Physics, 26.06.2020 15:01 lin550

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to: A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:. a) 19 s
b) 17 s
c) 21 s
d) 23 s
e) 15 s

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Answer from: KyiahDenise

The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)

Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:

$t = \frac{\omega - \omega_{o}}{\alpha}$ (1)

Where:

$\omega_{o}$ - Initial angular velocity, in radians per second.

$\omega$ - Final angular velocity, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

- Time, in seconds.

If we know that $\omega = 0\,\frac{rad}{s}$ , $\omega_{o} = 25\,\frac{rad}{s}$ and $\alpha = -1.5\,\frac{rad}{s^{2}}$ , then the time taken by the grinding wheel to stop:

$t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }$

$t = 16.667\,s$

The time interval of angular deceleration is 16.667 seconds. ( B)

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Answer from: rurbanok12

Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration $\alpha$ is

$25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}$

For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time ,

$0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s$

which makes B, approximately 17 s, the correct answer.

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Answer from: Quest

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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel a...

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