Question

At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until a circuit breaker trips at time t= 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.a.) Through what total angle did the wheel turn between t= 0 and the time it stopped? answer in radb.) At what time does the wheel stop? (secs)c.)What was the wheel's angular acceleration as it slowed down?Express your answer in radians per second per second.

Answer 1

Ф_T =544

The wheel will stop after 10 s.

α_z = -11.15 rad/s^2

Step-by-step explanation:

The angular acceleration is constant. Thus, we will apply the equations of rotation with constant angular acceleration model.

(a) In order to calculate the total angle, we will divide the entire interval from t = 0 to the time the wheel stops into two intervals.

From t = 0 to t = 2 s:

Ф-Ф_o =1/2(w_0z+w_z)t (1)

We will calculate w_z first:

w_z = w_0x +α_xt

w_z = 24 + (35)(2.5)

w_z = 111.5

Substitute w_x into Eq.1

Ф-Ф_o = 1/2(24+111.5)(2)

Ф-Ф_o = 136 rad

We can calculate it directly from the following equation:

Ф-Ф_o = w_0x*t+1/2a_xt^2

Ф-Ф_o = 24*2.5+1/2*35*2.5

Ф-Ф_o =103.75 rad

Thus, the total angle the wheel turned between t = 0 and the time it stopped:

Ф_T =103.75 +440

Ф_T =544

(b)

We will take the interval from when the circuit breaker trips until the wheel comes to a stop.

w_oz = 111.5 rad/s

wz = 0 the wheel will stop at the end. 1

Ф-Ф_o =1/2(w_0z+w_z)t

440 = 1/2(111.5+0)*t

t = 8s

Adding t of the first interval to t of the second interval :

t_T = 2 + 8 =10

The wheel will stop after 10 s.

(c)

We will take the interval from when the circuit breaker trips until the wheel comes to a stop.

w_oz = 111.5 rad/s

w_z = 0

w_x =w_oz+α_z*t

α_z = -11.15 rad/s^2

Answer 2

a)
`\Delta \theta = 617.604\,rad`, b)
`\Delta t = 10.392\,s`, c)
`\alpha = -14.128\,(rad)/(s^(2))`

Step-by-step explanation:

a) The final angular speed at the end of the acceleration stage is:

`\omega = 24\,(rad)/(s) + \left(35\,(rad)/(s^(2))\right) \cdot (2.50\,s)`

`\omega = 111.5\,(rad)/(s)`

The angular deceleration is:

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

`\alpha = (\left(0\,(rad)/(s) \right)^(2)-\left(111.5\,(rad)/(s) \right)^(2))/(2\cdot (440\,rad))`

`\alpha = -14.128\,(rad)/(s^(2))`

The change in angular position during the acceleration stage is:

`\theta = (\left(111.5\,(rad)/(s) \right)^(2)-\left(0\,(rad)/(s) \right)^(2))/(2\cdot \left(35\,(rad)/(s^(2)) \right))`

`\theta = 177.604\,rad`

Finally, the total change in angular position is:

`\Delta \theta = 440\,rad + 177.604\,rad`

`\Delta \theta = 617.604\,rad`

b) The time interval of the deceleration interval is:

`\Delta t = (0\,(rad)/(s) - 111.5\,(rad)/(s) )/(-14.128\,(rad)/(s^(2)) )`

`\Delta t = 7.892\,s`

The time required for the grinding wheel to stop is:

`\Delta t = 2.50\,s + 7.892\,s`

`\Delta t = 10.392\,s`

c) The angular deceleration of the grinding wheel is:

`\alpha = -14.128\,(rad)/(s^(2))`