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 59 rad/s. The wheel is run at that angular velocity for 26 s and then power is shut off. The wheel decelerates uniformly at 1.6 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:

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Physics

19 November, 17:45

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 59 rad/s. The wheel is run at that angular velocity for 26 s and then power is shut off. The wheel decelerates uniformly at 1.6 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:

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Enrique · Answer 1

Time interval; Δt ≈ 37 seconds

Explanation:

We are given;

Angular deceleration; α = - 1.6 rad/s²

Initial angular velocity; ω_i = 59 rad/s

Final angular velocity; ω_f = 0 rad/s

Now, the formula to calculate the acceleration would be gotten from;

α = Change in angular velocity/time interval

Thus; α = Δω/Δt = (ω_f - ω_i) / Δt

So, α = (ω_f - ω_i) / Δt

Making Δt the subject, we have;

Δt = (ω_f - ω_i) / α

Plugging in the relevant values to obtain;

Δt = (0 - 59) / (-1.6)

Δt = - 59/-1.6

Δt = 36.875 seconds ≈ 37 seconds