Two Blocks, Massless Rope. Pulley

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SUMMARY

The discussion focuses on a physics problem involving two blocks (m1 = 3.3 kg and m2 = 4.2 kg) connected by a massless rope over a pulley with a diameter of 12 cm and a mass of 2.0 kg. The friction at the axle exerts a torque of 0.35 Nm. To determine the time it takes for the 4.2 kg block to reach the floor from a height of 1.0 m, participants suggest using torque equations, moment of inertia, and kinematic equations. Key steps include writing separate force equations for each mass and a torque equation for the pulley, leading to a system of equations to solve for acceleration and time.

PREREQUISITES
  • Understanding of Newton's second law (f = ma)
  • Knowledge of torque and its relation to angular acceleration
  • Familiarity with moment of inertia calculations
  • Ability to apply kinematic equations in physics problems
NEXT STEPS
  • Study the relationship between torque and angular acceleration in rotational dynamics
  • Learn how to calculate moment of inertia for different shapes
  • Explore the effects of friction on pulley systems in mechanics
  • Practice solving systems of equations in physics problems involving multiple objects
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for problem-solving strategies in pulley systems.

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Homework Statement



The two blocks, m1 = 3.3 kg and m2 = 4.2, in the figure below are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.35 Nm. If the blocks are released from rest, how long does it take the 4.2 kg block to reach the floor from a height of h = 1.0 m?


Homework Equations



torque= r * F
sum of torques = moment of inertia * angular acceleartion
angular acceleration * radius = acceleration
x=v0t + 1/2 a t^2



The Attempt at a Solution



I tried summing torques, saying that m2 is on the left and has torque m2*g*radius. Following this, m1 acts opposite, so subtract m1*g*radius. Friction acts opposite of motion, so acts the direction of m1. So (m2*g*r) - (m1*g*r) - torque(friction) = alpha * 1/2m*r^2 (m is mass of pulley, r is .06m, m1 = 3.3, m2=4.2)

from here, i used alpha*r=a to find linear acceleration

then i used x=v0t + 1/2at^2 to find time, v0 is 0 because it starts at rest, a is what i found before

Any suggestions?
 
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You need to examine each part separately... the mass on one end of the pulley... the mass on the other end... and the pulley itself...

Write the f = ma equations for each of the 2 masses...

Write the torque = I*alpha equation for the pulley...

The two tensions cannot be assumed to be the same... so assume two variables T1 and T2...

The acceleration of one mass upward will equal the acceleration of the other mass downward.

alpha = a/0.06

you'll have 3 equations with 3 unknowns. a, T1, T2...
 

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