Amachinist 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 the wheel is run at that angular velocity for 37 s and then power is shut off. the wheel decelerates uniformly at until the wheel stops. in this situation, the time interval of angular deceleration (slowing down) is closest to:

Complete Question:

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 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to

t= 16.7 sec.

Explanation:

As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:

γ = (ωf -ω₀) / t

If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:

γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².

When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:

γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec

d.7.5cm is your answer