Wheel A with a mass of 5 kg and radius of 10 cm is connected to blade B with a m

Question

Wheel A with a mass of 5 kg and radius of 10 cm is connected to blade B with a mass of 0.6 kg and length L = 50 cm. A constant force F = 12 N is applied by a rope wrapped around the wheel. Treat the wheel as a uniform disk rotating about its center, with moment of inertia formula (1/2)MR^2, while the blade is a long thin rod rotating about its own center, with moment of inertia formula (1/12)ML^2.
a) What is the moment of inertia of the system A+B? Be careful of units!
b) What is the magnitude of the torque being applied to the system about the axis of rotation?
c) What is the magnitude of the angular acceleration experienced by the system?
d)How many times around will the system rotate in a time of t = 2s, if the rotation starts from rest? Be careful of units!

Explanation / Answer

a.Moment of Inertia of A = Mr^2/2 = 5 x 0.1^2/2= 0.025 kg.m^2

Moment of inertia of B =ML^2/12 = 0.6 x 0.5^2/12= 0.0125 kg.m^2

Total moment of inertia = 0.025 +0.0125=0.0375 kg.m^2

b. Torque = force x radius = 12 x 0.1= 1.2 N.m

c. Angular acceleration = torque/MI = 1.2/0.0375= 32 rad/s^2

d. angle covered = 1/2 x angular acceleration x time^2 = 1/2 x 32 x 2^2 = 64 radians= 10.18 revolutions

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