Question

5. A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest? (a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive? (b) How many revolutions did the wheel make during the time it was coming to rest?

Answer 1

Answer:

a) -1.57 rad/s²

b) 800 revolutions

Explanation:

a)In here you can use the equations of velocity as if it were a linear movement. In this case:

wf = wo + at

wo is the innitial angular velocity, that we can get this value using the fact that a revolution is 2π so:

wo = 20 * 2π = 125.66 rad/s

We have the time of 80 seconds, and the final angular speed is zero, because it's going to a rest so:

0 = 125.66 + 80a

a = -125.66 / 80

a = -1.57 rad/s²

b) In this part, we will use the following expression:

Ф = Фo + wo*t + 1/2 at²

But as this it's coming to rest then:

Ф = 1/2at²

solving we have:

Ф = 0.5 * (-1.57)*(80)²

Ф = 5,024 rad

Ф = 5024 / 2π

Ф = 800 revolutions