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Thread: Particle moving on a sphere mechanics help required

  1. #1
    Member ssadi's Avatar
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    Particle moving on a sphere mechanics help required

    A particle P is projected horizontally with speed u, where u^2 < 2ag, from the highest point A of a fixed smooth sphere of radius a and whose centre is at O. The particle slides on the outer surface of the sphere. Show that it leaves the sphere when PO makes an angle arccos ($\displaystyle \frac{2}{3}+\frac{u^2}{3ag}$) with the vertical.

    I figured out that when speed=2ag, reaction will be 0 and the particle will leave the sphere. And the change in speed have something to do with the energy change mgh=.5 mv^2. But all my workings to find out h lead to very complicated expressions. So how do i prove the angle is arccos ($\displaystyle \frac{2}{3}+\frac{u^2}{3ag}$) ?
    Please help me associate the change in speed with the angle.
    (vertical circular motion)
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  2. #2
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    Quote Originally Posted by ssadi View Post
    A particle P is projected horizontally with speed u, where u^2 < 2ag, from the highest point A of a fixed smooth sphere of radius a and whose centre is at O. The particle slides on the outer surface of the sphere. Show that it leaves the sphere when PO makes an angle arccos ($\displaystyle \frac{2}{3}+\frac{u^2}{3ag}$) with the vertical.

    I figured out that when speed=2ag, reaction will be 0 and the particle will leave the sphere. And the change in speed have something to do with the energy change mgh=.5 mv^2. But all my workings to find out h lead to very complicated expressions. So how do i prove the angle is arccos ($\displaystyle \frac{2}{3}+\frac{u^2}{3ag}$) ?
    Please help me associate the change in speed with the angle.
    (vertical circular motion)
    let O be the zero point for gravitational potential energy.

    $\displaystyle \theta $ = vertical angle between O and P

    total initial mechanical energy ...

    $\displaystyle mga + \frac{1}{2}mu^2$

    at point P ...

    $\displaystyle mga\cos{\theta} + \frac{1}{2}mv^2$ , $\displaystyle v > u$

    since energy is conserved ...

    $\displaystyle
    mga + \frac{1}{2}mu^2 = mga\cos{\theta} + \frac{1}{2}mv^2
    $

    $\displaystyle 2ga + u^2 = 2ga\cos{\theta} + v^2$

    $\displaystyle v^2 = u^2 + 2ga(1 - \cos{\theta})$

    at point P, the normal force of the sphere is zero. centripetal acceleration ...

    $\displaystyle \frac{v^2}{a} = g\cos{\theta}$

    $\displaystyle \frac{u^2}{a} + 2g - 2g\cos{\theta} = g\cos{\theta}$

    $\displaystyle \frac{u^2}{a} + 2g = 3g\cos{\theta}$

    $\displaystyle
    \frac{u^2}{3ga} + \frac{2}{3} = \cos{\theta}
    $
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