 x pi/60 rad/s = 28.42 rad/s

2. Deceleration of the flywheel = (271.4 - 500) x (2.pi)/(60x28) rad/s^2 = -0.855 rad/s^2
Use equation of motion: v = u + at
with v = 0 rad/s, u = 271.4 rad/s, a = -0.855 rad/s^2, t =?
hence, 0 = 271.4 + (-0.855)t
t = 317.4 s

3. Use equation of motion: v^2 = u^2 + 2as
with v = 0 rad/s, u = 271.4 rad/s, a = -0.855 rad/s^2, s =?
hence, 0 = 271.4^2 + 2.(-0.855)s
s = 43075 rad = 6856 rev

2013-07-09 19:44:20 補充：
Q2: sorry, I have used the wrong value for u. It should be 500 rpm, or 52.36 rad/s
Hence, t = (v-u)/a = 52.36/0.855 s = 61.23 s

2013-07-09 19:52:11 補充：
Q3: use u = 52.36 rad/s,
s = 52.36^2/(2x0.855) rad = 1603 rad
since 1 rev = 2.pi rad
1603 rad = 1603/(2.pi) rev = 255 revolutions

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