A uniform 5.50 kg square solid wooden gate 2.00 m on each side hangs vertically

Question

A uniform 5.50 kg square solid wooden gate 2.00 m on each side hangs vertically from a frictionless pivot at its upper edge. A 1.00 kg raven flying horizontally at 4.50 m/s flies into this door at its center and bounces back at 2.50 m/s in the opposite direction.

Part A

What is the angular speed of the gate just after it is struck by the unfortunate raven?

QUESTION #2

A 55.0-kg grindstone is a solid disk 0.570 m in diameter. You press an ax down on the rim with a normal force of 140 N(Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50Nm between the axle of the stone and its bearings.

Part A

How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 8.00 s ?

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Part B

After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min?

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Part C

How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone?

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Explanation / Answer

1) Before collision
I = mr*Vr1 = 1.0*4.5 = 4.5 Nsec
after collision
I = 4.5 Nsec
4.5 = -mr*Vr2+mg*Vg
Vg = (4.5+1.0*2.5)/5.5 = 1.27 m/sec
= Vg/r = 1.27/2 = 0.635 rad/sec

2)   I = ½mr² = ½ * 55kg * (0.285m)² = 2.23 kg·m²
= 120rev/min * 2 rad/rev * 1min/60s = 12.6 rad/s
so = / t = 12.6 rad/s /8s = 1.58 rad/s²

A) net torque = I = 2.23kg·m² * 1.58rad/s² = 3.52 N·m
But also = F*r = 3.52 = F*0.5m - 0.6 * 140N * 0.285m - 6.50N·m
F = 33.96N·m / 0.5m = 67.92 N

B) Now 0 N·m = F*0.5m - 0.6 * 140N * 0.285m - 6.50N·m
F = 30.44 N·m / 0.5m = 60.88 N·m

C) = 6.50 N·m = I = 2.23kg·m² *
= 2.91 rad/s²
t = / = 12.6rad/s /2.91 rad/s² = 4.33 s

Hope this helps!

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