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Torque/Rotational Dynamics Help Needed

  1. Mar 12, 2009 #1
    Basically, I've missed a week straight of school due to a family illness, and I get back today to find that our normal physics teacher is sick, and we have a substitute. I was able to read a bit on my own and understand most of it, but these are some that have stumped me a bit.

    I do not know if I'm allowed to post more than one question, but I have four. Please check out my method (make sure I am going in the right direction, or if I need help). Sorry in advance for my bad latex

    1. The problem statement, all variables and given/known data

    Compare the linear and angular velocities of the moon around the earth between 1912 and now if the average distance is 3.84x105km, average linear velocity is 3.68x103kph

    and the distance from the moon to the Earth in 1912 was 3.56x105km, and the 2009 distance is 4.07x105km

    (I don't even know if these are accurate numbers, but they're the ones I was given)

    2. Relevant equations

    [tex]v_{T}= r\omega[/tex]

    3. The attempt at a solution
    This was the only equation that seemed to pertain to the question. The thing I am confused about is finding the different variables for different dates

    To find the angular velocities in the different years, I did

    [tex]\frac{v_{T}}{r_{1912}}= \omega_{1912} }[/tex] for 1912 and I got 1.03x10-2kph

    and

    [tex]\frac{v_{T}}{r_{2009}}= \omega_{2009} }[/tex] for 2009 and I got 9.04x10-3kph

    They seem like really small numbers...


    For the linear velocities, I assumed that [tex]\frac{v_{Tavg}}{r_{avg}}=\omega_{avg}[/tex] and got 9.58x10-3kph

    With the number from that, I plugged that into

    [tex]v_{T1912}=r_{1912}\omega_{avg} [/tex] and got 3410.5kph

    and

    [tex]v_{T2009}=r_{2009}\omega_{avg} [/tex] and got 3899.1 kph



    1. The problem statement, all variables and given/known data

    A ball with a radius of 2m and mass of 1x104kg rolls down an incline plane with an angular acceleration of 6.13rad/s2

    What is the torque acting on the ball, and the linear/angular velocities at the bottom of the plane?

    2. Relevant equations

    see below

    3. The attempt at a solution

    I honestly have not tried this problem, and I do not know which equations to use. Seems to me, that I would need an angle of the inclined plane to even do anything with this problem, but there literally is none. If somebody could give me a hint, I have my paper and pencil ready to go, as well as a calculator.


    1. The problem statement, all variables and given/known data

    A string has an 89N tensile strength. One end of a 1m piece of this string is tied to a 4.54kg toy car. The other end is attached to a ring slipped over a pole so that the car can rotate around it freely. Calculate the maximum angular speed of the car without breaking the string. How many revolutions does the car travel in 1 minute (friction neglected)?

    2. Relevant equations

    [tex]F_{c}=mr\omega^{2}[/tex]

    [tex]\omega=\frac{d\theta}{dt}[/tex]


    3. The attempt at a solution

    Using the first equation, I solved for omega getting
    [tex]\omega=\sqrt\frac{F_{c}}{mr}[/tex]

    [tex]\omega=\sqrt\frac{89N}{4.54}[/tex]

    Getting 4.42rad/s

    I wasn't quite sure on the revolutions part, although it should be simple. I'm fairly sure I'm incorrect. I did

    [tex]\omega=\frac{d\theta}{dt}[/tex] so (4.42)(60s)=revolutions, and I got 265.2

    Seems like too much, so I converted it into radians, and got 4.63 which is in my opinion a more feasible answer, but I thought I was in radians already...

    1. The problem statement, all variables and given/known data

    A 10m centrifuge can spin at a linear velocity of 17.3 m/s and brake to a stop in 80 seconds

    a)What is the torque produced?
    b)What is the centripetal acceleration?
    c)What is the centripetal force an an 80kg astronaut?

    2. Relevant equations

    torque, see below

    [tex]a_{c}=\frac{v_{T}^{2}}{r}[/tex]

    [tex]F_{c}=\frac{mv_{T}^{2}}{r}[/tex]

    3. The attempt at a solution

    This is another one that I have not tried. I still feel like I'm not given enough information. How can I find the torque without a force (a mass to find the force). I can't use the 80kg astronaut, as far as I know, otherwise it would have been the first question asked.

    For the others, it looks like I could just plug and chug into the equations.


    Please and thank you!
     
  2. jcsd
  3. Mar 12, 2009 #2
    For the first problem, when you use Vt = r*omega, the units of omega are in rad/s. Thus if r is in km, Vt in kps, then you get
    kps [=] km*rad/s [=] km/s OK
    Better check your units on these calcs; you may have some problems there.

    Second problem: Set it up in symbols as though you had the angle. You are given the angular acceleration which will probably give you the key to backing out the angle (I have not worked it either, but this is where I would start). Write down what you can, using symbols, even where you don't have certain pieces of data, you still know that there is an angle so give it a name and work with it.

    Third problem: Your analysis gave you 265 radians. You needed to divide this by 2*pi to get revolutions, just roughly 44 rev. It is not clear whether this car is supported by something else, of if the string must both support the car and resist centrifugal loading; the problem statement is unclear.

    Fourth problem: I think you are short some inertia data for the centrifuge. Also, you are short some information about the way in which the braking is done. -- is it constant deceleration, or some other pattern?
     
  4. Mar 12, 2009 #3
    First Problem: How would omega have to be in rad/s? I thought that because the units were all in kilometers, and kph, then the units would at least be radians per hour...? As long as the units are all the same. The radians would just be on a larger scale due to the large radius of the "circle" made by the orbit.

    Second problem: I still don't quite know where to begin with it. I could use

    tau= Fd(sin<) for the torque

    And all of the equations for Rotational Motion with Constant acceleration require a change in angle, or a change in time -- both of which are unknowns.

    Third problem: Ah, of course. Thank you. I think the car is just supported by frictionless ground.

    Fourth problem: I would assume it's constant deceleration just for the sake of simplicity. Unfortunately, all of these problems are straight off of the worksheet, written as is. These are the only 4 out of the 13 we were given that I just didn't understand very well.


    Thank you for your reply.
     
  5. Mar 12, 2009 #4
    OK, yes, you can use units of rad/hr, but your first calculated result, which you stated as 1.03x10-2kph should actually be 1.03x10e-1 rad/hr because it is a value for omega.

    For the second problem, draw a FBD for the ball, and that will include several forces. Then write Newton's second law, in one case parallel to the ramp and in the other case perpendicular to the ramp. To do this, you don't have to say anything at all about constant acceleration, only that there is an acceleration.

    I don't know what else to say about the 4th problem.
     
  6. Mar 12, 2009 #5
    Alright, but wouldn't the numbers I calculated still be correct? I just have to change the units?

    Like, if I used the same equation vT = rw (w will be omega) for vT= 2 m/s, and r= 1m

    w would equal 2 rad/s, right? The units come out of the equation... Or do I still have to convert them?


    So do F=ma, and F=mg? I'm still confused. I drew the FBD... There's no friction, and there's just force due to gravity, and the normal force along with it's opposite force. I still don't see how you can do it without an angle.



    And don't worry about the 4th problem.



    Thank you again.




    Edit: I think I had a revelation with the second problem and what you told me to do.

    F=mg(sin <)
    F=ma

    So I could say that ma=mg(sin <) and (sin <) = ma/mg = a/g

    Which is 0.624872

    So now I can go back and say that torque= Fd(sin <), plug in mg(sin <) for F and plug and chug, right?

    So torque = (10000kg)(9.81m/s2)(0.624872)(2m)(0.62487)= 7.7x104 Nm

    Aha! I think.......
     
    Last edited: Mar 12, 2009
  7. Mar 12, 2009 #6
    Didn't I say that your numbers were correct but that your units needed to be fixed?

    The problem statement you gave for the ball coming down the plane said that the ball rolls down the plane. Now you say that there is no friction. It cannot be both ways. Without friction, the ball will not roll, and the ball will have no angular acceleration; it will simply slide down the plane without rotation at all without friction.

    I think it is more useful to write Sum F = ma and W = mg. Newton's second law talks about the sum of ALL forces acting on the body. The weight is one particular force acting on the body. Taking too many short cuts is the cause of your confusion.
     
  8. Mar 12, 2009 #7
    Ah, good point on the friction.


    Did I still do it correctly, though?


    Oh, and I misread your number that you said for problem 1. I thought you not only changed the units, but the number as well. I'm sorry.



    Edit: Oops, I just noticed that my previous post's edit was after your post!
     
  9. Mar 13, 2009 #8
    You never said what d was.l
     
  10. Mar 13, 2009 #9
    Wouldn't it be the radius of the ball because it's the center of the ball's mass, which is 2m?
     
  11. Mar 13, 2009 #10
    That's a nice idea, but you never said so when you wrote it down. Frequently d stands for diameter, not radius.
     
  12. Mar 13, 2009 #11
    I had the radius in the first post, the first time I posted it. I never edited my first post.

    We use d as the distance to the center of rotation (which is the radius) so that's just how I wrote it.


    Thank you for your help, though.
     
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