Question

A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 56 rad/s. The wheel is run at that angular velocity for 21 s and then power is shut off. The wheel slows down uniformly at 1.7 rad/s2 until the wheel stops. In this situation, the time interval of deceleration is closest to:

Answer 1

t = 32.94 sec

Step-by-step explanation:

From 1st equation of motion, we know that:

Vf = Vi + at

So, for angular motion it will become:

ωf = ωi + at

ωf - ωi = αt

t = (ωf - ωi)/α

where,

ωi = initial angular velocity = 56 rad/s

ωf = final angular velocity = 0 rad/s (since, wheel finally stops)

α = deceleration = - 1.7 rad/s²

t = time interval of deceleration = ?

Therefore,

t = (0 rad/s - 56 rad/s)/(-1.7 rad/s²)

t = 32.94 sec