My Math Forum A particle moving on the smooth inside of a hollow sphere.

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November 6th, 2016, 01:08 PM   #1
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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 $$v=\sqrt\frac{ag}{2}$$
Attached Files
 Hollow sphere of radius a.zip (21.7 KB, 2 views)

Last edited by Statistics132; November 6th, 2016 at 01:54 PM.

November 6th, 2016, 04:22 PM   #2
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Quote:
 Originally Posted by Statistics132 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 $$\sqrt (ga(\sqrt3 -1))$$
energy conservation ...

$\dfrac{1}{2}mu^2 = \dfrac{1}{2}mga = mgh \implies h = \dfrac{a}{2}$

half-way 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:
 (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.
at height $\dfrac{3a}{2}$, normal force = 0, so centripetal force is given by the radial component of the particle's weight, $mg\cos(60^\circ)$ ...

$F_c = mg\cos(60^\circ) = \dfrac{mv^2}{a} \implies v^2 = \dfrac{ga}{2}$

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, 12:35 PM #3 Member   Joined: Jun 2015 From: Warwick Posts: 33 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 12:39 PM.

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