Max Mass for Cylinder to Roll Without Slipping | Cylinder and Mass System

AI Thread Summary
The discussion focuses on determining the maximum mass m2 that can be attached to a solid cylinder without causing it to slip, given its mass and the coefficient of static friction. Key equations include torque, moment of inertia, and Newton's second law, which are applied to analyze the forces and accelerations involved. The participants emphasize the importance of using the condition for maximum static friction to find the maximum acceleration. A suggestion is made to derive m2 in terms of known quantities rather than tension. The conversation highlights the need for careful attention to signs in the equations to arrive at the correct solution.
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Homework Statement



The axle of a uniform solid cylinder of mass m1=7.1 Kg is attached to a string that runs over a massless, frictionless pulley. The other end of the string is attached to a mass m2. The coefficient of static friction between the cylinder and the table is U=0.202. What is the max mass m2 (in Kg) for which the cylinder will roll without slipping?

Homework Equations



Torque=I*alpha
Icylinder=1/2*mass*radius^2
F=ma
a=alpha*radius

The Attempt at a Solution


m2
Tension-m2*g=m2*a

a=(Tension-m2*g)/m2

m1
Torque=U*m1*g*radius=1/2*m1*radius^2*(a/radius)

a=2*U*g

System

2*U*g=(Tension-m2*g)/m2
See diagram:
 

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I think you're making this more difficult than it needs to be! The question, worded in another way, is asking for the maximum force that can be exerted on the cylinder so that it rolls and does not slip. [hint: limiting friction]
 
ecksee said:
m2
Tension-m2*g=m2*a

a=(Tension-m2*g)/m2
Be careful with your signs: The tension is up, but the acceleration is down.
m1
Torque=U*m1*g*radius=1/2*m1*radius^2*(a/radius)

a=2*U*g
Good. You've applied the condition for maximum static friction to find the maximum acceleration. Now use this to find the corresponding value of m2 that will produce such an acceleration.

You need one more equation: Apply Newton's 2nd law to the translation of m1. Your goal is find m2 in terms of known quantities (not in terms of tension).
 
Doc Al said:
You need one more equation: Apply Newton's 2nd law to the translation of m1. Your goal is find m2 in terms of known quantities (not in terms of tension).


I get this equation to be F=ma=T=m1*a

This doesn't really get me anywhere. Am I missing something?
 
ecksee said:
I get this equation to be F=ma=T=m1*a
Right. Now put it to work.

You found the acceleration using your second equation. Combine that result with your first equation and this one to solve for m2. (You'll need to correct the sign error in your first equation.)
 
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