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Torque with a pulley of radius R and mass M

  • Thread starter holmeskaei
  • Start date
  • #1

Homework Statement



Blocks of mass m_1 and m_2 are connected by a massless string that passes over the pulley in the figure (Intro 1 figure) . The pulley turns on frictionless bearings, and mass m_1 slides on a horizontal, frictionless surface. Mass m_2 is released while the blocks are at rest.

http://session.masteringphysics.com/problemAsset/1073792/5/12.P71.jpg

I figured out A and B.
A. Assume the pulley is massless. Find the acceleration of m_1.
a=(m2g)/(m2+m1)
B. Find the tension in the string.
T=m1(m2g/m2+m1)
C. Suppose the pulley has mass m_p and radius R. Find the acceleration of m_1. Verify that your answers agree with part A if you set m_p=0.

D. Find the tension in the upper portion of the string. Verify that your answers agree with part B if you set m_p=0.

E.Find the tension in the lower portions of the string. Verify that your answers agree with part B if you set m_p=0.

Homework Equations


torque=Fxd
torque net=sum of all forces
T1=m1a+m1g
T2=m2a+m2g
torque net=T1r-T2r=(-a/r)I

The Attempt at a Solution


I know that the tensions change because of the mass and radius of the pulley, but I don't know what it does. I made a FBD, but I don't know what to do with my equations... I don't get it.
 

Answers and Replies

  • #2
Doc Al
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T1=m1a+m1g
The weight acts vertically; just consider horizontal forces on m1. (Vertical forces cancel.)
T2=m2a+m2g
Assuming "a" is a positive number, it looks like you have the signs wrong.
torque net=T1r-T2r=(-a/r)I
OK. But what's I for the pulley? (Maybe you can treat it like a uniform disk.)

Correct these equations and solve simultaneously.
 

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