Solving Solid Disk Rolling Up Incline: Find Height h

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SUMMARY

A solid disk of mass m and radius R rolls up an incline without slipping, converting its kinetic energy into potential energy. The relationship between linear velocity v and angular velocity ω is crucial for solving the problem. The final height h reached by the disk is determined to be h = (5/4)(v^2)/g. This conclusion is derived from the energy conservation equation, where the initial kinetic energy equals the potential energy at height h.

PREREQUISITES
  • Understanding of rotational inertia (I = 0.5mr²)
  • Knowledge of kinetic energy (KE = 0.5mv²)
  • Familiarity with potential energy (PE = mgh)
  • Concept of rolling motion and the relationship between linear and angular velocity
NEXT STEPS
  • Study the relationship between linear velocity and angular velocity in rolling objects
  • Explore energy conservation principles in mechanical systems
  • Learn about different forms of energy and their conversions in physics
  • Investigate the dynamics of rolling motion and frictionless surfaces
USEFUL FOR

Students in physics, particularly those studying mechanics and energy conservation, as well as educators looking for examples of rotational dynamics in real-world applications.

NathanLeduc1
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Homework Statement


A solid disk of mass m and radius R rolls without slipping with a velocity v. Assuming it doesn't slip, how far vertically will it roll up an incline?

Homework Equations


I=0.5mr2
E=0.5Iω2
KE=0.5mv2
PE=mgh

The Attempt at a Solution


I'm thinking that we need to find the height h when the kinetic energy is converted to potential energy. So:
0.5Iω2+mv2=mgh
0.5*0.5mr2ω2+mv2=mgh
0.25r2ω2+v2=gh
h=(0.25r2ω2+v2)/g
Is that right? I feel weird because I still have ω in there.
 
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NathanLeduc1 said:

Homework Statement


A solid disk of mass m and radius R rolls without slipping with a velocity v. Assuming it doesn't slip, how far vertically will it roll up an incline?

Homework Equations


I=0.5mr2
E=0.5Iω2
KE=0.5mv2
PE=mgh

The Attempt at a Solution


I'm thinking that we need to find the height h when the kinetic energy is converted to potential energy. So:
0.5Iω2+mv2=mgh
0.5*0.5mr2ω2+mv2=mgh
0.25r2ω2+v2=gh
h=(0.25r2ω2+v2)/g
Is that right? I feel weird because I still have ω in there.
What's the relationship between v and ω ?
 
Ah, either you're a genius or I'm dumb. Probably both. Thank you.

I got a final answer of (1.2*v^2)/g = h.
 
Oops, 1.25... ((5/4)(v^2))/g = h
 
NathanLeduc1 said:
Ah, either you're a genius or I'm dumb. Probably both. Thank you.
Probably neither.

I considered a lengthier reply to your Original Post, but then though I'd see what you could do with a fairly subtle hint/question. From that, you completed the exercise. I commend you for that.

You might be surprised at how many people need to be led by the nose, step-by-step to an answer.

I got a final answer of (1.2*v^2)/g = h.
 

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