Solving Force on Rotating Bar at Pivot

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A 60-cm-long, 500 g bar rotates on an axle, with air jets providing thrust, reaching an angular velocity of 150 rpm in 10 seconds. The conversion of rpm to rad/s yields 15.71 rad/s, and the angular acceleration is calculated as 0.3142 rad/s². Using the force equation F = m(rα), the force exerted by each air jet is determined to be approximately 0.04713 N. There is confusion regarding the equations for angular velocity and angular displacement, with clarification needed on their correct applications. The discussion highlights the importance of accurately applying rotational dynamics equations to solve for forces in rotating systems.
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Homework Statement



A 60-cm-long, 500 g bar rotates in a horizontal plane on an axle that passes through the center of the bar. Compressed air is fed in through the axle, passes through a small hole down the length of the bar, and escapes as air jets from holes at the ends of the bar. The jets are perpendicular to the bar's axis. Starting from rest, the bar spins up to an angular velocity of 150 rpm at the end of 10 s.

How much force does each jet of escaping air exert on the bar?

Homework Equations



w1=w0t+1/2alpha(t^2)
f=ma
a=r(alpha)


The Attempt at a Solution



change velocity into rad/s

150rpm=15.71rad/s

F=ma
a=r(alpha)

so i subbed in a into the force equation

F=m(r(alpha))
and i sub in the velocities to find alpha

w1=1/2(alpha)t^2
15.71=1/2(alpha)(10)^2

alpha=0.3142

sub all the known #s into F=mr(alpha)
F=.5(.3)(.3142)
F=0.04713N

what did i do wrong??
 
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Isn't the equation for angular velocity
\omega=\omega_{0}+\alpha t?

I think you're confusing it with
\phi=\phi_{0}+\omega t+\frac{1}{2}\alpha t^{2}
 
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