Fall 2017 (20 points) 6) A 4.20 kg flat disk rotates in a circle, radius 0.300 m
ID: 1778267 • Letter: F
Question
Fall 2017 (20 points) 6) A 4.20 kg flat disk rotates in a circle, radius 0.300 m, on a frictionless tabletop.It is held in this orbit by a light cord that is connected to a dangling 2.50 kg block through a hole in the center of the table. a) Draw a free-body diagram for the hanging mass and the disk. b) Write down Newton's 2d Law for the hanging mass and the disk c) Solve these to find the speed of the disk Now consider the same problem but with a coefficient of kinetic friction, A-0250. The disk in this case will spiral in. Assume that at - 0 the velocity is tangent to the cirele and has the magnitude that you calculated in part c), find the net force, F. acting d) on the disk at 1 = 0.Explanation / Answer
a) Free body diagram for the disc and hanging mass is,
Fc = centripetal force, N = normal reaction, w = weight of the disc, W = weight of the hanging block, T = tension in the rope due to hanging mass
b) As per the FBD, we have
CPF on the disc gets balanced by the tension in the rope due to hanging mass.
So, Fc = T
mv²/r = M*g
c) m = 4.2 kg, M = 2.5 kg, r = 0.3 m
After solving for v, we get,
v = Mgr/m = (2.5*9.8*0.3)/4.2 = 7.35/4.2 = 1.75 = 1.323 m/s (Answer)
d) It is the force of friction which provides the centripetal force to the disc, at t = 0, v = 1.323 m/s.
Force of friction Fr = Fc = µ*N = µmg = 0.25 * 4.2 * 9.8 = 10.29 N (Answer)
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