
Physics Physics Forum 
 LinkBack  Thread Tools  Display Modes 
November 3rd, 2019, 04:01 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  How do I find the centripetal acceleration when vectors are given?
The problem is as follows: A lactose solution is put in a test tube inside a centrifuge as shown in the graph from below where a small dot represents the tube with the sugar (view from the top). Assuming that the sugar sample along the test tube behaves as a particle it spins following a rotation with constant angular acceleration. The radius of the apparatus is $32\pi\,m$ and its angular acceleration is $\frac{\pi}{6}\,\hat{k}\,\frac{rad}{s^{2}}$. When $t=0\,s$ the tube is at point labeled $A$ and it has an angular speed of $\frac{\pi}{12}\,\hat{k}\,\frac{rad}{s}$. Find the centripetal acceleration when $t=4\,s$ The alternatives given on my book are: $\begin{array}{ll} 1.&9\pi\left(\hat{i}+\sqrt{3}\hat{j}\right)\\ 2.&4\pi\left(\hat{i}+3\hat{j}\right)\\ 3.&9\pi\left(\hat{i}\sqrt{3}\hat{j}\right)\\ 4.&4\pi\left(\hat{i}3\hat{j}\right)\\ 5.&3\pi\left(\hat{i}\sqrt{3}\hat{j}\right)\\ \end{array}$ For this particular problem the only relevant equation for position which I can recall for rotation with constant angular acceleration is: $\theta (t)=\theta_{o}+\omega_{0} t + \frac{1}{2}\alpha t^2$ But from then is where I'm stuck. I don't know how to relate this equation with the values that are given other than: When $t=0$, then $\omega_{0}=\frac{\pi}{12}$ $\theta (t)=\theta_{o}+ \frac{\pi}{12} t + \frac{1}{2}\left(\frac{\pi}{6}\right) t^2$ $\theta (t)=\theta_{o}+ \frac{\pi}{12} t + \frac{\pi}{12}t^2$ $\theta (0)=\theta_{o}+ \frac{\pi}{12} (0) + \frac{\pi}{12}(0)^2$ $\theta(0)=\theta_{0}$ But since I don't know that value or the condition I don't know what further thing I can do. What about the other quantities given. How can I relate them? Can somebody help me here?. 
November 3rd, 2019, 04:35 PM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 
$\omega = \omega_0 + \alpha t$ If the radius is constant, then $a_{centripetal} = \omega^2 r$. 
November 3rd, 2019, 05:25 PM  #3  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
$\omega_{f} = \omega_{0} + \alpha t$ $\omega_{f} = \frac{\pi}{12} + \left(\frac{\pi}{6}\right)(4)$ $\omega_{f} = \frac{9\pi}{12}= \frac{3\pi}{4}$ From there: $a_{c}=\left(\frac{3\pi}{4}\right)^2\left(\frac{32 }{\pi}\right)$ $a_{c}=18\pi$ But that's how far I went needless to say, how about the vectors shown in the alternatives?. Does it imply that I must obtain the position? From which should I use the tangent function to obtain the coordinates?. Now for this part I'm assuming when the object is at: $A\,\theta(0)=0$ $\theta(t)=\theta_{0}+\omega_{0} t + \frac{1}{2}\alpha t^2$ $0=\theta(0)=\theta_{0}+\omega_{0}(0) + \frac{1}{2}\alpha (0)^2$ $\theta_{0}=0$ Then this is simplified to: $\theta(t)=\omega_{0} t + \frac{1}{2}\alpha t^2$ The initial speed is given: $\theta(t)=\frac{\pi}{12} t + \frac{1}{2}\frac{\pi}{6} t^2$ $\theta(t)=\frac{\pi}{12} t + \frac{\pi}{12} t^2$ Then by pluggin in $t=4$ there should be: $\theta(t)=\frac{\pi}{12} (4) + \frac{\pi}{12} (4)^2=\frac{5\pi}{3}$ Which gives: $\theta=\frac{5\pi}{3}$ which is in the fourth quadrant. Since what is being asked is the position all that would be left to do is to take the sine and cosines functions to obtain the direction of such vector using the known radius which is $\frac{32}{\pi}$. Horizontal: $\frac{32}{\pi}\cos\left(\frac{5\pi}{3}\right)$ $\frac{32}{\pi}\cos\left(\frac{5\pi}{3}\right)$ $\frac{32}{\pi}\times\frac{1}{2}=\frac{16}{\pi}$ Vertical: $\frac{32}{\pi}\sin\left(\frac{5\pi}{3}\right)$ $\frac{32}{\pi}\times\frac{\sqrt{3}}{2}=\frac{16\sqrt{3}}{\pi}$ Therefore the position vector would be: $\frac{16}{\pi}\hat{i}\frac{16\sqrt{3}}{\pi}\hat{j}$ But that's how far I went: A factorization led me to: $\frac{16}{\pi}\left( \hat{i}\sqrt{3}\hat{j}\right )$ Therefore the acceleration would be: $18\pi \times \frac{16}{\pi}\left( \hat{i}\sqrt{3}\hat{j}\right )$ However this does not appear within the alternatives. Could it be that I'm not getting the right picture? Last edited by Chemist116; November 3rd, 2019 at 06:18 PM.  
November 3rd, 2019, 05:56 PM  #4 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle 
What does power ranger hold in his hand? It sure looks like a "cylinder with cap".

November 3rd, 2019, 06:34 PM  #5  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
Okay going back to my question, can you perhaps take a look into it. Did I misunderstood any of the steps?.  
November 3rd, 2019, 07:19 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
Is $r=32\pi$, or $\dfrac{32}{\pi}$ ? The original problem statement says $32\pi$, yet your calculations use $\dfrac{32}{\pi}$ I'll assume $r=\dfrac{32}{\pi}$ $\theta(4) = 0 + \dfrac{\pi}{12} \cdot 4 + \dfrac{\pi}{12} \cdot 16 = \dfrac{5\pi}{3}$ $\omega(4) = \dfrac{\pi}{12} + \dfrac{\pi}{6} \cdot 4 = \dfrac{3\pi}{4}$ $a_c = r \omega^2 = \dfrac{32}{\pi} \left(\dfrac{3\pi}{4}\right)^2 = 18\pi$ at $t=4$, the particle is at position $\theta = \dfrac{5\pi}{4}$, indicating the direction of $a_c$ (toward the center) is $\dfrac{2\pi}{3} = \dfrac{1}{2} \hat{i} + \dfrac{\sqrt{3}}{2} \hat{j}$ $a_c = 18\pi\left(\dfrac{1}{2} \hat{i} + \dfrac{\sqrt{3}}{2} \hat{j}\right) = 9\pi\left(\hat{i}+\sqrt{3} \hat{j} \right)$ 

Tags 
acceleration, centripetal, find, vectors 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How can I find the speed and the displacement of an object from acceleration graph?  Chemist116  Physics  6  November 1st, 2019 05:14 PM 
Find acceleration ... velocity is t sin 3t ms^1  DomB  Calculus  4  May 16th, 2017 11:39 AM 
Trouble comparing angular acceleration to linear acceleration :  SophiaRivera007  Physics  2  December 29th, 2016 10:56 PM 
Centripetal force  SRAKER123  Physics  5  December 18th, 2016 03:56 PM 
Centripetal force  hoyy1kolko  Physics  3  May 24th, 2011 08:17 AM 