Finding the radius based on time and velocit

  • Thread starter Thread starter ogehsim
  • Start date Start date
  • Tags Tags
    Radius Time
AI Thread Summary
The discussion revolves around calculating the radius of a circular path based on the motion of a particle at two different times. The particle's velocities and accelerations at these times are given, but the user struggles with the formulas for centripetal acceleration and angular velocity. They attempt various calculations to find the radius, using the distance traveled and the relationship between linear and angular motion. Despite multiple approaches, the user is uncertain about their results and seeks verification of their calculations. The conversation highlights the complexity of circular motion problems and the importance of correctly applying physics principles.
ogehsim
Messages
4
Reaction score
0

Homework Statement


A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.70 s, it is at point (4.30 m, 5.50 m) with velocity (3.10 m/s)j and acceleration in the positive x direction. At time t2 = 13.0 s, it has velocity (–3.10 m/s)i and acceleration in the positive y direction. What are the (a)x and (b)y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.


Homework Equations


I honestly do not know... Anything to do with circular motion I guess.


The Attempt at a Solution


physics.jpg


270degrees=4.712 radians

w=angular velocity
w=theta/t
w=4.712/(13-3.7)
w=.5066rad/s

ac=centripetal acceleration
ac=v^2/r
r=v^2/ac
r=3.10^2/ac
r=9.61/ac

ac=r*alpha
ac=r*(w/t)
ac=r*(.5066/(13-3.7))

r=9.61/(.5066r/9.3)
r=89.373/.5066r
r^2=176.41
r=13.28

so 4.3+13.28=17.58 and y would remain 5.5 y (add radius to x, y would remain same as it just is to the direct right of point 1)
17.58 is WRONG
5.5 is RIGHT

I don't get how to solve for r when you don't know ac.
 
Physics news on Phys.org
With regards to your approach:
w*r = v
Here you are confusing centripetal acceleration with angular acceleration.
Although there is a force (and thus an acceleration) maintaining the centripetal motion there is no acceleration of the speed of this motion as the acceleration is always tangent (in this case).
 
Ohhh. So I could only use w*r=v if v at point 1 was 3.10 and if it was then different at point 2?

Is there any way to solve this numerically, or is it purely a logic-out question like what I did for y? It doesn't give me the location of the 2nd point, so I don't know x2, and therefore can't logic my way to the x-coordinate of the origin.
 
Additional Info on Problem (enough wrong answers allows you a hint): At each time, consider where the center of the circle is, and where the particle is along its path around the circle. From the known speed and times, the distance the particle travels can be found. This allows you to calculate the radius of the circle.

13.0s-3.7s=9.3s
9.3s*3.10m/s=28.83m

... can I assume that even though 3.7*4=14.8, not 13, that 3.7s is approximately the time to travel 1/4 of the circle's circumfrence, and 13.0s is the time it takes to travel all of it?
In which case: 13.0s*3.10m/s=40.4m
c=pi*2r
40.4m=pi*2r
r=6.43m

x=4.3m+6.43m=10.73m

10.73 is WRONG, and I only have 1 try left.
I guess this means my assumption is wrong, but I don't understand how if my drawing is correct... So the particle must start b/w point 1 and point 2 and not at point 2... so how do I find the time to travel the entire circle... do I need to find that itme?
 
13.0s-3.7s=9.3s
9.3s*3.10m/s=28.83m

c=pi*2*r
.75c=.75*pi*2*r
28.83m=.75*pi*2*r
r=6.1179m

x=4.3m+6.1179m=10.4179m

(To see if this is within the 2% margin of error for my other answer. If it is, this is also wrong.
10.4179*.02=.208
10.4179+.208=10.626
This does NOT encompass my previous guess of 10.73, making it a valid new answer.)

Can anyone verify my math on this?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top