Motion on an Incline

  • Thread starter amcavoy
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  • #1
amcavoy
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I don't know a lot about physics so please excuse me. Let's say you have a marble rolling down a frictionless plane of an incline θ degrees from the horizontal. It begins at a height ho with an initial velocity of 0. I want to find the velocity of the marble at any height along the way. What I've done is below:

[tex]mgh=\frac{1}{2}mv^{2}\implies v=\sqrt{2gh}[/tex]

Since g is downward, I worked it out that the force in the direction of motion would be [itex]g\sin{\theta}[/itex]. Thus the velocity at any given height would be [itex]\sqrt{2gh\sin{\theta}}[/itex]. However, this doesn't account for the fact that the ball begins at rest. Is this incorrect or am I on the right track?

Thanks for your help.
 

Answers and Replies

  • #2
FluxCapacitator
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You would be completely on the right track if it was a block, not a rolling ball.

If you've covered angular mechanics, then this becomes a much more complicated problem because a rolling ball stores energy (the rotation is a form of kinetic energy). The kinetic energy of a spinning object of any sort, or, angular kinetic energy is given by [tex]KE=1/2*I\omega ^{2}[/tex], where I is you moment of inertia and [tex]\omega[/tex] is your angular velocity. In addition to this you have your normal [tex]1/2mv^{2}[/tex] for your kinetic energy.

Now, assuming that you haven't covered angular mechanics, then we can ignore the whole rotational aspect of the problem, and then you're completely right thus far.

Your expression for the velocity at a height, h, [tex]v= \sqrt{2gh}[/tex] is right.

If you need an expression for the velocity at time, t, then you need to use a force approach, where the net force on the object is [tex]F=mg\sin\theta[/tex]. From there you find acceleration, then integrate to find velocity at time t.

I hope I've been of help, and if you haven't covered angular mechanics yet, ignore that little section I posted about it, you'll cover that in due time.
 
  • #3
amcavoy
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Great thanks a lot for the response I appreciate it.
 

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