My Math Forum Vertical circular motion of a particle on a sphere resting against a wall.

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 November 4th, 2016, 02:42 PM #1 Member   Joined: Jun 2015 From: Warwick Posts: 34 Thanks: 1 Vertical circular motion of a particle on a sphere resting against a wall. A smooth sphere with centre $O$ and radius $a$ is fixed with a point $B$ of its surface in contact with a vertical wall. A particle $P$ of mass $m$ rests at the highest point $A$ of the sphere. It is slightly disturbed so that it moves from rest towards the wall in the plane $OAB$. When $P$ makes an angle $θ$ with $OA$ the velocity of $P$ is $\sqrt2ag(1-\cosθ)$ and $P$ leaves the sphere when $\cos θ = \frac{2}{3}$ when its speed is $\sqrt\frac{2ag}{3}$. Show that P hits the wall at a height $\frac{1}{8}a(5\sqrt5 -9)$ above $B$. (Treat the particle as a projectile moving freely under gravity). Last edited by skipjack; November 5th, 2016 at 04:44 PM.
November 4th, 2016, 04:27 PM   #2
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Quote:
 Originally Posted by Statistics132 A smooth sphere with centre $O$ and radius $a$ is fixed with a point $B$ of its surface in contact with a vertical wall. A particle $P$ of mass $m$ rests at the highest point $A$ of the sphere. It is slightly disturbed so that it moves from rest towards the wall in the plane $OAB$. When $P$ makes an angle $θ$ with $OA$ the velocity of $P$ is $\sqrt2ag(1-\cosθ)$ and $P$ leaves the sphere when $\cos θ = \frac{2}{3}$ when its speed is $\sqrt\frac{2ag}{3}$. Show that P hits the wall at a height $\frac{1}{8}a(5\sqrt5 -9)$ above $B$. (Treat the particle as a projectile moving freely under gravity).
particle enters freefall at position ...

$x = a\sin{\theta} = \dfrac{a\sqrt{5}}{3}$ units to the right of point A, and $y = a\cos{\theta} = \dfrac{2a}{3}$ above point B.

to hit the vertical wall ...

$\Delta x = a - a\sin{\theta} = a\left(\dfrac{3-\sqrt{5}}{3}\right)$

$v_x = v\cos{\theta} = \sqrt{\dfrac{2ag}{3}} \cdot \dfrac{2}{3}$

$\Delta x = v_x \cdot t \implies t = \dfrac{\Delta x}{v_x} = \dfrac{a\sqrt{3}(3-\sqrt{5})}{2\sqrt{2ag}}$

$\Delta y = v(-\sin{\theta}) \cdot t - \dfrac{1}{2}gt^2$

$y_f - \dfrac{2a}{3} = \sqrt{\dfrac{2ag}{3}} \cdot \left(-\dfrac{\sqrt{5}}{3}\right) - \dfrac{g}{2} \cdot \bigg[\dfrac{a\sqrt{3}(3-\sqrt{5})}{2\sqrt{2ag}}\bigg]^2$

now it's just a matter of cleaning up this algebraic mess to get $y_f = \dfrac{a}{8}\left(5\sqrt{5}-9\right)$

happy simplifying ...

Last edited by skipjack; November 5th, 2016 at 04:45 PM.

 November 4th, 2016, 06:08 PM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,302 Thanks: 1130 inadvertently omitted "t" in the linear term of the last equation ... $y_f - \dfrac{2a}{3} = \sqrt{\dfrac{2ag}{3}} \cdot \left(-\dfrac{\sqrt{5}}{3}\right) \cdot \bigg[\dfrac{a\sqrt{3}(3-\sqrt{5})}{2\sqrt{2ag}}\bigg]- \dfrac{g}{2} \cdot \bigg[\dfrac{a\sqrt{3}(3-\sqrt{5})}{2\sqrt{2ag}}\bigg]^2$ now it's even messier than before ... don't you love it?
 November 5th, 2016, 01:15 PM #4 Member   Joined: Jun 2015 From: Warwick Posts: 34 Thanks: 1 Thanks for your response. I had the $x$ and $y$ components before, but did not have $(-\sinθ)$ for the $y$-component, so here we are using the convention that as the movement of $P$ is downwards so we have a negative $y$-component. (We also have $-\sinθ$ instead of $-\cosθ$ (or $cosθ$) for the $y$-component in the first term of the $Δy$ equation.) Your answer simplifies to $a(\frac{5\sqrt5}{8} - \frac{43}{24})$ I tried this by hand and with wolframalpha: goto the link https://www.wolframalpha.com/example...ification.html Copy and paste the following onto the bar at the top of the simplification page ((sqrt((2xy)/3))*-(sqrt(5)/3) * (((xsqrt(3)(3-sqrt(5)))/(2sqrt(2xy)))) - ((y/2))(((xsqrt(3)(3-sqrt(5)))/(2sqrt(2xy)))^2 Last edited by skipjack; November 5th, 2016 at 04:52 PM.
November 5th, 2016, 03:43 PM   #5
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$y_f - \dfrac{2a}{3} = \sqrt{\dfrac{2ag}{3}} \cdot \left(-\dfrac{\sqrt{5}}{3}\right) \cdot \bigg[\dfrac{a\sqrt{3}(3-\sqrt{5})}{2\sqrt{2ag}}\bigg]- \dfrac{g}{2} \cdot \bigg[\dfrac{a\sqrt{3}(3-\sqrt{5})}{2\sqrt{2ag}}\bigg]^2$

$y_f = \dfrac{2a}{3} - \sqrt{\dfrac{\cancel{2ag}}{\cancel{3}}} \cdot \left(\dfrac{\sqrt{5}}{3}\right) \cdot \bigg[\dfrac{a\cancel{\sqrt{3}}(3-\sqrt{5})}{2\sqrt{\cancel{2ag}}}\bigg]- \dfrac{\cancel{g}}{2} \cdot \dfrac{3a^2 (14-6\sqrt{5})}{8 a \cancel{g}}$

$y_f = \dfrac{2a}{3} - \dfrac{3a\sqrt{5} - 5a}{6} - \dfrac{3a(14-6\sqrt{5})}{16}$

$y_f = \dfrac{2a}{3} - \dfrac{3a\sqrt{5} - 5a}{6} - \dfrac{42a-18a\sqrt{5}}{16}$

$y_f = \dfrac{2a}{3} - \dfrac{3a\sqrt{5} - 5a}{6} - \dfrac{21a-9a\sqrt{5}}{8}$

$y_f = \dfrac{16a}{24} - \dfrac{12a\sqrt{5} - 20a}{24} - \dfrac{63a-27a\sqrt{5}}{24}$

$y_f = \dfrac{16a - 12a\sqrt{5} + 20a - 63a + 27a\sqrt{5}}{24}$

$y_f = \dfrac{15a\sqrt{5} - 27a}{24}$

$y_f = \dfrac{3a(5\sqrt{5} - 9)}{24}$

$y_f = \dfrac{a(5\sqrt{5}-9)}{8}$

TI Nspire simplification attached ...
Attached Images
 IMG_1413.JPG (95.7 KB, 5 views)

Last edited by skeeter; November 5th, 2016 at 04:11 PM.

 Tags circular, motion, particle, resting, sphere, vertical, wall

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