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

In summary, the question asks for the maximum mass m2 that can be attached to a string connected to a solid cylinder without causing it to slip, given the cylinder's mass m1, coefficient of static friction U, and a pulley system. Using torque and Newton's 2nd law, the maximum acceleration is found, and by combining this with the equation for tension and acceleration, the maximum mass m2 can be determined.
  • #1
ecksee
2
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

  • Doc1.doc
    26.5 KB · Views: 235
Physics news on Phys.org
  • #2
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]
 
  • #3
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).
 
  • #4
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?
 
  • #5
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.)
 

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

What is rolling motion?

Rolling motion is a type of motion in which an object moves along a surface while also rotating around its own axis.

What are examples of objects that exhibit rolling motion?

Some common examples of objects that exhibit rolling motion include balls, wheels, and cylinders.

How is rolling motion different from sliding motion?

The main difference between rolling motion and sliding motion is that in rolling motion, the object rotates around its own axis while moving, whereas in sliding motion, the object does not rotate and simply moves along a surface.

What factors affect the speed of rolling motion?

The speed of rolling motion is affected by the size and shape of the object, the surface it is rolling on, and any external forces acting on the object.

What is the principle of conservation of angular momentum?

The principle of conservation of angular momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. This means that in the absence of external forces, the angular momentum of a rolling object will remain constant.

Similar threads

  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
977
Replies
39
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
3K
Back
Top