Question
a light string 4 meters long is wrapped around a solid cylindrical spool with a radius of 0.075 m and a mass of .5 kg.a 5kg mass is then attached to the end of the string, causing the string to unwind from the spool. a. what is the angular acceleration of the spool? b. how fast will the spool be rotating after all of the string has unwound?
Answers
You will need to calculate the moment of inertia of the spool, which is
I = (1/2)M R^2 = (0.5)(0.5 kg)(.075m)^2
= 1.406*10^-3 kg m^2
The mass m on a string applies a torque of
T = m g R = 5*9.8*.075 = 3.675 N-m
(a) Angular acceleration of the spool is
(alpha) = T/I
Compute the number, in radians/s^2
(b) The angle rotated through when the string is fully unwound is
theta = L/(2 pi R)
The time t to unwind is given by
theta = (1/2)*(alpha)*T^2
Solve for T and then compute the angular velocity at that time:
w = alpha*T
I = (1/2)M R^2 = (0.5)(0.5 kg)(.075m)^2
= 1.406*10^-3 kg m^2
The mass m on a string applies a torque of
T = m g R = 5*9.8*.075 = 3.675 N-m
(a) Angular acceleration of the spool is
(alpha) = T/I
Compute the number, in radians/s^2
(b) The angle rotated through when the string is fully unwound is
theta = L/(2 pi R)
The time t to unwind is given by
theta = (1/2)*(alpha)*T^2
Solve for T and then compute the angular velocity at that time:
w = alpha*T
A spool has a radius of 5cm . A string is 8.8 m long. How many times can the string be wound around the spool?
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