Acceleration of Falling Mass Attached to 5cm Pulley

  • Thread starter hannibal
  • Start date
  • Tags
    Pulleys
In summary, the tension/force method works if there is no moment of inertia, however my phys prof had to skip over the last chapter in the book due to time constraints.
  • #1
hannibal
2
0
I have a question where there's two masses attached to a pully that has a radius of 5 cm. one mass is heavier than the other one and the first part of the question is asking you to find the acceleration of the mass that's falling (it falls 75 cm in 5 sec). Now the way that i figured it should be done is by using the tension formulas (Mg - T = Ma for the mass falling) however this give the wrong answer, the equation you have to use is Y = Yo +(1/2)at^2. I can;t seem to figure out why you can't use the first method to figure this one out.
 
Physics news on Phys.org
  • #2
hannibal said:
I have a question where there's two masses attached to a pully that has a radius of 5 cm. one mass is heavier than the other one and the first part of the question is asking you to find the acceleration of the mass that's falling (it falls 75 cm in 5 sec). Now the way that i figured it should be done is by using the tension formulas (Mg - T = Ma for the mass falling) however this give the wrong answer, the equation you have to use is Y = Yo +(1/2)at^2. I can;t seem to figure out why you can't use the first method to figure this one out.

To use Newton's second law for such a problem you would have to consider the moment of inertia of the pulley. However, since the acceleration is uniform and you are given the time interval and the displacement, you can just use plain old kinematics [itex]\Delta s = at^2/2 [/itex] and solve for a.
 
  • #3
Ok I know what the moment of inertia is, however my phys prof had to skip over the last chapter in the book due to time constraints. If there was no moment of inertia then the tension/force method would work right?
 

1. What is the formula for calculating the acceleration of a falling mass attached to a 5cm pulley?

The formula for calculating acceleration in this scenario is a = (m1 - m2)g / (m1 + m2), where m1 is the mass of the falling object, m2 is the mass of the object attached to the pulley, and g is the acceleration due to gravity (9.8 m/s²).

2. How does the mass of the object attached to the pulley affect the acceleration of the falling mass?

The mass of the object attached to the pulley has a direct impact on the acceleration of the falling mass. The greater the mass of the object attached to the pulley, the slower the acceleration of the falling mass will be. This is because the heavier object attached to the pulley will have a greater force pulling it down, causing a smaller difference in weight between the two objects and resulting in a smaller acceleration.

3. Does the distance of the pulley affect the acceleration of the falling mass?

No, the distance of the pulley does not have an effect on the acceleration of the falling mass. The only factor that affects the acceleration is the mass of the object attached to the pulley.

4. What is the significance of the 5cm pulley in this scenario?

The 5cm pulley serves as a point of attachment for the two masses in this experiment. It allows for the masses to move freely and for the tension in the string to remain constant, making it easier to measure and calculate the acceleration of the falling mass.

5. Can this experiment be used to study the effects of air resistance on falling objects?

No, this experiment does not account for the effects of air resistance on falling objects. To study air resistance, the setup would need to include a way to measure and account for the resistance, such as using a more controlled environment like a vacuum chamber.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
784
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
350
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
962
Back
Top