1) A machinist turns the power on to a grinding wheel, at rest, at time t=0s. Th
ID: 1352730 • Letter: 1
Question
1) A machinist turns the power on to a grinding wheel, at rest, at time t=0s. The wheel accelerates uniformly for 10s and reaches the operating angular velocity of 96 rev/s. The wheel is run at that angular velocity for 40 s and then power is shut off. The wheel slows down uniformly at 1.5 rev/s^2 until the wheel stops. In this situation, the angular displacement in Revolutions during the slowdown is: a) 680 rev b) 489 rev c) 2560 rev d) 410 rev e) 3072 rev
2) 1) Alex turns the power on to a grinding wheel, at rest, at time t=0s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 96 rad/s. The wheel is run at that angular velocity for 40 s and then power is shut off. The wheel slows down uniformly at 1.5 rad/s^2 until the wheel stops. The total angular displacement in revolutions from start to stop is close to: a) 2400 rev b) 1200 rev c) 489 rev d) 680 rev e) 7500 rev
Explanation / Answer
As per guide lines I am working first problm
from the rotational kinematic equation
w2 = w1 + alpha t
= 0 + alpha t
alpha = w2/t = 96 rev/s/10 s = 9.6rad/s^2
angaul displacment is
theta = w1t + 1/2 * alpha t^2
= 0 + 1/2 *9.6rev/s^2 ( 10 s)^2
=480 rev aprroximately 489 rev
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