Two radii pulley lifting a pack

  • Thread starter Thread starter getty102
  • Start date Start date
  • Tags Tags
    Lifting Pulley
Click For Summary
SUMMARY

The discussion focuses on calculating the linear acceleration of a pack attached to a pulley system with a moment of inertia of 676 kgm². The outer radius of the pulley is 0.8 m, and the inner radius is 0.56 m, with a tension of 477.9 N applied to the rope. The mass of the pack is 30.8 kg. The derived formula for linear acceleration is at = [R(Tyr) - r(mp*g)] / [(I/R) - (r*mp)], but the initial calculation of 24.233 m/s² was incorrect, indicating a need to reassess the forces and torques involved.

PREREQUISITES
  • Understanding of rotational dynamics, specifically torque and moment of inertia.
  • Familiarity with Newton's second law of motion (ƩF = ma).
  • Knowledge of the relationship between linear and angular acceleration.
  • Basic algebra for manipulating equations involving multiple variables.
NEXT STEPS
  • Review the principles of torque and moment of inertia in rotational systems.
  • Study the relationship between linear acceleration and angular acceleration in pulley systems.
  • Practice problems involving multiple forces acting on a pulley to solidify understanding.
  • Explore the effects of varying mass and radius on the acceleration of connected objects.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking examples of pulley systems in action.

getty102
Messages
38
Reaction score
0

Homework Statement



You are pulling on a rope attached to the outer radius of a pulley with a moment of inertia I=676 kgm2 as shown. The tension in the rope that you are pulling is 477.9 N. A pack of mass 30.8 kg is attached to the pulley's inner radius as shown. If the inner radius of the pulley is 0.56 m and the outer radius is 0.8 m, what is the magnitude of the linear acceleration of the pack?

Homework Equations



Ʃtorque=(I)([itex]\alpha[/itex])
ƩF=ma

The Attempt at a Solution



I found the tension due to the mass of the pack on the pulley.
Tps=mp(g-at)
p=pack, at=linear acceleration of pack

I then summed the torques, and inserted the tension due to the pack on the pulley
R(Tyr)-r(mp(g-at))=I(at/R)

solved for at=[R(Tyr)-r(mp*g)]/[(I/R)-(r*mp)]

my final at=24.233 m/s/s which is not right
 
Physics news on Phys.org
You might want to check your expression for Tps. Make sure that the implied forces result in an acceleration in the desired direction for the pack.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
7K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
Replies
4
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K