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m_a_m_a [10]
3 years ago
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 pow

er 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?
Physics
1 answer:
finlep [7]3 years ago
7 0

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

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