Rotational Inertia: Mass m, Radius R, h, v, g

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SUMMARY

The discussion focuses on calculating the rotational inertia of a body with mass m and radius R that rolls without slipping up a hill to a maximum height h, defined by the equation h = 3v² / 4g. The solution involves applying the principles of energy conservation, where the initial kinetic energy (both translational and rotational) equals the gravitational potential energy at maximum height. The final result for the rotational inertia is derived as I = (1/2)MR², identifying the body as a solid cylinder.

PREREQUISITES
  • Understanding of rotational inertia and its calculation using I = ∫r² dm
  • Knowledge of energy conservation principles in physics
  • Familiarity with translational and rotational kinetic energy equations
  • Basic concepts of rolling motion and the relationship between translational and rotational velocities
NEXT STEPS
  • Study the derivation of rotational inertia for various shapes, focusing on cylinders and spheres
  • Learn about energy conservation in rolling motion scenarios
  • Explore the relationship between angular velocity and linear velocity in rolling objects
  • Investigate the implications of rolling without slipping in different physical contexts
USEFUL FOR

Students studying classical mechanics, physics educators, and anyone interested in understanding the dynamics of rolling objects and energy conservation principles.

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Homework Statement


A body radius R and mass m is rolling horizonttally without slipping with speed v. Then is rolls up a hill to maximum hieght h. If h=\frac{3v^2}{4g}
a) what is the body's rotational inertia?
b) What body might it be?


Homework Equations


h=\frac{3v^2}{4g}


The Attempt at a Solution


I have no clue!
 
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Show some effort, you just copied the problem under 2. Granted this is one of the tougher problems posted recently, but add some info re eqns for energy conservation in this instance.
 
ok so
energy of conservation in this problem is K_{i} =U_{f}\longrightarrow\frac{1}{2}mv^2=mgh
and rotational inertia for a body is
I=\int r^2 dm
 
Call the rotational inertia I. Make use of the fact that it rolls without slipping. (Hint: The KE will be part translational and part rotational.)
 
Some questions:
  • What are the components of energy with which you need to be concerned? You have equations for translational kinetic energy and gravitational potential energy. Are you missing anything?
  • What is the total energy of the body at the bottom of the hill (before it starts rolling up the hill)?
  • What is the total energy of the body when it reaches the maximum height?
 
so \frac{1}{2}I_{center of mass}w^2+\frac{1}{2}Mv_{center of mass}^2=Mgh This is the potenial energy once it reaches its maximum hieght.
The intial KE is \frac{1}{2}I_{center of mass}w^2+\frac{1}{2}Mv_{center of mass}^2 since it starts from rolling
Components of energy are translational+rotational=rolling
and I=\int r^2 dm
 
Last edited:
The body is rolling without slipping. That means there is a relationship between the rotational velocity \omega and the translational velocity v. You were also given a relationship between the height h and the velocity. Use these relationships to simplify the equations.
 
so you saying: v_{translational}=\omega R\longrightarrow\omega=\frac{v_{center of mass}}{R}
right?
I plugg that into my equation and the h=\frac{3v^2}{4g}
 
Last edited:
So what do you get?
 
  • #10
After simplifying every thing I got
I=\frac{M}{2}R^2
which happens to be the Rotational Inertia of a cylinder! AHH!
 
  • #11
Well done.
 

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