Please assist with part C: Thank you so much? A string is wrapped around a unifo
ID: 1618418 • Letter: P
Question
Please assist with part C: Thank you so much?
A string is wrapped around a uniform disk of mass M = 1.6 kg and radius R = 0.07 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.15 m, each with a small mass m = 0.8 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 31 N. At the instant when the center of the disk has moved a distance d = 0.039 m, a length w = 0.012 m of string has unwound off the disk.
(a) At this instant, what is the speed of the center of the apparatus?
v = .709 m/s
(b) At this instant, what is the angular speed of the apparatus?
1 = 3.13 radians/s
(c) You keep pulling with constant force 31 N for an additional 0.030 s. Now what is the angular speed of the apparatus?
radians/s
Explanation / Answer
a = F / m = 31 / (1.6 + 4x0.8) = 6.46 m/s^2
Net torque = I alpha
I = (1.6 x 0.07^2 / 2) + (4 x 0.8 x 0.15^2) = 0.07592 kg m^2
0.07 x 31 = 0.07592 alpha
alpha = 28.6 rad/s^2
vf^2 - vi^2 =2 a d
v^2 - 0^2 = 2 x 6.46 x 0.039
v = 0.710 m/s ..........Ans(a)
(b) 0.710 = 0 + 6.46t
t = 0.110 sec
wf = wi + alpha t
w = 0 + (0.110 x 28.6) = 3.15 rad/s
(c) now t = 0.110 + 0.03 = 0.140 s
w = 0 + (0.140 x 28.6) = 4 rad/s