Solving for Kinetic Energy with Circular Wheel and Weight: Equation Included

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SUMMARY

The discussion focuses on calculating the total kinetic energy (K) of a system consisting of a circular wheel and a weight attached to it. The moment of inertia (I) of the wheel is defined as I = kmR^2, where k ranges from 0.5 to 1.0. The total kinetic energy is derived by combining the rotational kinetic energy, E_{k(rotational)} = \frac{1}{2} I \omega^2, and the translational kinetic energy, K = 1/2 mv^2. The user also seeks to determine the speed (v) of the weight after descending a vertical distance (h) when the system is released from rest.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly kinetic energy.
  • Familiarity with moment of inertia calculations.
  • Knowledge of rotational motion equations and their application.
  • Basic algebra for solving equations involving variables like mass (m), radius (R), and speed (v).
NEXT STEPS
  • Study the derivation of the total kinetic energy for systems involving both rotational and translational motion.
  • Learn about the relationship between linear speed (v) and angular velocity (ω) using the equation V = Rω.
  • Explore the implications of different values of k on the moment of inertia and kinetic energy calculations.
  • Investigate energy conservation principles in mechanical systems to solve for speed after a descent.
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to kinetic energy and rotational dynamics.

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Homework Statement



Consider a circular wheel with a mass m, and a radius R. The moment of inertia about the center of the wheel is I = kmR^2, where k is a constant in the range between 0.5<k<1.0. A rope wraps around the wheel. A weight of mass 2m is attached to the end of this rope. At some moment, the weight is falling with a speed v. The total kinetic energy K of the system at this moment is given by what mathematical equation?

Homework Equations



K = 1/2 mv^2

The Attempt at a Solution



I know that the mass of the wheel will have to be considered to find the final kinetic energy...can someone please give me a hint what to do next?
 
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Use E_{k(rotational)} = \frac{1}{2} I \omega^2, then add the translational kinetic energy term to it. This would give the total kinetic energy. Energy terms in classsical mechanics are hardly ever multiplied or divided, so what's left to do is add or take away. In this case, since it's a total one wants, one adds.
 
Last edited:
thank you so much!
 
V=rw. Net T=T(rotational) + T(translational)
 
How about this?? It's the second part to the question...

Assume that k=1/2

QUESTION...If this system is released from rest, find the speed, v, at the moment when the weight has descended a vertical distance h. Any help would be nice.

Thanks
 
Momentum09 said:

Homework Statement



Consider a circular wheel with a mass m, and a radius R. The moment of inertia about the center of the wheel is I = kmR^2, where k is a constant in the range between 0.5<k<1.0. A rope wraps around the wheel. A weight of mass 2m is attached to the end of this rope. At some moment, the weight is falling with a speed v. The total kinetic energy K of the system at this moment is given by what mathematical equation?

Homework Equations



K = 1/2 mv^2

The Attempt at a Solution



I know that the mass of the wheel will have to be considered to find the final kinetic energy...can someone please give me a hint what to do next?

obviously your relevant equations 2 are incomplete, otherwise you would have your answer... what other relevant equations are there?
 

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