Solving a Frictionless Pulley Problem: Acceleration of m2

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SUMMARY

The discussion focuses on solving the acceleration of a block (m2) in a frictionless pulley system involving a 29.3 kg block (m1) and a 6.70 kg block (m2). A force of 202.5 N is applied at an angle of 29.7° on m1, which is on a horizontal surface. The moment of inertia of the pulley is 0.100 kg·m², and its radius is 0.055 m. To find the upward acceleration of m2, participants emphasize applying Newton's 2nd law to both masses and the pulley, and combining the resulting equations to solve for acceleration.

PREREQUISITES
  • Understanding of Newton's 2nd law of motion
  • Familiarity with concepts of tension in a pulley system
  • Knowledge of moment of inertia and its impact on rotational motion
  • Basic trigonometry to resolve forces at an angle
NEXT STEPS
  • Study the application of Newton's 2nd law in multi-body systems
  • Learn about calculating tension in pulley systems with multiple masses
  • Explore the relationship between torque, moment of inertia, and angular acceleration
  • Investigate the effects of applied forces at angles on motion dynamics
USEFUL FOR

Physics students, educators, and anyone interested in understanding dynamics in mechanical systems, particularly those involving pulleys and forces at angles.

physicsbro
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Homework Statement


A 29.3 kg block, m1, is on a horizontal surface, connected to a 6.70 kg block, m2, by a massless string as shown. The frictionless pulley has a radius R = 0.055 m and a moment of inertia I = 0.100 kg·m2. A force F = 202.5 N acts on m1 at an angle θ = 29.7°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

prob24.gif

Homework Equations





The Attempt at a Solution



My textbook says that the net tension divided by the moment of inertia would give me the \alpha and that multiplied by the radius would give me the acceleration, but it didnt work.
 
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physicsbro said:
My textbook says that the net tension divided by the moment of inertia would give me the \alpha and that multiplied by the radius would give me the acceleration, but it didnt work.
That's certainly true, but you'll need the tensions to make use of it. Hint: Apply Newton's 2nd law to both masses and the pulley. Combine those three equations and you can solve for the acceleration.
 

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