
Applied Math Applied Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 4th, 2016, 02:42 PM  #1 
Member Joined: Jun 2015 From: Warwick Posts: 33 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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,211 Thanks: 1052  Quote:
$x = a\sin{\theta} = \dfrac{a\sqrt{5}}{3}$ 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: 2,211 Thanks: 1052 
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: 33 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)(3sqrt(5)))/(2sqrt(2xy))))  ((y/2))(((xsqrt(3)(3sqrt(5)))/(2sqrt(2xy)))^2 Last edited by skipjack; November 5th, 2016 at 04:52 PM. 
November 5th, 2016, 03:43 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,211 Thanks: 1052 
$y_f  \dfrac{2a}{3} = \sqrt{\dfrac{2ag}{3}} \cdot \left(\dfrac{\sqrt{5}}{3}\right) \cdot \bigg[\dfrac{a\sqrt{3}(3\sqrt{5})}{2\sqrt{2ag}}\bigg] \dfrac{g}{2} \cdot \bigg[\dfrac{a\sqrt{3}(3\sqrt{5})}{2\sqrt{2ag}}\bigg]^2$ $y_f = \dfrac{2a}{3}  \sqrt{\dfrac{\cancel{2ag}}{\cancel{3}}} \cdot \left(\dfrac{\sqrt{5}}{3}\right) \cdot \bigg[\dfrac{a\cancel{\sqrt{3}}(3\sqrt{5})}{2\sqrt{\cancel{2ag}}}\bigg] \dfrac{\cancel{g}}{2} \cdot \dfrac{3a^2 (146\sqrt{5})}{8 a \cancel{g}}$ $y_f = \dfrac{2a}{3}  \dfrac{3a\sqrt{5}  5a}{6}  \dfrac{3a(146\sqrt{5})}{16}$ $y_f = \dfrac{2a}{3}  \dfrac{3a\sqrt{5}  5a}{6}  \dfrac{42a18a\sqrt{5}}{16}$ $y_f = \dfrac{2a}{3}  \dfrac{3a\sqrt{5}  5a}{6}  \dfrac{21a9a\sqrt{5}}{8}$ $y_f = \dfrac{16a}{24}  \dfrac{12a\sqrt{5}  20a}{24}  \dfrac{63a27a\sqrt{5}}{24}$ $y_f = \dfrac{16a  12a\sqrt{5} + 20a  63a + 27a\sqrt{5}}{24}$ $y_f = \dfrac{15a\sqrt{5}  27a}{24}$ $y_f = \dfrac{3a(5\sqrt{5}  9)}{24}$ $y_f = \dfrac{a(5\sqrt{5}9)}{8}$ TI Nspire simplification attached ... Last edited by skeeter; November 5th, 2016 at 04:11 PM. 

Tags 
circular, motion, particle, resting, sphere, vertical, wall 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A book resting in equilibrium on a horizontal shelf and against a vertical wall.  Statistics132  Applied Math  2  September 20th, 2015 01:31 PM 
Particle motion problem  slabbxo  Calculus  4  February 23rd, 2014 07:21 PM 
Motion of a particle  musicboy  Calculus  2  January 7th, 2014 07:21 AM 
Vertical motion  Rumblefish  Physics  1  June 23rd, 2012 09:07 PM 
Motion in vertical circle  hoyy1kolko  Physics  1  June 2nd, 2011 10:45 PM 