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

Click For Summary

Homework Help Overview

The problem involves a uniform solid cylinder connected to a mass via a pulley system, where the goal is to determine the maximum mass that can be attached without causing the cylinder to slip. The scenario includes parameters such as the mass of the cylinder, the coefficient of static friction, and the forces acting on the system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between tension, acceleration, and the forces acting on both the cylinder and the mass. There are attempts to apply equations related to torque and static friction, as well as considerations of the signs in the equations. Some participants suggest that the problem may be simpler than initially perceived.

Discussion Status

The discussion is active, with participants exploring various equations and relationships. Some guidance has been offered regarding the application of Newton's second law and the need to express mass in terms of known quantities. There is an ongoing examination of the conditions for maximum static friction and how they relate to the system's dynamics.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the mass of the cylinder and the coefficient of static friction. There is an emphasis on ensuring correct sign conventions in the equations being used.

ecksee
Messages
2
Reaction score
0

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:
 

Attachments

Physics news on Phys.org
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.)
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
5K
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
39
Views
4K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
3K