- #1
Oribe Yasuna
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A string is wrapped around a uniform disk of mass M = 1.3 kg and radius R = 0.05 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 = 0.08 m, each with a small mass m = 0.3 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 = 33 N. At the instant when the center of the disk has moved a distance d = 0.034 m, a length w = 0.023 m of string has unwound off the disk.
(a) At this instant, what is the speed of the center of the apparatus?
v =
(b) At this instant, what is the angular speed of the apparatus?
omega1 =
(c) You keep pulling with constant force 33 N for an additional 0.039 s. Now what is the angular speed of the apparatus?
omega2 =
Idisk = 1/2 MR^2
I = m1r1^2 + m2r2^2 ...
torque = r x F
I = 1/2 MR^2 + m1r1^2 + ... m4r4^2
I = 1/2 (1.3)(0.05)^2 + 4*(0.3)(0.08)^2
I = 0.009305 kg*m^2
torque = r x f
torque = 1.65 N*m
I have no idea where to go from this.
(a) At this instant, what is the speed of the center of the apparatus?
v =
(b) At this instant, what is the angular speed of the apparatus?
omega1 =
(c) You keep pulling with constant force 33 N for an additional 0.039 s. Now what is the angular speed of the apparatus?
omega2 =
Homework Equations
Idisk = 1/2 MR^2
I = m1r1^2 + m2r2^2 ...
torque = r x F
The Attempt at a Solution
I = 1/2 MR^2 + m1r1^2 + ... m4r4^2
I = 1/2 (1.3)(0.05)^2 + 4*(0.3)(0.08)^2
I = 0.009305 kg*m^2
torque = r x f
torque = 1.65 N*m
I have no idea where to go from this.