When do vcm=rw and a=gsintheta/(1+I/(MR)^2) apply?

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SUMMARY

The equations vcm=rw and a=(gsintheta)/(1+I/(MR)^2) specifically apply to scenarios involving rolling without slipping. When slipping occurs, such as a sphere sliding without rotation, the first equation does not yield accurate results. The second equation requires a clear understanding of the system's setup, particularly the role of torque, static friction, and moment of inertia in determining acceleration down an inclined plane.

PREREQUISITES
  • Understanding of rotational dynamics, including torque and moment of inertia.
  • Familiarity with the concept of rolling motion and conditions for rolling without slipping.
  • Knowledge of basic physics principles related to forces on inclined planes.
  • Ability to manipulate equations involving angular acceleration and linear acceleration.
NEXT STEPS
  • Study the principles of rolling motion and the conditions for rolling without slipping.
  • Learn about the relationship between torque, static friction, and angular acceleration in rotational systems.
  • Explore examples of objects on inclined planes and analyze their motion using Newton's laws.
  • Investigate the implications of slipping on the equations of motion for rolling objects.
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Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of rolling objects and the effects of slipping on motion equations.

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Homework Statement


Do the equations below only apply for rolling without slipping?

Homework Equations


vcm=rw, a=(gsintheta)/(1+I/(MR)^2)
cm=center of mass

3. The Attempt at a Solution [/B]
I know that they both apply for rolling without slipping, but do they apply when there is slipping? Thank you.
 
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Consider some simple examples of slipping - like a sphere sliding and not rotating at all. Does the first equation give the right answer?
I don't recognise the second equation. You'll need to describe the set up it applies to.
 
I see now. The equation vcm=rw would not apply if the ball was moving but not having rotational motion.

This is how I set up the equation:

Consider an object rolling down a inclined plane.

Torque=RFs
R=radius, Fs=static friction
Also,
Torque=I*alpha
I=moment of inertia
alpha=angular acceleration
This leads to Torque = I*alpha=I*(a/R)
where a=acceleration centripetal

Setting RFs=I*(a/R) gives us Fs=(I*a)/R^2

So we have Fnet=ma=gravitational force down the incline plane-force of friction
which is=(mgsintheta)-(Ia)/R^2

Solving for a and manipulating the equation gives that a=(gsintheta)/(1+(I*alpha)/MR^2)
 
Last edited:

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