Rolling resistance force for a 75gram vehicle

In summary, the homework statement is asking for a calculation of the frictional torque caused by the axle and wheel friction.
  • #1
Isti0503
6
1

Homework Statement


Estimate the rolling resistance force for a 75gram vehicle with axle friction coefficient of 0.2 if the wheel is 40mm diameter and the axle 2mm diameter.

I am having trouble with applying the axle friction coefficient. I could not find a similar problem online.
Basically we know that the smaller the radius of the circle the more tangential force is needed to produce the same torque (axle and the wheel). However, after I am not sure what to do with the axle friction coefficient.
 
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  • #2
What is the normal force at the axle?
What is the frictional force at the axle?
What is the frictional torque that creates?
What force at the rim will overcome that torque?
 
  • #3
Ok so the normal force would be just mg, the frictional torque that creates would be the tangential force at the axle times the radius of the axle, the force that will overcome that torque is basically the radius of the wheel times the tangential force to the rim right?
What do you mean by the frictional force at the axle? Would that just be the coefficient of axle friction given (0.2)?
 
  • #4
Isti0503 said:
What do you mean by the frictional force at the axle? Would that just be the coefficient of axle friction given (0.2)?
The coefficient of friction is a pure number. A ratio. It does not have the right units to be a force.

You've said that the normal force at the axle is mg. If the coefficient of friction at the axle is 0.2, what is the force of friction at the axle?
 
  • #5
Well the force of friction will be the coefficient times mg. Right?
 
  • #6
Isti0503 said:
Well the force of friction will be the coefficient times mg. Right?
Right.
 
  • #7
Alright but then out off those two forces (at the rim, and the frictional force), which force am I looking for?
 
  • #8
Isti0503 said:
Alright but then out off those two forces (at the rim, and the frictional force), which force am I looking for?
You tell us. It is your homework problem.

You claimed to be looking to determine frictional torque. You have all the information needed to calculate it.
 
  • #9
So if I use those formulas I will get:
F=Uk*mg*(r1/r2).

So r1 is the radius of the axle and r2 radius of the wheel.
 
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  • #10
Isti0503 said:
So if I use those formulas I will get:
F=Uk*mg*(r1/r2).

So r1 is the radius of the axle and r2 radius of the wheel.
Yes.
 
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  • #11
Alright thank you.
 

1. What is rolling resistance force?

Rolling resistance force is the force that opposes the motion of a vehicle as it rolls on a surface. It is caused by the deformation of the tires and the friction between the tires and the surface.

2. How is rolling resistance force calculated?

The rolling resistance force for a 75gram vehicle can be calculated using the formula F = c * m * g, where F is the rolling resistance force, c is the rolling resistance coefficient, m is the mass of the vehicle, and g is the acceleration due to gravity.

3. What factors affect rolling resistance force?

The factors that affect rolling resistance force include the type and condition of the tires, the weight and shape of the vehicle, the type of surface the vehicle is rolling on, and the speed and acceleration of the vehicle.

4. How can rolling resistance force be reduced?

To reduce rolling resistance force, one can use tires with lower rolling resistance coefficients, maintain proper tire pressure, reduce the weight and streamline the shape of the vehicle, and choose smooth and flat surfaces to drive on.

5. Why is it important to consider rolling resistance force for a 75gram vehicle?

Rolling resistance force is important to consider for a 75gram vehicle because it can affect the overall performance and efficiency of the vehicle. A higher rolling resistance force will require more energy to overcome, resulting in lower speed and poorer fuel efficiency. By understanding and minimizing rolling resistance force, the vehicle's performance and energy efficiency can be improved.

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