1- The flywheel of a steam engine begins to rotate from rest with a constant ang
ID: 1352580 • Letter: 1
Question
1- The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.43 radians/s2. It accelerates for 34.3 s, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 45.7 s after it began rotating.
2- A flywheel comprises a uniform circular disk with a moment of inertia of 96.33 Kg m2. It rotates with an angular velocity of 1085. rpm. A constant tangential force is applied at a radial distance of 0.600 m. What energy (in Kilo Joule) must this force dissipate to stop the wheel?
Explanation / Answer
First method
1) for t = 34.3 seconds,
= w0t + 1/2at2
where = angular angle
w0 = starting angular velocity = 0
a = angular acceleration = 1.43 radians/s2
=> angular angle = 0+1/2*1.43*(34.3)2 = 841.19 radians
w1 = w0 + at
here w0 = 0
a = 1.43 radians/s2
t = 34.3s
=> w1 = 0+1.43*34.3 = 49.05radians/s
for t = 45.7-34.3 = 11.4s
= w1*t = 49.05*11.4 = 559.17radians
total angle covered = 841.19 + 559.17 = 1400.36 radians
Second method
1) after t = 34.3s
the angular velocity is (34.3 s)*(1.43 radians/s2) = 49.05 radians/s
The angle through which it rotates in those first 34.3 seconds is 34.3 s times
the mean velocity (49.05/2 = 24.525 radians/s),
so angle = 24.525*34.3 = 841.2 radians.
In the next 45.7-34.3= 11.4 seconds, it goes through 49.05*11.4 radians = 559.17 radians,
so the total angle covered in the 45.7 seconds is 1400.37 radians
2)
The rotational energy of an object = 1/2 I w2
where I is the moment of inertia and w is the angular velocity
convert 1085 rpm to rad/sec:
1085 rev/min = 1085rev/min x 2 pi rad/rev x 1 min/60s = 113.67 rad/s
KE rot = 1/2* (96.33kgm2) *(113.67rad/s)2= 6.22x105 J
To stop the flywheel, this much energy must be “dissipated”.
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