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
Answer:
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
