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

Answer:

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

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 = $\alpha$

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

$\alpha =\frac{\omega _2-\omega _1}{t}=\frac{0.00 rps-20.0 rps}{80.0 s}$

$\alpha =-0.25$ revolutions per second square

$\alpha =-0.25\times 2\pi =-1.57 rad/s^2$

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

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

$\omega _1= 20.0 rps = 20.0\times 2\pi =125.66 rad/s$

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

$\Delta \theta=\omega _1t+\frac{1}{2}\alpha t^2$

$=125.66 rad/s\times 80.0 s+\frac{1}{2}(-1.57 rad/s^2)\times (80.0)^2$

$\Delta \theta=5028.8 revolutions$

Number of revolution = $\frac{5028.8 }{2\pi }=800.35 \approx 800$

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