Torque with a pulley of radius R and mass M

[holmeskaei]

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.

 

[Doc Al]

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.

 

[nlightner]

Hello!

Torque is a measure of the force that causes an object to rotate. In this scenario, the pulley has a radius R and a mass M, which will affect the acceleration and tensions in the system.

To find the acceleration of m_1, we can use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. In this case, the net force acting on m_1 is the tension in the string, which is equal to the weight of m_2. So, the equation for the acceleration becomes a=T/m_1 = m_2g/m_1.

When considering the mass and radius of the pulley, we need to take into account the torque it produces. The torque net on the pulley is equal to the torque produced by the tension in the string, which is equal to the torque produced by the weight of m_1. So, the equation for the torque net becomes T1R-T2R = (-a/R)I, where I is the moment of inertia of the pulley. This equation can be rearranged to solve for a, which will give us the same result as in part A if we set m_p=0.

For part D, we need to consider the tension in the upper portion of the string. This can be found by setting up a free body diagram for the pulley and using the equation T1=m_1a+m_1g. Again, if we set m_p=0, we will get the same result as in part B.

For part E, we need to consider the tension in the lower portions of the string. This can be found by setting up a free body diagram for m_2 and using the equation T2=m_2a+m_2g. Again, if we set m_p=0, we will get the same result as in part B.

I hope this helps to clarify the role of the pulley's mass and radius in the system. Keep in mind that these equations assume a frictionless and massless pulley, so the actual values may differ slightly in real-world scenarios.