Rolling without slipping problem

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SUMMARY

The discussion focuses on calculating the acceleration of a solid sphere rolling without slipping down a ramp inclined at 32 degrees. The initial approach using the equation for torque, sum of torques = I(alpha), was flawed due to incorrect application of the torque axis. The correct moment of inertia for the sphere about the point of contact is I = (7/5)mr^2, leading to the realization that gravity does exert a torque at this point. The final acceleration calculation must consider both translational and rotational dynamics, confirming that the acceleration cannot exceed gravitational acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and moment of inertia
  • Knowledge of the concept of rolling without slipping
  • Basic trigonometry, particularly sine functions
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  • Study the application of the parallel axis theorem in rotational dynamics
  • Learn about the conditions for rolling motion and frictional forces
  • Explore the derivation of acceleration for objects rolling down inclines
  • Investigate the effects of different coefficients of friction on rolling motion
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Physics students, mechanical engineers, and anyone interested in the dynamics of rolling motion and torque analysis.

ph123
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A solid sphere rolls without slipping down a ramp that is at an angle of 32 above horizontal. The magnitude of the acceleration of the center of mass of the sphere as it rolls down the ramp is?


sum of torques = I(alpha)

rmgsin32 = (2/5)mr^2(a_tan/r)

the radii drop out as one would expect with no radius given. the masses also drop out since they weren't provided.

gsin32 = (2/5)a_tan

(9.8 m/s^2)sin32 = (2/5)a_tan

a = 12.98 m/s^2

This result clearly makes no sense because it is greater than the accelertion due to gravity. But that was the only approach I could think to use since I was only given the angle of the incline. Any ideas?
 
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Hints: Does gravity exert a torque on the sphere? What other force acts on the sphere that exerts the torque?
 
friction. but how can I calculate friction without mass or coefficient of friction?
 
ph123 said:
friction. but how can I calculate friction without mass or coefficient of friction?
Just try it--maybe you won't need that information. :wink:

Apply Newton's law for translation and rotation. And the condition for rolling without slipping.
 
omg duh. i forgot about the forces in the x-direction. i always do that. thanks.
 
ph123 said:
sum of torques = I(alpha)

rmgsin32 = (2/5)mr^2(a_tan/r)

The second line is not correct. On the left hand side, you wrote the torque around the center which is the point where the sphere touches the slope, so you need to use inertial momentum I corresponding to the axis through above point.

On the right-hand side, it should be

I=2/5mr^2+mr^2=7/5 mr^2 (axis-parallel theorem, or Steiner's theorem)
 
That's perfectly OK as well: Since the sphere rolls without slipping, you can view it as being in pure instantaneous rotation about the point of contact. With this approach, you need to use torques and rotational inertia about the point of contact, not about the center. Note that gravity does exert a torque about the point of contact. (I find the two step approach--analyzing translation and rotatation separately--to be more instructive. But it's all good! :smile: )
 
Yes Doc Al, that's what I meant.
 

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