Rigid Body Reel and hanging mass.

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SUMMARY

The discussion focuses on a physics problem involving a rigid body reel and a hanging mass. The mass of the object is 5.4 kg, and it is attached to a light string wrapped around a solid disk reel with a radius of 0.265 m and mass of 3.00 kg. Key equations used include ƩFy = ma for vertical forces and τ = Iα for torque, with the moment of inertia calculated as I = 0.5(M)(R^2). The user seeks clarification on calculating tension in the string and the acceleration of the object, indicating a need for a systematic approach to solve the problem.

PREREQUISITES
  • Understanding of Newton's second law (ƩFy = ma)
  • Familiarity with rotational dynamics (τ = Iα)
  • Knowledge of moment of inertia for solid disks (I = 0.5MR^2)
  • Basic kinematic equations for linear motion
NEXT STEPS
  • Calculate the tension in the string using torque and rotational dynamics.
  • Determine the acceleration of the object using the relationship between tension and gravitational force.
  • Apply kinematic equations to find the final speed of the object just before it hits the floor.
  • Explore the implications of varying mass and radius on the system's dynamics.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of rigid body dynamics and kinematics in real-world applications.

JJRKnights
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Homework Statement



An object with a mass of m = 5.4 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.265 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 5.50 m above the floor.

(a) Determine the tension in the string.
(b) Determine the magnitude of the acceleration of the object.
(c) Determine the speed with which the object hits the floor.

Homework Equations



ƩFy = ma
τ = Iα

The Attempt at a Solution



ƩFy = mg - T = ma
T = m(g-a)
a = -10/t^2
I = .5(M)(R^2)
τ = Iα = .5MRa
τ=.3975a

I found that to find Tension in this way, I would need acceleration first, but there must be another way to find Tension because b) asks for acceleration while a) wants Tension.

Could I say that the Torque = Tension x Radius?

I think I'm on the right tracks, but can anybody help please?
 
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Also for part c,

I looked at it and can see relevant equations:

v^2 =(v_0)^2 + 2a(x-(x_0))
which would come out to
v^2 = -10a
 

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