Rotational Dynamics of a Bicycle wheel

In summary, a bicycle wheel with a diameter of 63.8 cm and a mass of 1.79 kg is subjected to a resistive force of 117 N applied tangent to the rim of the tire. The task is to calculate the force needed to give the wheel an acceleration of 4.53 rad/s^2 when using a 9.01-cm-diameter sprocket and then when using a 5.60-cm-diameter sprocket. Using equations for torque and moment of inertia, the force needed for part (a) is 847 N, while the force needed for part (b) is 1473 N.
  • #1
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Homework Statement



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 9.01-cm-diameter sprocket in order to give the wheel an acceleration of 4.53 rad/s2?

(b) What force is required if you shift to a 5.60-cm-diameter sprocket?

R=wheel radius
R= .319m
m= wheel mass
m=1.79kg
F=Resistive force tangent to tire
F=117N
r=radius of sprocket
r=.0901m
α=angular acceleration of the wheel
α=4.53rad/s^2
τf= Torque from the bike wheel
τext=external torque for sprocket


Homework Equations


Torque
τ=rF
Moment of Interia of a hoop
I=MR^2
Torque
τ=Iα

The Attempt at a Solution


First calculate torque on the bike wheel

τf=.319m(117N)= 37.3Nm

Due to laws of rotational dynamics and why the wheel is spinning relate the torque of the wheel to torque of sprocket to moment of interia of the wheel

τext-τf=Iα

Break down each term into quantities we already know

τext=rF
τf=37.3Nm
I=MR^2
α=4.53rad/s^2

Thusly
rF-37.3Nm=MR^2(4.53Rad/s^2)

Rearrange and solve for F

F=MR^2(α)+τf/r

Input values
F=(1.79kg)(.319)^2(4.53rad/s^2)+37.3Nm/.0901m

F=423N

Maybe I messed up somewhere but I've tried twice already submitting the answer online and this is wrong but that was my shot, thank you for helping.
 
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  • #2
I forgot to mention that to solve part b you would use the same equation just replace r with .0560m
 
  • #3
You used the diameter of the sprocket where you needed the radius. Using your equations, I'm getting for part a:
F = 847 N
 
  • #4
Oops, should have checked the date before replying.
 
  • #5




Hello there, it seems like you have a good understanding of rotational dynamics and the equations involved. However, there are a few errors in your calculations and approach. Let's go through each part and correct them.

(a) First, let's calculate the moment of inertia of the wheel. Since it is a hoop with all the mass concentrated on the outer radius, the moment of inertia can be calculated as I = MR^2 = (1.79 kg)(0.319 m)^2 = 0.182 kgm^2. This will be the same for both sprockets, so we can use this value for both parts.

Next, let's calculate the torque from the resistive force applied tangent to the tire. This can be done as τ = rF = (0.319 m)(117 N) = 37.3 Nm. This is the same for both sprockets as well.

Now, let's set up the equation τext - τf = Iα and input the values for the first part.

τext - 37.3 Nm = (0.182 kgm^2)(4.53 rad/s^2)

Solving for τext, we get τext = 37.3 Nm + (0.182 kgm^2)(4.53 rad/s^2) = 38.8 Nm. This is the torque required from the chain passing over the 9.01-cm sprocket.

To find the force required, we can use the equation F = τext/r. Plugging in the values, we get F = (38.8 Nm)/(0.0901 m) = 431 N. This is the force required from the chain passing over the 9.01-cm sprocket to give the wheel an acceleration of 4.53 rad/s^2.

(b) For the second part, the only difference is the radius of the sprocket. So, we can use the same values for the moment of inertia and torque from the resistive force. We just need to plug in the new radius of 5.60 cm = 0.056 m.

τext - 37.3 Nm = (0.182 kgm^2)(4.53 rad/s^2)

Solving for τext, we get τext = 37.3 Nm + (0.182 kg
 

Related to Rotational Dynamics of a Bicycle wheel

1. What is rotational inertia and how does it relate to a bicycle wheel?

Rotational inertia is the measure of an object's resistance to changes in its rotational motion. In the case of a bicycle wheel, it is the tendency of the wheel to resist changes in its spinning motion. The larger the wheel's rotational inertia, the harder it is to change its rotational speed or direction.

2. How does the mass distribution of a bicycle wheel affect its rotational dynamics?

The mass distribution of a bicycle wheel plays a crucial role in its rotational dynamics. A wheel with most of its mass concentrated towards the outer edge will have a larger rotational inertia compared to a wheel with the same mass distributed evenly. This means that it will be harder to change the rotational motion of the wheel with more mass towards the outer edge.

3. Can the rotational dynamics of a bicycle wheel be affected by the friction between the tire and the ground?

Yes, the friction between the tire and the ground can affect the rotational dynamics of a bicycle wheel. When the wheel is rolling, the friction between the tire and the ground creates a torque that can either increase or decrease the wheel's rotational speed. This is why it is important to have properly inflated tires and a good grip on the ground for optimal performance.

4. How does the angular momentum of a bicycle wheel change when turning?

When a bicycle wheel is turning, its angular momentum remains constant unless an external torque is applied. This means that as the wheel turns and changes direction, its angular velocity may change, but its angular momentum will remain the same.

5. Can the rotational dynamics of a bicycle wheel be affected by external forces?

Yes, external forces such as air resistance, bumps on the road, and the rider's movements can all affect the rotational dynamics of a bicycle wheel. These forces can create torques that can change the wheel's rotational speed and direction, making it important for riders to maintain balance and control while riding.

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