A cylinder and sphere rolling without slipping

  • Thread starter scavok
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A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 5 m. The cylinder arrives at the bottom of the plane 2.9 s after the sphere. Determine the angle between the inclined plane and the horizontal.

Ok, I think I'm close, but I'm getting really nasty equations at the end so I'm not very confident.

The acceleration of the sphere is 5*g*sin(x)/7
The acceleration of the cylinder is g*sin(x)/2

I'm pretty confident these are correct.

With the accelerations I should be able to use regular kinematics.

[tex]\Delta x=\frac{1}{2}at^2[/tex]
Where [tex]\Delta x=5[/tex] and a=the acceleration of either the sphere or cylinder.

For the cylinder I get [tex]5=\frac{1}{2}(\frac{gsin\Theta}{2})(t+2.9)^2[/tex]
For the sphere I get [tex]5=\frac{1}{2}(\frac{5gsin\Theta}{7})t^2[/tex]

But when I solve for [tex]sin\Theta[/tex] or t, and plug it into the opposite equation, I get nasty looking things like:
http://home.comcast.net/~andykovacs/equation.GIF [Broken]

Am I doing this right?
 
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Answers and Replies

  • #2
NateTG
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Since they roll (rather than sliding) you need to account for the energy that goes into making them spin.
 
  • #3
Doc Al
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NateTG said:
Since they roll (rather than sliding) you need to account for the energy that goes into making them spin.
The equations already take that into consideration.

To scavok:

First solve for t by setting [itex]a_1 t^2 = a_2 (t + 2.9)^2[/itex] (the [itex]g \sin\theta[/itex] will drop out).

Then use t to solve for [itex]\sin\theta[/itex].
 

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