Physics: Angular Acceleration of Rotational Objects

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A bicycle wheel has a diameter of 63.8 cm and a mass of 1.79 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 117 N is applied tangent to the rim of the tire.

(a) What force must be applied by a chain passing over a 8.92-cm-diameter sprocket in order to give the wheel an acceleration of 4.49 rad/s2? (b) What force is required if you shift to a 5.65-cm-diameter sprocket? 

Jul 6th, 2015

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Torque = Inertia x alpha (angular acceleration rad/s^2) 
Torque also equals force x radius 

First of all lets find the inertia of the wheel. 

hoop = mr^2 = 1.79 x 0.319^2 = 0.18215kg-m^2 

Now, Torque = 0.18215 x 4.49= 0.8178/radius= force

0.8178 x 0.0446 =36.47N

What force must be applied to attain acc of 4.49 rad/s^2? 

17.33 N 

117 N x 0.316 m = 36.972 Nm 

36.972/ 0.0446 = 828.96 N 

Total = 17.33 + 828.96= 846.29 N (answer) 

What force is required if the chain shifts to a 5.65 cm diameter sprocket? 

the torque is the same and divide by the radius

The answer comes out to be 1390.9 N

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 6th, 2015

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