Does Applying Excess Torque to a Wheel Cause It to Slip?

  • Thread starter Thread starter koolraj09
  • Start date Start date
  • Tags Tags
    Rolling Torque
AI Thread Summary
Excess torque applied to a wheel can lead to slipping if the available friction force is insufficient to match the required force for linear acceleration. When torque exceeds the frictional force, the wheel will not only accelerate but may also lose grip, resulting in slip. The relationship between torque, angular acceleration, and linear acceleration is crucial in determining this behavior. For a solid uniform cylinder, the total inertia is significantly higher than that of a non-rolling cylinder, affecting the angular acceleration. Understanding these dynamics is essential for predicting whether a wheel will slip under excessive torque.
koolraj09
Messages
166
Reaction score
5
Hi guys.
I had a l'll confusion regarding rolling. Suppose you have a wheel and we set it into motion. If we give it more torque than required to just start rolling it'll accelerate..right? But will it slip greatly if we supply more torque?
Thanks in Advance.
 
Physics news on Phys.org
hi koolraj09! :smile:

use torque to find the angular acceleration, from that find the linear acceleration if there's no slipping, then use F = ma to find the applied force needed to produce that linear acceleration …

the only (horizontal) applied force is the available friction force, so if that's less than F, the wheel will slip :wink:
 
Torque is opposed by both angular and linear inertia, assuming no slippage, the rate of acceleration is affected by the total inertia, both angular and linear.

If all the mass is at the center than you only have linear factors:

F = m a = T / r
a = T / (m r)
α = a / r = T / (m r2) = T / Il
Il = m r2

If the mass is not at the center, then total inertial equals the sum of angular and linear related components:

I = Ia + Il
c = constant for angular component
Ia = c m r2
I = (c + 1)m r2

For a solid uniform cylinder, c = 1/2
I = Ia + Il = 1/2 m r2 + m r2
Total inertial is triple that of a non-rolling cylinder (zero friction)

α = T / (3/2 m r2) = 2/3 T / (m r2) = 1/3 T / (1/2 m r2)
Angular acceleration is 1/3 of a non-rolling cylinder (zero friction)
 
Last edited:
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top