How Long Does It Take for m1 to Reach the Floor in an Atwood Machine?

  • Thread starter Thread starter petereater
  • Start date Start date
  • Tags Tags
    Atwood Time
Click For Summary
SUMMARY

The discussion focuses on calculating the time interval for mass m1 (27.0 kg) to reach the floor in an Atwood machine setup, where m2 (13.6 kg) is resting on the floor and the pulley has a mass of 5.00 kg. The initial height of m1 is 4.70 m. Key equations utilized include the conservation of mechanical energy and the relationship between angular acceleration and torque. The user derived the final angular velocity (omega[final] = 119) but encountered difficulties in determining the angular acceleration and subsequently the time taken for m1 to hit the floor.

PREREQUISITES
  • Understanding of Atwood machines and their mechanics
  • Familiarity with conservation of mechanical energy principles
  • Knowledge of angular motion equations and torque
  • Ability to calculate moment of inertia for a pulley
NEXT STEPS
  • Learn how to derive angular acceleration from torque and moment of inertia
  • Study the application of free-body diagrams in solving dynamics problems
  • Explore the relationship between linear and angular motion in mechanical systems
  • Investigate the effects of friction and mass distribution in Atwood machines
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to enhance their understanding of Atwood machines and related calculations.

petereater
Messages
2
Reaction score
0

Homework Statement



Consider the system shown in the image attached with m1 = 27.0 kg, m2 = 13.6 kg, R = 0.290 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.70 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley.

Calculate the time interval required for m1 to hit the floor.



Homework Equations


6be716e191ceca92e0712ac4b3254a8c.png

I*Angular acceleration=torque
Conservation of Mechanical Energy( K+U, final=K+U, initial)



The Attempt at a Solution


I played around with the formulas starting with the mechanical energy formula. I got this:

omega [final] = (1/R) [(2*(m2-m1)gh)/(m1+m2+(I/R^2))]^(1/2)
I=.5MR^2, M=mass of pulley

So I plugged in the values and got omega[final]= 119

Now, I got to find time, but I'm stuck. I plug in 119 in the formala:

Omega[final] = omega[initial] +angular acceleration*time, omega[initial]= 0

I have to determine the angular acceleration (equals torque/Inertia)

Inertia equals to .21075 (.5*5kg*.29m^2). To find torque, I could use Tau=rFSin(theta), but I end trying to find Force and theta. I don't really know where to go from here...
 
Physics news on Phys.org
Draw a free-body diagram first.

ehild
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 97 ·
4
Replies
97
Views
17K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
8
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
1
Views
4K