As shown in the figure, a given force is applied to a rod in several different ways. In which case is the torque about the pivot
P due to this force the smallest?
1 answer:
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Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:
is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following
Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:
638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>