Rotational motion, Pulley system

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SUMMARY

The discussion focuses on a physics problem involving a pulley system with a 5kg mass and a 10kg mass connected by a rope over a disc-shaped pulley weighing 5kg and having a radius of 10cm. The calculated speed of the masses at the moment they pass each other is approximately 1.11 m/s, correcting an initial miscalculation of 1.8 m/s. The torque exerted on the pulley is determined to be 4.9 Nm, based on the difference in gravitational forces acting on the two masses. The analysis emphasizes the importance of equating mechanical energy to solve for speed in such systems.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with rotational dynamics and torque calculations
  • Knowledge of angular momentum and its equations
  • Basic principles of energy conservation in mechanical systems
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  • Study the principles of rotational motion and torque in detail
  • Learn about the conservation of mechanical energy in pulley systems
  • Explore advanced topics in angular momentum calculations
  • Investigate the effects of friction in pulley systems and how to account for it
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Students studying physics, particularly those focusing on mechanics, as well as educators and tutors looking to enhance their understanding of rotational motion and pulley systems.

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Homework Statement


A 5kg mass and a 10kg mass are connected by a rope and hung over a pulley. The pulley is a disc with a mass of 5kg and a radius of 10cm. the two masses are initially at rest with the 10kg mass 50cm higher than the 5kg mass.
Find, The speed of each mass at the instant they pass one another
The torque exerted on the pulley by the masses
The angular momentum of the pulley at the instant the masses pass one another.

Homework Equations


L = m(r * v)
T = r * F
I(disc) = 1/2 mv^2


The Attempt at a Solution


Part (1)
F=(10-5)*9.8 = 49N
Mtotal = (10+5)=15
a =49/15 =3.26 ms^-2
v^2 = 0^2 + 2 * 0.5 * 3.26
v = 1.8 m/s

Part (2)
The only external forces acting on the disc are m1*g and m2*g
T = r * F
T = r (m2*g-m1*g)
T = 0.1 (10*9.8-5*9.8)
T = 4.9 Nm
 
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Your answer to part 1 does not look correct. Assuming no frictional losses, you may find it helpful to equated the mechanical energy (i.e. the sum of potential and kinetic energy) of the three masses before they are released with the mechanical energy of the system when the two masses pass each other. Doing that, you should arrive at a speed around 1.11 m/s.

Your answer to part 2 looks correct.
 

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