Rotational Motion Problem (a bike)

In summary: So I think it's a bit more elegant this way.In summary, the angular speed of the rear wheel (Wr) of a bicycle is directly proportional to the number of teeth on the pedals and front sprocket (Nf) and inversely proportional to the number of teeth on the rear sprocket (Nr). The formula for Wr/Wf is Nf/Nr. When the front and rear sprockets have 52 and 13 teeth respectively, the ratio of Wr/Wf is 4:1 (52/13=4). The distance between teeth is not a factor in this formula.
  • #1
Carpe Mori
19
0

Homework Statement



How is the angular speed of the rear wheel (Wr) of a bicycle related to that of the pedals and front sprocket (Wf)? That is derive a formula for Wr/Wf. Let Nf and Nr be the number of teeth on the front and rear sprockets respectively. The teeth are spaced equally on all sprockets so that the chain meshes properly. Then evaluate the ratio of Wr/Wf when fron and rear sprockets have 52 and 13 teeth respectively


Homework Equations



W = v/r

Circumference (c) = 2pi*r (i am not sure if this is relevant)

The Attempt at a Solution



v for both front and back sprocket is the same (note W is not)

Wf = v/Rf

Wr = v/Rr

Since distances between teeth are same in back and front:

C = 2pi*Rr/Nr = 2pi*Rf/Nf

so...

Wr = v/(Nr*Rf/Nf*Rr)

Wf = v/(Nf*Rr/Nr*Rf)

Wr/Wf = Nf^2*Rr^2/Nr^2*Rf^2

but i do not think that is right because the question implies that your only variable should be N...

HELP! (if i seem unclear about anything...do tell)
 
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  • #2
Your equations for angular velocity look sound. I think your confusion starts when you substitute: C = 2pi*Rr/Nr. That to me doesn't make much sense. I would agree that the circumference C = kNr, where k is the spacing or pitch (some constant) of the teeth. And C is also equal to C=2pi*Rr. Which would lead to Rr being expressed in terms of Nr.

Give that a try.
 
  • #3
mezarashi said:
Your equations for angular velocity look sound. I think your confusion starts when you substitute: C = 2pi*Rr/Nr. That to me doesn't make much sense.

Nr/(2pi*Rr) is just the number of teeth per unit length, and so would be the same on both. He has just taken the reciprocal, which is constant C.

Carpe Mori said:
Wr = v/(Nr*Rf/Nf*Rr)

Wf = v/(Nf*Rr/Nr*Rf)

Wr/Wf = Nf^2*Rr^2/Nr^2*Rf^2
What is all this? Number of teeth on each is proportional to the circumference and hence the radius. Find Wr/Wf as some ratio of Rr and Rf and then put Nr and Nf in the ratio.
 
  • #4
well i took my C = 2pi*Rr/Nr = 2pi*Rf/Nf and solved for both Rr and Rf and then plugged those values into my initial angular velocity equations.

and i resulted with

Wr/Wf = Nf^2*Rr^2/Nr^2*Rf^2

is this correct yes or no?? it means angular velocity depends on both Radius and number of teeth of both sprockets...which makes sense to me
 
  • #5
Carpe Mori said:
well i took my C = 2pi*Rr/Nr = 2pi*Rf/Nf and solved for both Rr and Rf and then plugged those values into my initial angular velocity equations.

and i resulted with

Wr/Wf = Nf^2*Rr^2/Nr^2*Rf^2

is this correct yes or no?? it means angular velocity depends on both Radius and number of teeth of both sprockets...which makes sense to me

I don't know. You check it once more. But I'll show you the easy and natural way.

v = Rf*Wf = Rr*Wr =>
Wf/Wr = Rr/Rf = 2pi*Rr/2pi*Rf = C*Nr/C*Nf = Nr/Nf =>
Wr/Wf = Nf/Nr.

There was no need to introduce the C, since it's obvious that the number of teeth is directly proportional to the radius.
 

Related to Rotational Motion Problem (a bike)

1. How does the angular velocity of a bike wheel affect its motion?

The angular velocity of a bike wheel is directly related to its linear velocity, or speed. The faster the wheel is rotating, the faster the bike will be moving forward. This is because the linear velocity is equal to the product of the angular velocity and the radius of the wheel.

2. How does the mass distribution of a bike affect its rotational motion?

The mass distribution of a bike affects its rotational motion by changing its moment of inertia. A bike with a more concentrated mass will have a smaller moment of inertia and will rotate more easily, while a bike with a more spread out mass will have a larger moment of inertia and will be more resistant to rotational motion.

3. How does friction affect the rotational motion of a bike?

Friction can have a significant impact on the rotational motion of a bike. When there is friction between the tires and the ground, it can cause the bike to slow down and eventually come to a stop. Friction can also affect the stability of the bike, making it more difficult to maintain balance while riding.

4. What is the role of the center of mass in the rotational motion of a bike?

The center of mass of a bike is the point where its mass is evenly distributed. It plays a crucial role in rotational motion by determining the bike's stability and the direction in which it will rotate. The lower the center of mass, the more stable the bike will be and the easier it will be to maintain balance.

5. How does the force applied to the pedals affect the rotational motion of a bike?

The force applied to the pedals of a bike affects its rotational motion by increasing or decreasing its angular velocity. When more force is applied, the bike will accelerate and its wheels will rotate faster. On the other hand, if less force is applied, the bike will decelerate and its wheels will rotate slower.

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