Which arrives at the bottom first, (a) a ball rolling without sliding down a cer

Question

Which arrives at the bottom first, (a) a ball rolling without sliding down a certain incline A, (b) a solid cylinder rolling without sliding down incline A, or (c) a box of the same mass as the ball sliding down a frictionless incline B having the same dimensions as A? Assume that each object is released from rest at the top of its incline. A horizontal disk with moment of inertial I1 rotates with angular speed omega 1 about a vertical frictionless axle. A second horizontal disk, with moment of inertial I2 and initially not rotating, drops onto the first. Because their surfaces are rough, the two eventually reach the same angular speed omega. The ratio omega/omega 1 is equal to (a) I1/I2 (b) I2/I1 (c) I1/(I1 + I2) (d) I2/(I1 + I2) If global warming continues, it's likely that some ice from the polar ice caps of the Earth water will melt and the water will be distributed closer to the Equator. If this occurs, would the length of the day (one revolution (a) increase, (b) decrease, or (c) remain the same? A device revolving at 1100 rpm speeds up to 2200 rpm in 4s. Calculate alpha of the motor and the number of revolutions it makes in this time. A disk revolves with constant acceleration alpha = 5rad/s^2. a) How many turns does it make in 8 seconds from rest? b) During the 3rd second? A wheel revolving at 900 rpm slows down to 300 rpm in 50rev. Find i) alpha, and ii) time required for these 50rev. A wheel rolls down a smooth inclined plane 50' high. i) What are GPEi and KEi? i) As the wheel rolls down the incline, it gives up GPE for KE. What are the two forms of KE being gained? ___ and __, iii) What is the equation that shows this relationship? iv) Understanding that PE lost = KE gained, show that the v of the wheel when it is at the foot of the plane is v = 40 ft/s. Assume the weight of the wheel is concentrated at the rim.

Explanation / Answer

18. Applying angular momentum conservation,

I1 w1 + 0 = (I1 + I2)w

w/w1 = I1 / (I1 + I2)

ans(c)

19. I - will increase

w - will decrease

day length will increase

(a)

20. wf = wi + alpha t

2200 = 1100 + alpha(4/60)

alpha = 16500 rev/m^2

wf^2 - wi^2 = 2alpha rev

2200^2 - 1100^2 = 2(16500) rev

rev = 110 rev

21. (a) angle revolved = wi t + alpha t^2 / 2

= 0 + (5 x 8^2 / 2)

= 160 rad

revolutions = 25.5 rev

(b) angle = 5 x (3^2 - 2^2) /2 = 12.5 rad

turns = 12.5 / 2pi = 2 turns

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