An object with a weight of 50.0 N is attached to the free end of a light string
ID: 1290493 • Letter: A
Question
An object with a weight of 50.0 N is attached to the free end of a light string wrapped around a reel of radius 0.250 m and mass 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through the center. Find the tension in the string and the acceleration of the object.?
An object with a weight of 50.0 N is attached to the free end of a light string wrapped around a reel of radius 0.250 m and mass 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through the center. Find the tension in the string and the acceleration of the object.?Explanation / Answer
Torque causes the pulley to rotate from its radius. The force is acting a distance to keep the pulley spinning and thus the object moving along. Torque is the moment of inertia x angular acceleration where I = (1/2)MR^2. M stands for the mass of the pulley. Solving for alpha, alpha = 2T/MR. This comes from setting the two torque equations together, (1) T = R*F and (2) T = (1/2)MR^2*alpha.
T = -ma + mg
= - m(Radius x Angular Acceleration) + mg
Substitute it back into alpha = 2T/MR and you get alpha = 2(-mR*alpha + mg)/MR and angular accleration is = (2mg)/(MR + 2mR)
1) Tension is T = -mR*alpha + mg
? = 2mg/(Mr + 2mr) = (2*5.1020)(9.81)/((3)(.25) + 2(5.1020)(.25)) = 30.2938 radians/s^2
T = 50 - (5.102)(.250)(30.2938) = 11.36 N
2) Acceleration is related to angular acceleration by
a = R*alpha
a = (.250)(30.2938) = 7.57 m/s^2
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