Sphere rolling down a ramp linear acceleration

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SUMMARY

A solid sphere of radius R and mass M rolls down a ramp from a height h, experiencing linear acceleration due to gravity. The linear acceleration of the sphere is determined to be \(\frac{5}{7}g\sin\theta\), where g is the acceleration due to gravity and \(\theta\) is the angle of the ramp. The moment of inertia for the sphere is given as \(I = \frac{2}{5}MR^2\). The discussion emphasizes the importance of applying Newton's Second Law to both linear and rotational motion to derive the correct equations for acceleration without needing to calculate the frictional force directly.

PREREQUISITES
  • Understanding of Newton's Second Law for linear and rotational motion
  • Familiarity with the moment of inertia, specifically \(I = \frac{2}{5}MR^2\)
  • Knowledge of rolling motion and the relationship between linear and angular velocity
  • Basic trigonometry to analyze forces on an inclined plane
NEXT STEPS
  • Study the derivation of linear acceleration for rolling objects using Newton's Second Law
  • Learn about the dynamics of rolling motion and the role of friction
  • Explore the concept of torque and its application in rotational dynamics
  • Investigate the effects of different ramp angles on the acceleration of rolling spheres
USEFUL FOR

Physics students, educators, and anyone interested in understanding the mechanics of rolling motion and the application of Newton's laws in real-world scenarios.

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Homework Statement


"A solid sphere of radius R and mass M is initially at rest at the top of a ramp. The lowest point of the sphere is a vertical h above the base of the ramp. It is released and rolls without slipping down the ramp. Determine the linear acceleration while the sphere is anywhere on the ramp.

M (mass), R (radius), h (height), g (gravity), theta

Homework Equations


conservation of momentum
I = 2/5MR^2
w = v/r


The Attempt at a Solution



I ended up finding the linear velocity anywhere on the ramp to be square root of 10gh/7. How would I be able to use that though to find acceleration? I seriously don't know what else to do.
 
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Apply Newton 2nd Law twice, for linear and rotational motion. Friction is providing a net torque.
 
That requires that I find the frictional force, but that's the next question so there has to be some way I can do it without knowing the friction force.
 
You don't need to calculate the friction force. Just call it F and get the value from one of the equations and substitute in the other equation.
 
The right answer is \frac{5}{7}g\sin\theta. Did you find it? :smile:
 
Yes, I did. Thanks for your help. Would you be able to explain how I would get the frictional force then? Is it just F = m times that or is it -mgsin(theta) because it's on a ramp?
 
You got two equations before. Solve one to get the value of F. :smile:
 
pmp! is correct.

It'd be helpful to draw a force body diagram to really grasp the idea of this =).

There's a normal force, a force of a friction, a force of gravity pulling it down in the x and y direction. There's an incline on the ramp (theta).

We know in the Y direction it's not accelerating; thus we can set that summination in the Y axis to 0. However in the X axis we do have a mass that's accelerating; thus we can set that summination in the X direction to mass * acceleration.

After you setup your forces you can then figure it all out mathematically.
 
Last edited:
Oooh okay. That really helps. (I didn't originally use force equations to find the linear acceleration). I used vf^2 = vi^2 + 2ad. Is the correct answer:
(2mgsin(theta))/7?
 

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