Rolling Motion Question

  • Thread starter ecksee
  • Start date
  • #1
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: 159

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,122
74
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
Doc Al
Mentor
45,180
1,506
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
2
0
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
Doc Al
Mentor
45,180
1,506
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 Threads on Rolling Motion Question

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
7K
  • Last Post
Replies
6
Views
4K
Top