Hollow sphere rolling down a slope

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SUMMARY

A hollow spherical shell with a mass of 2.50 kg rolls down a slope at an angle of 35.0 degrees. To determine the acceleration, friction force, and minimum coefficient of friction to prevent slipping, the equations of motion and energy conservation principles are applied. The moment of inertia for the hollow sphere is calculated using the formula I = (2/3)MR². The energy conservation equation mgh = (1/2)mv² + (1/2)Iw² is utilized to derive the necessary values, confirming that the radius does not affect the final results.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of moment of inertia for hollow spheres (I = (2/3)MR²)
  • Familiarity with energy conservation principles in mechanics
  • Ability to relate linear and angular motion (v = rω)
NEXT STEPS
  • Calculate the acceleration of a hollow sphere on an inclined plane using specific mass and angle values.
  • Determine the friction force acting on a rolling object using the derived equations.
  • Explore the concept of rolling without slipping and its implications in physics.
  • Investigate the effects of varying the angle of inclination on the motion of rolling objects.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of rolling motion and energy conservation principles.

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Homework Statement


A hollow, spherical shell with mass 2.50kg rolls without slipping down a slope angled at 35.0 degrees.
i) Find the acceleration
ii) Find the friction force
iii) Find the minimum coefficient of friction to prevent slipping


Homework Equations



F=ma

I=\frac{2MR^2}{3}

Energy must stay constant:
E_{p}+E_{kr}+E_{k} = E_{p}+E_{kr}+E_{k}

mgh+\frac{1}{2}Iw^{2}+\frac{1}{2}mv^{2}=mgh+\frac{1}{2}Iw^{2}+\frac{1}{2}mv^{2}

The Attempt at a Solution


I really have no idea where to start, because a radius is not given, so I cannot find the moment of inertia. I was thinking if I used the energies, and said that

mgh=\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}

mgh=\frac{1}{2}mv^{2}+\frac{1}{2}((\frac{2}{3}mR^2)w^{2}

because we could use a pretend example, say that the sphere starts at rest at the top of the ramp, and in the end, all the energy is kinetic. But from here, I'm not really sure where to go. Thanks!
 
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Who knows, maybe it will turn out that the radius doesn't matter! If not, then an R in the answers will be expected. Go ahead and do it with an R and see. You will need v = r*ω, too.

You energy approach looks good. No doubt it could also be done with force, acceleration, etc.
 
Example.
http://www.feynmanlectures.info/solutions/roll_without_slipping_sol_1.pdf
 
I think I figured it out. Thanks!
 

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