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November 6th, 2016, 12:08 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 12:54 PM. November 6th, 2016, 03:22 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, 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. Tags hollow, inside, moving, particle, smooth, sphere Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Statistics132 Applied Math 4 November 5th, 2016 03:43 PM mathkid Calculus 9 November 11th, 2012 06:26 PM Chikis Physics 10 September 8th, 2012 04:53 PM vlamers Calculus 1 January 27th, 2010 07:12 PM djzero3 Real Analysis 3 October 5th, 2007 04:10 AM

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