At time t =0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a c

Question

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 28.0 rad/s2 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.

Part A

Through what total angle did the wheel turn between t=0 and the time it stopped?

Express your answer in radians.

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Part B

At what time does the wheel stop?

Express your answer in seconds.

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Part C

What was the wheel's angular acceleration as it slowed down?

Express your answer in radians per second per second.

Explanation / Answer

here,

initial angular velocity , w0 = 30 rad/s

angular accelration , a1 = 28 rad/s^2

t1 = 2.5 s

velocity at t= 2.5 s , w1 = 30 + 28 * 2.5

w1 = 100 rad/s

(a)

theta1 = w0 * t1 + 0.5 * a1 * t1^2

theta1 = 30 * 2.5 + 0.5 * 28 * 2.5^2

theta1 = 162.5 rad

total angle did the wheel turn between t=0 and the time it stopped = theta1 + theta

total angle did the wheel turn between t=0 and the time it stopped is 602.5 rad

(b)

let the wheel deaccelration be a

using third equation of motion

0 - w1^2 = 2 * theta * a

- 100^2 = 2 * 440 * a

a = - 11.36 rad/s^2

using first equation of motion

0 = w1 + a * t2

0 = 100 - 11.36 * t2

t2 = 8.8 s

the time at which the wheel srops = t1 + t2

the time at which the wheel stops is 11.3 s

(c)

the angular accelration of the wheel while slowing down is - 11.36 rad/s^2

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