1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Direction of vector acceleration in circular motion

  1. Apr 18, 2017 #1
    There is a problem in my Physics textbook which says:

    1. The problem statement, all variables and given/known data
    "A car runs counter-clockwise in a circular lane of [itex]1 km[/itex] of diameter, going through the south extreme at [itex]60 km/h[/itex] on the instant [itex]t = 0[/itex]. From that point onwards, the driver accelerates the car uniformely, reaching [itex]240 km/h[/itex] in [itex]10[/itex] seconds.

    Determine the vector average acceleration between [itex]t = 0[/itex] and [itex]t = 10s[/itex]"

    P.S: Sorry for any typos, I had to translate it from a textbook written in Portuguese and I'm not too familiar with english technical language yet. Anyways...

    2. Relevant equations

    3. The attempt at a solution

    So what we know:
    Diameter of the circunference: [itex]1km[/itex];
    Radius of the circunference: [itex]\frac{D}{2} = 500m[/itex];
    Initial position: [itex]0[/itex];
    Initial velocity: [itex]60km/h[/itex] or [itex]16.7m/s[/itex];
    Velocity after [itex]10s[/itex]: [itex]240km/h[/itex] or [itex]66.7m/s[/itex];
    Acceleration in the interval: [tex]\vec a = \frac{\Delta \vec{V}}{\Delta t} = \frac{66.7-16.7}{10-0} = \frac{50}{10} = 5m/s^2 [/tex]Total displacement: [tex]S = S_0 + \vec{V_0}t + \frac{at^2}{2} \Rightarrow S = 0 + 16.7 \times 10 + \frac{5 \times 10^2}{2} = 167 + 250 = 417m[/tex]Now, vectors contain sense, magnitude and direction, so, as it is a circular motion problem, we should find out the angle of the arc inscribed by the displacement:[tex]\theta = \frac{360^\circ}{2\pi r} \times 417 \Rightarrow \theta = \frac{360^\circ}{2\pi \times 500} \times 417 = \frac{360^\circ}{1000 \pi} \times 417 \simeq 47.8^\circ[/tex]Now, I'm not really sure what to do next, but I've tried Law of Cosines:[tex]\vec{V_r}^2 = \vec{V_1}^2 + \vec {V_2}^2 - 2 \vec{V_1} \vec{V_2} \times \cos A \Rightarrow \vec{V_r}^2 = 16.7^2 + 66.7^2 - 2 \times 16.7 \times 66.7 \times \cos 47.8 = 278.89 + 4448.89 - 1496.4 = 3231.38 \Rightarrow \vec{V_r} = \sqrt{3231.38} \simeq 56.8m/s[/tex]Continuing:[tex]\vec{a_avg} = \frac{\Delta \vec{V}}{\Delta t} = \frac{56.8}{10} = 5.68m/s^2[/tex]OK. So that matches the answer in my textbook, in regards to the magnitude. But the problem is that my answer says nothing about the direction and the sense of the vector, which should be [itex]60.3 ^\circ[/itex] north from direction east, according to my textbook.

    And now I have no idea about what to do to be able to calculate that. My understanding of the circular motion is still quite blurry, and I have some other questions too:

    1) How was I able to solve this problem using a linear movement formula for displacement and acceleration? Why didn't I need to use the angular formulas and all of that?

    2) I couldn't grasp what velocity and acceleration meant in this problem. Were they both tangencial? Was acceleration centripetal...?

    I've invested some time writing this so if someone could help me I'd be really grateful. I have nobody else to ask. x_x
     
    Last edited: Apr 18, 2017
  2. jcsd
  3. Apr 18, 2017 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You could have used angular formulas, if you wanted to. But you were given the tangential velocities, so it makes sense to use the linear motion formulas to calculate the displacement along the path and thus the arc traveled.

    To answer the problem, use the basic definition of acceleration: The change in the velocity over time. What is the initial velocity? (Magnitude and direction!) What is the final velocity? (Magnitude and direction.) Find the change in velocity by subtracting those velocity vectors. (That way you'll get the magnitude and direction of the average acceleration.

    In this problem, the acceleration has both a centripetal and a tangential component. (If it moved with constant speed, the acceleration would be purely centripetal.)
     
  4. Apr 18, 2017 #3
    Thank you for the input. I'm able to grasp that better now.

    As for the other part, I was already able to find the magnitude [itex](5.68m/s^2)[/itex], I just can't figure out a way to find the direction as it is given in the textbook ([itex]60.3^\circ[/itex] north from east).
     
  5. Apr 18, 2017 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You will find both magnitude and direction by subtracting the two velocity vectors. ##\vec{a}_{ave} = \Delta \vec{v}/\Delta t##.

    Are you familiar with vector subtraction?
     
  6. Apr 18, 2017 #5
    By definition, I know it is the same as addition by the opposite of the other vector. Is there anything else on that? D:
     
  7. Apr 18, 2017 #6

    Doc Al

    User Avatar

    Staff: Mentor

    That's all there is to it. I suggest you draw yourself a diagram of those vectors being subtracted. (Then you'll understand how you were able to find the magnitude using the law of cosines.)

    Another approach is to subtract the components of the initial and final velocities. That works too.
     
  8. Apr 18, 2017 #7
    Yes, I'm trying to make a diagram. As you said, the velocities are tangential, so they should be tangent to the circunference along the south (since it is where the motion started). Is that right?
     
  9. Apr 18, 2017 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Both velocities are tangential to the circular path. If you use x&y to represent east & north, then the initial velocity is purely in the x-direction. The final velocity is at some angle (which you figured out) with respect to the x-axis.
     
  10. Apr 18, 2017 #9
    After a few attempts, I seem to be lost. x_x

    I can't apply the pythagorean theorem, and I've already used law of cosines, so how should I do that vector sum?
     
  11. Apr 18, 2017 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Please tell me the magnitude and direction of both vectors.

    Then find the components and subtract.
     
  12. Apr 18, 2017 #11
    OK. So I think the magnitudes and directions were [itex]16.7[/itex] to the east and [itex]66.7[/itex], [itex]47.8^\circ[/itex] to the northeast.

    So their components ended up being <16.7, 0> and <44.8, 49.4>. Subtracting them I get <28.1, 49.4>.

    Oh my god. I did it! Thank you very much!
    That was so dumb of me, hahaha.
     
    Last edited: Apr 18, 2017
  13. Apr 18, 2017 #12

    Doc Al

    User Avatar

    Staff: Mentor

    Perfect!

    Now find the magnitude and angle of that vector.
     
  14. Apr 18, 2017 #13
    Yes! I already did that the moment I saw the triangle. :33
     
  15. Apr 18, 2017 #14

    Doc Al

    User Avatar

    Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted