# Physics

I am having trouble with the last portion of this question. If someone can help me, I would sincerely appreciate it.

A bicycle wheel has a radius R = 32.0 cm and a mass M = 1.82 kg which you may assume to be concentrated on the outside radius. A resistive force f = 147 N (due to the ground) is applied to the rim of the tire. A force F is applied to the sprocket at radius r such that the wheel has an angular acceleration of a = 4.50 rad/s^2. The tire does not slip.

a) If the sprocket radius is 4.53 cm, what is the force F (N)?

(mr^2)(4.5) = (r)(F)- (147)(.32)
(1.82)(.32^2)(4.5) = (.0453)(F)-(147)(.32)
= 1056.9

b) If the sprocket radius is 2.88 cm, what is the force F(N)?

(1.82)(.32^2)(4.5) = (.0288)(F)-(147)(.32) = 1662.45

c) What is the combined mass of the bicycle and rider (kg)?
I am not sure how to do this last part

The correct answer is 102 kg
No one has answered this question yet.

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asked by Matt
1. Ok, the last part, you have a force on the sprocket producing a net force and acceleration.
assume the force is transmitted as a lever to the ground...

F(sprocketradius/wheel radius)-147*.32= mass*acceleration

but acceleration= 4.5*wheel radius

F(sprocketradius)-147*wheelradius=mass*
4.5*wheel radius

or mass= (F*sproketradius/wheelradius-147)/4.5 = mass
mass= (1662*.0288/.32 -147)/4.5=99.7kg

check all calculations

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2. F = ma for the whole bike as a unit. The force F that is causing the bike to move is your resistive force 147N. a = R*alpha, so you should have m=F/R*alpha so m = 147/(.32*4.5) = 102.083 which is the answer given by OP as being correct.

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posted by Jordan

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