Acceleration of a mass hanging from a pulley

In summary, The two masses are attached to a massless string by a tension in the string and the string does not slip on the pulley. The downward acceleration of mass 2 is given by the tension in the string and the rotational moment of inertia of the pulley.
  • #1
EzequielSeattle
25
0

Homework Statement


Two blocks, of masses m1 and m2 (with m1<m2) are attached by a massless string of fixed length that runs over a pulley which is free to rotate about its fixed center. The pulley has mass m3, radius r3, and rotational moment of inertia I = (2/3) m3 r3^2 about its center (it is a non-uniform disk). The string does not slip on the pulley. Give the downward acceleration of m2 in terms of the quantities above.

Homework Equations


F=ma
α=aR
τ=Iα

The Attempt at a Solution


I drew FBD's for each object. The forces acting on m2 are going to be m1g in the upward direction and m2g in the downward direction. I'm just not sure how to incorporate the pulley in here. The torque exerted on the pulley is Iα, so (2/3)m3r3^2*α. It will also be r x F, so r3 x (m2g-m1g). I'm just not sure where to go from here...
 
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  • #2
EzequielSeattle said:
The forces acting on m2 are going to be m1g in the upward direction
Not so.
Create unknowns for the tensions, bearing in mind that the string tensions may be different on the two sides of the pulley.
 
  • #3
OK, I've tried this and here's what I've got (T_2 is the tension in the string connecting m_2, T_1 is the tension in the string connecting m_1)

τ=T_2 * r - T_1 * r
a = αr

a will be the same as the acceleration of mass 2, because mass 2 is on a string connected to the pulley.

Iω = τ

ω = τ/I

ω = gr(m_2 - m_1) / I

Am I correct so far? So to find a, should I find the derivative of ω and then multiply by the radius, because a = αr?
 
  • #4
EzequielSeattle said:
Iω = τ
ω?
 
  • #5
Ok, yeah, I realize that that's not the equation and it ought to be α.

I decided to try a different approach. The torque on the pulley is equal to rg(m_2 - m_1).

The torque is also equal to Iα, or I(a/r)

Combining my two equations for the torque, I get

rg(m_2 - m_1) = (2/3)(m_pulley)r^2 * (a/r)

Solving for a gives

a = ((3/2)g (m_2 - m_1))/(m_pulley)

I'm pretty sure this is correct, but I could be wrong.
 
  • #6
EzequielSeattle said:
Ok, yeah, I realize that that's not the equation and it ought to be α.

I decided to try a different approach. The torque on the pulley is equal to rg(m_2 - m_1).

The torque is also equal to Iα, or I(a/r)

Combining my two equations for the torque, I get

rg(m_2 - m_1) = (2/3)(m_pulley)r^2 * (a/r)

Solving for a gives

a = ((3/2)g (m_2 - m_1))/(m_pulley)

I'm pretty sure this is correct, but I could be wrong.
No, you are neglecting that m1 and m2 are also accelerating.
Write the free body equations for the two hanging masses.
 

1. What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. How does a mass hanging from a pulley experience acceleration?

When a mass is hanging from a pulley, it experiences acceleration due to the force of gravity acting on it. This acceleration is known as the acceleration due to gravity, and it is constant at 9.8 meters per second squared on Earth.

4. What factors affect the acceleration of a mass hanging from a pulley?

The acceleration of a mass hanging from a pulley is affected by the mass of the object, the force of gravity, and any external forces acting on the object. The angle of the pulley and the friction between the pulley and the string can also affect the acceleration.

5. How is acceleration related to the tension in the string holding the mass?

According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it. In the case of a mass hanging from a pulley, the tension in the string provides the net force, so the acceleration is directly proportional to the tension in the string. This means that as the tension increases, the acceleration increases as well.

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