- #1
JJBladester
Gold Member
- 286
- 2
Homework Statement
It is known that 1500 revolutions are required for the 6000-lb flywheel to coast to rest from an angular velocity of 3000 rpm. Knowing that the radius of gyration of the flywheel is 36 in., determine the average magnitude of the couple due to kinetic friction in the bearings.
Answer:
[tex]\left |M \right |=87.8lb\cdot ft[/tex]
Homework Equations
[tex]KE_1+Work=KE_2[/tex]
[tex]Work_{1\rightarrow 2}=M\theta[/tex]
The Attempt at a Solution
It takes 1500 revolutions to come to rest and 1 revolution = 2[itex]\pi[/itex], so 1500 revolutions = 3000[itex]\pi[/itex].
[tex]W = 600lbs[/tex]
[tex]m=\frac{W}{g}=\frac{6000}{9.81}=612lbs[/tex]
[tex]\omega _1=300\frac{rev}{min}*\frac{2\pi}{1rev}*\frac{1min}{60sec}=10\pi\frac{rad}{sec}[/tex]
Radius of gyration (k) = 36 in = 3 ft
[tex]KE_1=\frac{1}{2}\bar{I}\omega_1^{2}=\frac{1}{2}\left (k^2m \right )\omega_1^{2}=\frac{1}{2}\left (3^2*612 \right )\left (10\pi \right )^{2}=2.72e^6J[/tex]
[tex]Work_{1\rightarrow 2}=M\theta=M(3000\pi)[/tex]
[tex]KE_1+Work=0[/tex]
[tex]2.72e^6=\left (-3000\pi \right )M[/tex]
[tex]M=-288.6lb\cdot ft[/tex]
I'm off... but where?