How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis o

Question

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 85, 000 g's? A bucket of mass 1.80 kg is whirled in a vertical circle of radius 1.30 m. At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N. (a) Find the speed of the bucket. (b) How fast must the bucket move at the top of the circle so that the rope does not go slack? A jet pilot takes his aircraft in a vertical loop. (a) If the jet is moving at a speed of 1600 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's. (b) Calculate also the 72 kg pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle. (c) Calculate the pilot's effective weight at the top of the circle. (Assume the same speed.) A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 310 km/h in a semicircular arc with a radius of 198 m. Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration. Determine the radial acceleration of the car at this time. If the curve were flat, what would the coefficient of static friction have to be between the tires and the roadbed to provide this acceleration with no

Explanation / Answer

2.
(b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

From g = a = u^2/R
u = sqrt(Rg)
= sqrt(1.30*9.8) = 3.57 m/s

3. V = 1600 km/h = 1600(1000/3600) = 444.44 m/s
Ac = Centripetal Acceleration = V²/R
5g = (444.44)^2/R
R = (444.44)^2/(5(9.81)) = 4027 m
At bottom of circle = Normal force = mAc + mg
= m(444.44)^2/4027 + m(9.81)
Normal force = 49.05m + 9.81m = 58.86(72) = 4237.9 N
At top of circle = Normal force
= mAc - mg = 49.05 m - 9.81m = 39.24m
Normal force = 39.24(72) = 2825.2 N

r = 198 m , v1=0 km/s=0m/s , v2=310 km/h=86.1m/s = 86.1m/s
tangeital acceleration is related to rotational acceleration or alpha.
a = r
and we know, 2^2 =1^2 + 2(b) where 1 , and 1 are angular velocities one and two, and is angular acceleration, and b is the angle in radans . so since we have v1, and v2 and we know that :
v = r ---------------we can find 2, and 1
1 = v1 / r = 0 radan/second
2 = v2/ r = 86.1 / 198 = 0.4348 radan/s
since the car moves in a semicircular arc it is 180 degrees so half way is a quarter of the circle or almost 90 degree which is equal to /2 radan so b=/2
0.4348^2 = 0^2 + 2(pi/2)
so = 0.06017 radan/s/s
now that we have we find a
a = r = 198 * 0.06017 = 11.91 m/s^2
if by radial acceleration u mean angular acceleration then we have : = .06017 radan/s/s
since we also have a Centripetal acceleration we need a centripetal force
F = mv^2/r
in this case the F or force is the force of friction between tires and the asfalt since there is no tilt to the path.
so we have
F(friction) = mv^2/r
and we know that F = mg
mg = m v^2/r
g = v^2 /r
= v^2 /r*g since the maximum speed reached by the car in our interval is v =310 km/g = 86.1 m/s we have
= (86.1)^2 / (9.8 * 198) = 3.204

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