Question

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a circuit breaker trips at time t = 2.40 s. From then on, it turns through an angle 435 rad as it coasts to a stop at constant angular acceleration. Part APart complete Through what total angle did the wheel turn between t=0 and the time it stopped? Express your answer in radians. θ = 570 rad Previous Answers Correct Part B At what time did it stop? Express your answer in seconds. t = nothing s Request Answer Part C What was its acceleration as it slowed down? Express your answer in radians per second squared.

Answer 1

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Step-by-step explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

`\theta = \theta_(o) + \omega_(o)\cdot t + (1)/(2)\cdot \alpha \cdot t^(2)`

`\omega = \omega_(o) + \alpha \cdot t`

`\omega^(2) = \omega_(o)^(2) + 2 \cdot \alpha \cdot (\theta-\theta_(o))`

Where:

`\theta_(o)`,
`\theta` - Initial and final angular position, measured in radians.

`\omega_(o)`,
`\omega` - Initial and final angular speed, measured in radians per second.

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

`t` - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

`\theta - \theta_(o) = \omega_(o)\cdot t + (1)/(2)\cdot \alpha \cdot t^(2)`

If
`\omega_(o) = 25\,(rad)/(s)`,
`t = 2.40\,s` and
`\alpha = 26\,(rad)/(s^(2))`, then:

`\theta-\theta_(o) = \left(25\,(rad)/(s) \right)\cdot (2.40\,s) + (1)/(2)\cdot \left(26\,(rad)/(s^(2)) \right)\cdot (2.40\,s)^(2)`

`\theta-\theta_(o) = 134.88\,rad`

The final angular angular speed can be found by the equation:

`\omega = \omega_(o) + \alpha \cdot t`

If
`\omega_(o) = 25\,(rad)/(s)`,
`t = 2.40\,s` and
`\alpha = 26\,(rad)/(s^(2))`, then:

`\omega = 25\,(rad)/(s) + \left(26\,(rad)/(s^(2)) \right)\cdot (2.40\,s)`

`\omega = 87.4\,(rad)/(s)`

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

`\Delta \theta = 134.88\,rad + 435\,rad`

`\Delta \theta = 569.88\,rad`

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

`\omega^(2) = \omega_(o)^(2) + 2 \cdot \alpha \cdot (\theta-\theta_(o))`

The angular acceleration is now cleared:

`\alpha = (\omega^(2)-\omega_(o)^(2))/(2\cdot (\theta-\theta_(o)))`

Given that
`\omega_(o) = 87.4\,(rad)/(s)`,
`\omega = 0\,(rad)/(s)` and
`\theta-\theta_(o) = 435\,rad`, the angular deceleration is:

`\alpha = ( \left(0\,(rad)/(s)\right)^(2)-\left(87.4\,(rad)/(s) \right)^(2))/(2\cdot \left(435\,rad\right))`

`\alpha = -8.780\,(rad)/(s^(2))`

Now, the time interval of the Deceleration Phase is obtained from this formula:

`\omega = \omega_(o) + \alpha \cdot t`

`t = (\omega - \omega_(o))/(\alpha)`

If
`\omega_(o) = 87.4\,(rad)/(s)`,
`\omega = 0\,(rad)/(s)` and
`\alpha = -8.780\,(rad)/(s^(2))`, the time interval is:

`t = (0\,(rad)/(s) - 87.4\,(rad)/(s) )/(-8.780\,(rad)/(s^(2)) )`

`t = 9.954\,s`

The total time needed for the grinding wheel before stopping is:

`t_(T) = 2.40\,s + 9.954\,s`

`t_(T) = 12.354\,s`

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.