Respuesta :
Answer:
a)[tex]-1.57 rad/s^2[/tex] 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 [tex]\omega _1[/tex]= 20.0 rps
Final angular velocity of grinding wheel[tex]\omega _2[/tex] = 0.0 rps
Time taken by wheel to come to rest = t = 80.0 s
a) Angular acceleration of the wheel = [tex]\alpha [/tex]
[tex]\omega _2=\omega _1+\alpha t[/tex]
[tex]\alpha =\frac{\omega _2-\omega _1}{t}=\frac{0.00 rps-20.0 rps}{80.0 s}[/tex]
[tex]\alpha =-0.25 [/tex] revolutions per second square
[tex]\alpha =-0.25\times 2\pi =-1.57 rad/s^2[/tex]
[tex]-1.57 rad/s^2[/tex] was the angular acceleration of the grinding wheel.
b) Initial angular velocity of grinding wheel [tex]\omega _1[/tex]= 20.0 rps
[tex]\omega _1= 20.0 rps = 20.0\times 2\pi =125.66 rad/s[/tex]
Number of revolution before coming to rest = [tex]\Delta \theta[/tex]
[tex]\Delta \theta=\omega _1t+\frac{1}{2}\alpha t^2[/tex]
[tex]=125.66 rad/s\times 80.0 s+\frac{1}{2}(-1.57 rad/s^2)\times (80.0)^2[/tex]
[tex]\Delta \theta=5028.8 revolutions[/tex]
Number of revolution = [tex]\frac{5028.8 }{2\pi }=800.35 \approx 800[/tex]
800 revolutions has made by the wheel during the time it was coming to rest
