A 68-cm-long, 490 g bar rotates in a horizontal plane on an axle that passes thr

Question

A 68-cm-long, 490 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 180 rpm at the end of 9.0 s

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

b)If the axle is moved to one end of the bar while the air jets are unchanged, what will be the bar's angular velocity at the end of 9.0 seconds?

Explanation / Answer

a) Moment about center of the bar:
M = I ang acel = F L
Then, the force F is:
F = I ang acel / L

I = moment of inertia = 1/12 m L^2
ang acel = w / t = (2pi/60 * n) / t = pi n / 30 t
Replacing
F = (1/12 m L^2) * ( pi n /30 t) / L
F = pi m L n / 360 t
F = 3.1416 * 0.49kg * 0.68m *180rpm / 360* 9s
F = 0.058N

b) Bar rotates around one of its ends:
I = 1/3 m L^2

M = F L = I pi n / 30 t
F L = 1/3 m L^2 pi n / 30 t
F = m L pi n / 90 t
n = 90 t F / m L pi
n = 90* 9s *0.058N / 0.49kg* 0.68m* pi
n = 44.88 rpm

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