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

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SUMMARY

Applying excess torque to a wheel can lead to slippage if the frictional force is insufficient to counteract the applied force. The relationship between torque, angular acceleration, and linear acceleration is governed by the equations F = ma and a = T / (m r). For a solid uniform cylinder, the total inertia is calculated as I = (1/2 m r² + m r²), resulting in an angular acceleration of α = 1/3 T / (1/2 m r²) when excess torque is applied. Thus, understanding the balance between torque and friction is crucial to prevent slippage.

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koolraj09
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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.
 
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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)
 
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