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Homework Help: Graphical analysis of angular displace v time

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data


    a) Calculate the magnitude of the constant average torque that acted on the system.
    b) Assuming constant torque how far did the object rotate between t = 2 and t = 20 seconds?

    2. Relevant equations

    3. The attempt at a solution

    Just checking to see if my answers are correct..

    part a) since the graph is the change in θ over change in t, I can calculate the slope to find the ω. And then to find the torque, I needed to calculate 2 ω so I can find the [itex]\alpha[/itex] which then leads me to the torque.

    So to find the 2 ω, I chose the points

    ω1 = (100 rad - 1 rad) / (10s - 1s) = 11 rad/s
    ω2 = (400rad - 100rad) / (20s - 10s) = 30 rad/s

    [itex]\alpha[/itex] = (30 rad/s - 11 rad/s) / (20s - 1s ) = 1 rad/s2

    [itex]\tau[/itex] = I[itex]\alpha[/itex] = (16kgm^2)(1 rad/s2) = 16 N * m

    part b) is the question just asking for the change in angular displacement between t = 2 and t = 20?

    if so I can figure that out by just looking at the graph 300 rad - 4 rad = 296 rad
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2


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    Sorry, but that approach is not valid. Calculating [itex] \omega [/itex] using [itex] \omega = \frac{\theta_2 - \theta_1}{t_2 - t_1} [/itex] is only valid for constant angular velocity. But here, the system is accelerating, so that approach is not valid. If your angular velocities are not valid, you angular acceleration will not be valid either. (Hint: it is not. The correct angular acceleration is not 1 rad/s2.)

    Instead, you your kinematics equation,

    [tex] \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2. [/tex]

    As of yet you don't know what [itex] \theta_0, [/itex] [itex] \omega_0 [/itex] and [itex] \alpha [/itex] are.

    But you can find them! :smile: You know what the relationships is between [itex] \theta [/itex] and [itex] t [/itex] (Hint: the given graph). So go ahead and plug in three sets of ([itex] t, \theta [/itex]) numbers into the kinematics equation.

    That will leave you with three simultaneous equations and three unknowns (read: "solvable").

    Boldface mine.

    You should look at the graph again: your "300" number is off by 100.

    Anyway, when you're finished there should be two ways to find part b), one by looking at the graph, and the other by using your kinematics equation. They should both agree. :wink:
  4. Apr 30, 2013 #3
    hmm, I'm having a hard time how to use the data points in the kinematic equation.

    So I am to choose 3 points, I can see the most easiest points to read at is at times 1, 10, and 50.
    so that gives me the 3 data points of this formation (t,θ) = (1, 1), (10, 100), (50, 2500)

    and the relationship between θ and t is that it is the angular velocity?

    I can't see how I am able to solve 3 equations with 3 unknowns using those points?

    so far my 3 equations look like..

  5. May 1, 2013 #4


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    If you were to find the instantaneous change in θ divided by an infinitesimal change in t, that would give you the instantaneous ω. But there is no easy way to do that by directly looking at the graph. (Not in this problem, anyway.) The value of ω is different for every point in time.

    I would have used the (2 sec, 4 rad) instead of the (50 sec, 2500 rad) point. But that's just me. Either way works. :approve:

    So you have three equations and three unknowns. Solve for θ0, ω0 and α.
    Last edited: May 1, 2013
  6. May 1, 2013 #5


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    Just for clarity, your equations are:

    1 [rad] = θ0 + ω0(1 ) + (1/2)α(1 )2
    100 [rad] = θ0 + ω0(10 ) + (1/2)α(10 )2
    2500 [rad] = θ0 + ω0(50 ) + (1/2)α(50 )2

    Multiplying a few things out gives,

    1 = θ0 + ω0 + (0.5)α
    100 = θ0 + (10)ω0 + (50)α
    2500 = θ0 + (50)ω0 + (1250)α

    There are several ways to solve sets simultaneous, linear equations such as this set: linear algebra, method of addition, method of substitution, Cramer's rule, etc.
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