Solving Force on Rotating Bar at Pivot

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SUMMARY

The discussion focuses on calculating the force exerted by air jets on a rotating bar, specifically a 60-cm-long, 500 g bar that reaches an angular velocity of 150 rpm in 10 seconds. The user correctly converts the angular velocity to radians per second (15.71 rad/s) and applies the equations of motion for rotational dynamics. The calculated angular acceleration (alpha) is 0.3142 rad/s², leading to a force of 0.04713 N using the formula F = m(r(alpha)). The confusion arises from mixing up angular velocity equations.

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  • Understanding of rotational dynamics and angular motion
  • Familiarity with the equations of motion for rotating bodies
  • Knowledge of converting units from rpm to rad/s
  • Basic principles of force and mass (F=ma)
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  • Explore the principles of fluid dynamics related to jet propulsion
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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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