Find Frictional Torque for Poorly Maintained Bicycle

In summary: You need to multiply the 0.30 by 9.8 m/s^2, but for units to cancel properly, you need to do it as (0.30 kg*m^2)(9.8 m/s^2)/(m/s^2) = 2.94 N*m.[Off to class now. Good luck!]In summary, the magnitude of the frictional torque on a poorly maintained bicycle wheel that is spun at 0.72 revolutions per second and stops in 12 seconds is 2.94 N*m. This is calculated using the equation ∑τ=Iα, where α is found to be 0.38 rad/s^2 and the moment of inertia I is 0.
  • #1
kenway
13
1
I honestly don't know how to find the friction, much less frictional torque, when given things like this.

1. Homework Statement

"If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72rev/s, friction in the bearings causes the wheel to stop
in just 12s.
If the moment of inertia of the wheel about its axle is 0.30kg\m2 , what is the magnitude of the frictional torque?"

2. Homework Equations

I know:
Στ=Iα
α=Δω/Δt
I=mr2

The Attempt at a Solution


Like I said, I have no idea how I am supposed to use the given variables to find friction.

This is an attempt I tried:
α=Δω/Δt
=0.72/12
=0.06m/s2

So I can now find torque with the angular acceleration:
∑τ=Iα
= (0.30kg/m2)(0.06m/s2)
= 0.018Nm

That's a small number that I'm pretty sure isn't right. Even if it is, I don't know where to go from here to find friction. If someone could just guide me/give me equations then I'd appreciate it

thanks!
 
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  • #2
kenway said:
This is an attempt I tried:
α=Δω/Δt
=0.72/12
=0.06m/s2

So I can now find torque with the angular acceleration:
∑τ=Iα
= (0.30kg/m2)(0.06m/s2)
= 0.018Nm
Couple things wrong there. You are given the angular speed in revolutions per second, not in radians per second (ω). And the units of α are 1/s^2, not m/s^2.

You are correct to be using this equation to solve this problem: [tex]\tau = I \alpha[/tex]

Once the wheel is set spinning, the only torque acting on it is from bearing friction.
 
  • #3
kenway said:
This is an attempt I tried:
α=Δω/Δt
=0.72/12
=0.06m/s2
If you put in the units that go with the 0.72 and the 12, do you get m/s2? Should you get m/s2 for α?

Also, is Δω positive or negative for this problem?

[Sorry, I posted just after berkeman. He's got you covered!]
 
  • #4
berkeman said:
Couple things wrong there. You are given the angular speed in revolutions per second, not in radians per second (ω). And the units of α are 1/s^2, not m/s^2.

You are correct to be using this equation to solve this problem: [tex]\tau = I \alpha[/tex]

Once the wheel is set spinning, the only torque acting on it is from bearing friction.

So .72rev/s = 1.44π(rad)/s or 4.52rad/s

So then I put that into my equations:

α=Δω/Δt
=4.52rad/sec / 12s
=0.38rad/sec2

∑τ=Iα
= (0.30kg/m2)(0.38rad/s2)
= 0.79Nm

And it doesn't state which direction the wheel was spun, so I am assuming the negative/positive is irrelevant
 
  • #5
kenway said:
So .72rev/s = 1.44π(rad)/s or 4.52rad/s

So then I put that into my equations:

α=Δω/Δt
=4.52rad/sec / 12s
=0.38rad/sec2

∑τ=Iα
= (0.30kg/m2)(0.38rad/s2)
= 0.79Nm

And it doesn't state which direction the wheel was spun, so I am assuming the negative/positive is irrelevant
Closer, but it looks like you copied the units of the moment of inertia I incorrectly in your original problem statement, and that carried through the calculations. The units of I are not kg/m^2...
 
  • #6
berkeman said:
Closer, but it looks like you copied the units of the moment of inertia I incorrectly in your original problem statement, and that carried through the calculations. The units of I are not kg/m^2...

Sorry it's kg*m2. But how does that affect my equation? Torque is measured in N*m... Do I have to multiply 0.30 by 9.8m/s2 because of gravity to put it in Newtons?
 
Last edited:
  • #7
kenway said:
Sorry it's kg*m2. But how does that affect my equation? Torque is measured in N*m... Do I have to multiply 0.30 by 9.8m/s2 because of gravity to put it in Newtons?

[I have class tonight so I may not respond to your next reply until later tonight ~3 hours]
It's just that the units of your answer were wrong, so that led me back to where the units were wrong in your OP.

All else looks good. Always check your units for consistency at each step. It can help to find other mathematical errors.
 
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Likes kenway

Related to Find Frictional Torque for Poorly Maintained Bicycle

1. What is frictional torque?

Frictional torque is the force that resists the rotation of an object due to the contact between its moving parts.

2. How does frictional torque affect a poorly maintained bicycle?

A poorly maintained bicycle may have parts that are not properly lubricated or are worn out, resulting in increased frictional torque. This can make it harder to pedal and decrease the efficiency of the bicycle.

3. How can I measure the frictional torque of my bicycle?

The frictional torque of a bicycle can be measured by attaching a torque sensor to the crank or pedal and measuring the force needed to rotate the pedals at a constant speed.

4. What factors can increase frictional torque in a bicycle?

Factors that can increase frictional torque in a bicycle include lack of lubrication, misaligned or worn out parts, and debris or dirt buildup in the moving parts.

5. How can I reduce frictional torque in my bicycle?

To reduce frictional torque in a bicycle, make sure to regularly maintain and lubricate all moving parts, replace any worn out parts, and keep the bicycle clean from debris and dirt. Using high-quality, well-maintained components can also help reduce frictional torque.

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