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November 6th, 2016, 12:08 PM  #1 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1  A particle moving on the smooth inside of a hollow sphere.
A particle $P$ is projected horizontally with speed $u$ from the lowest point $A$ of the smooth inside surface of a fixed hollow sphere of internal radius $a$. (i) In the case when $u^2 = ga$ show that $P$ does not leave the surface of the sphere. Show that, when $P$ has moved halfway along its path from $A$ towards the point at which it first comes to rest, its speed is $$\sqrt (ga(\sqrt3 1))$$ (ii) Find $u^2$ in terms of $ga$, in the case when $P$ leaves the surface at a height $\frac{3a}{2}$ above $A$, and find, in terms of $a$ and $g$, the speed of $P$ as it leaves the surface. So far, for the first part I have (see the attachment): $$cosθ = \frac{u^2 + 2ga}{3ga}$$ For (ii), the equations should be: $$u^2 = \frac{5ag}{2}$$ and $$v=\sqrt\frac{ag}{2}$$ Last edited by Statistics132; November 6th, 2016 at 12:54 PM. 
November 6th, 2016, 03:22 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521  Quote:
$\dfrac{1}{2}mu^2 = \dfrac{1}{2}mga = mgh \implies h = \dfrac{a}{2}$ halfway point, $h = a\left(\dfrac{2\sqrt{3}}{2}\right)$ $\dfrac{1}{2}mga = mga\left(\dfrac{2\sqrt{3}}{2}\right) + \dfrac{1}{2}mv^2$ $v = \sqrt{ga(\sqrt{3}1)}$ Quote:
$F_c = mg\cos(60^\circ) = \dfrac{mv^2}{a} \implies v^2 = \dfrac{ga}{2}$ energy conservation again ... $\dfrac{1}{2}mu^2 = mg \cdot \dfrac{3a}{2} + \dfrac{1}{2} \cdot \dfrac{mga}{2}$ $u^2 = \dfrac{7ga}{2}$  
November 7th, 2016, 11:35 AM  #3 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 
Also, using conservation of mechanical energy: $$\frac{1}{2}mv^2 = \frac{1}{2}mu^2  mgh$$ $$\frac{1}{2}mv^2 = \frac{1}{2}m .\frac{7}{2}ga  mg.\frac{3a}{2}$$ $$ v^2 = ga(\frac{7}{2} 3)$$ So $$v^2 = \frac{ga}{2}$$ So $$v = \sqrt\frac{ga}{2}$$ Thanks. Last edited by Statistics132; November 7th, 2016 at 11:39 AM. 

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