Question

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

a)
`-1.57 rad/s^2` was the angular acceleration of the grinding wheel.

b)800 revolutions has made by the wheel during the time it was coming to rest

Step-by-step explanation:

Initial angular velocity of grinding wheel
`\omega _1`= 20.0 rps

Final angular velocity of grinding wheel
`\omega _2` = 0.0 rps

Time taken by wheel to come to rest = t = 80.0 s

a) Angular acceleration of the wheel =
`\alpha`

`\omega _2=\omega _1+\alpha t`

`\alpha =(\omega _2-\omega _1)/(t)=(0.00 rps-20.0 rps)/(80.0 s)`

`\alpha =-0.25` revolutions per second square

`\alpha =-0.25* 2\pi =-1.57 rad/s^2`

`-1.57 rad/s^2` was the angular acceleration of the grinding wheel.

b) Initial angular velocity of grinding wheel
`\omega _1`= 20.0 rps

`\omega _1= 20.0 rps = 20.0* 2\pi =125.66 rad/s`

Number of revolution before coming to rest =
`\Delta \theta`

`\Delta \theta=\omega _1t+(1)/(2)\alpha t^2`

`=125.66 rad/s* 80.0 s+(1)/(2)(-1.57 rad/s^2)* (80.0)^2`

`\Delta \theta=5028.8 revolutions`

Number of revolution =
`(5028.8 )/(2\pi )=800.35 \approx 800`

800 revolutions has made by the wheel during the time it was coming to rest