Calculate Torque on Unicycle for Homework Statement

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Homework Help Overview

The problem involves calculating the maximum torque generated by a woman riding a unicycle, given her mass, the mass of the unicycle, the diameter of the wheel, and the length of the crank. The context is centered around the application of torque in a mechanical system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the woman's weight and the torque generated on the unicycle pedal, questioning the direction of the force and the angle involved in the torque calculation. There is a focus on understanding the cross product and the implications of the angle between the force and the radius.

Discussion Status

Some participants have provided clarifications regarding the angle of force application and the relationship between the crank length and the force exerted on the pedal. There is an ongoing exploration of the concepts involved, with no explicit consensus reached yet.

Contextual Notes

Participants are considering the implications of the force direction and the geometry of the system, including the potential confusion regarding the angle in the torque equation. The discussion reflects a need for visual aids to enhance understanding.

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Homework Statement


A woman is riding a unicycle. The mass of the woman is: m(w)=66.0 kg,
the mass of the unicycle is: m(u)=5.40 kg, the diameter of the unicycle
wheel is: d = 55.0 cm, and the length of the unicycle crank is L =
11.0 cm

What is the maximum torque that the woman is able to
generate just by using her weight to push on the unicycle
pedal?


Homework Equations


τ = r x F

The Attempt at a Solution


So, the maximum force that she can exert on the unicycle is her weight i.e. her mass times g. To find the torque all you have to do is multiply the length of the crank by the weight. I'm just wondering though why this is correct. I mean her weight acts vertically downwards, directly above the centre of the unicycle wheel. By the cross product formula r x F = |r||F|Sin(θ). But isn't theta in this case 0? Implying that the torque is 0, which can't be correct. I may just be visualizing this incorrectly though. If someone could explain this it'd be great. I'll attach a diagram of what I'm visualizing. Thanks.
 
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ImageUploadedByPhysics Forums1366653977.001940.jpg
 
Am I visualising this incorrectly?
 
Your diagram did not show the force acting on the pedal but other than that I think you are correct. I think the angle is between th radius and the force and this would be 90 deg and sin 90 = 1 so T = F X r
 
Yeah, I was thinking that initially but our lecturer seemed pretty sure that it was the length of the crank times the applied force. This makes I sense to me as the rider is applying the force to the pedals and not the crank itself.
 
It is the length of the crank times the force on the pedal. There is nothing magic about the pedal other than a way to get the force from the foot to the end of the crank.
 
Ohhh, ok. I see. Thanks man.
 

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