Why Do Two Falling Objects Have the Same Acceleration in This Physics Problem?

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SUMMARY

The discussion centers on the physics problem involving a block and a cylinder, both with mass m, connected by a massless string over a pulley. The conclusion drawn is that both objects experience the same downward acceleration of (2/3)g, derived from the equations of motion and torque. The relationship between tension (T), mass (m), and acceleration (a) is established through the equations F=ma and torque equations, confirming that the accelerations are equal without needing to calculate the actual values. The conservation of energy approach also supports this conclusion, indicating that both objects fall with the same acceleration due to the forces acting on them.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Basic knowledge of rotational dynamics and torque (T*R=I*α)
  • Familiarity with conservation of energy principles in physics
  • Concept of angular acceleration and its relation to linear acceleration
NEXT STEPS
  • Study the relationship between linear and angular acceleration in rotational systems
  • Explore the derivation of acceleration in pulley systems using both force and energy methods
  • Learn about the moment of inertia and its impact on rotational motion
  • Investigate more complex pulley systems and their dynamics
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of pulley systems and the relationship between linear and angular motion.

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Homework Statement



One pulley, on one side we have a block with mass m.
On the other side we have a cylinder with mass m.
Cylinder (radius R) has unlimited string (massless, negligible thickness, no slippage)

So you can imagine two blocks falling as more string unravels from the cylinder.

I am told that these two objects have the same acceleration downwards. Why?

Homework Equations



F=ma
ma=mg-T

torque? T*R=I*[itex]\alpha[/itex]

The Attempt at a Solution



The problem made it sound like this was a quick, obvious argument.
I proceeded to a longer argument, finding that the accelerations of both masses were (2/3)g downward:

starting with the cylinder-
TR=I*[itex]\alpha[/itex]
T*R2 = (1/2)M*R2 * a
Finding T= (1/2)ma, then plugging T back into the standard F=ma equations.
Then I find the acceleration of the block to be (2/3)g as well.

Is this even right? If it is, was there a way to show that the objects' accelerations were the same without finding the actual accelerations?

The reason I don't think this is right is because I did the problem using conservation of energy using the assumption that the objects fell down at the same acceleration.
With conservation of energy I got acceleration=(1/2)g (conserving KE, PE, angular KE)
 
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I think you don't need to use torque.
By F= m * a(center of mass) you get that their accel will only depend on forces applied to them. In both cases block and cylinder you have F = T - mg.
Then => m * aBlock = m * aCylinder => aBlock = aCylinder
 

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