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Pivoting Stickstuck on two smaller parts

  1. Nov 24, 2003 #1
    A picture has been included for your viewing pleasure.

    A stick of uniform density with mass M = 7.7 kg and length L = 0.6 m is pivoted about an axle which is perpendicular to its length and located 0.16 m from one end. Ignore any friction between the stick and the axle.

    =======================
    a) What is the moment of inertia of the stick about this axle?

    The stick is held horizontal and then released.
    b) What is its angular speed as it passes through the vertical
    c) What is its angular acceleration as it passes through the vertical position?
    d) What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical?

    =============
    part a) Not too bad. I used the parallel axis theorem, and ended up with the correct answer of .38192 kg*m^2
    b) Not too bad either. Used energy of conservation where mgh=.5I*w^2
    where w is omega, and I is moment of inertia.
    Got correct answer of 7.4379 rad/s

    c) Now I'm stuck. I know that the forumula for calculating net torque is:
    NetTorque=I*alpha

    I know what (I) is, alpha I don't know, but I don't know how to calculate torque though for this problem. I know that in general, torque = r*F sin phi

    So I need some help figuring out this torque business. :p

    part d) Also stuck. Not sure really what to do here. :p
     

    Attached Files:

  2. jcsd
  3. Nov 24, 2003 #2

    Doc Al

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    What are the forces acting on the stick? What torque do they produce?
     
  4. Nov 24, 2003 #3

    NateTG

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    You need to calculate the net torque on the stick about the pivot caused by gravity. You can either use calculus and integrate the torque along the stick, or determine the center of gravity of each side, and use them to determine the torque from each side.

    For part d, there are two componets to the force exerted by the stick, one due to gravity, and the other due to the centripetal acceleration of the stick.
    Remeber that the torque exerted by gravity will change as the stick drops, so you should probably use energy to calculate the angular speed of the stick. You could also do the integral of angular acceleration, but it's a bit tricky.
     
  5. Nov 24, 2003 #4

    Doc Al

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    Remember, they are asking about the vertical position. Which way does gravity act? The "calculation" should be trivial.
     
  6. Nov 24, 2003 #5
    Ok. So yea, gravity is the only force in the vertical position. I am probably messing up the calculation of the net torque. I'm still new to this!

    So for net torque, is it going to be r*m*g*sin 90?
    or r*7.7kg*9.8m/sec^2*sin 90?

    but I'm not sure what my r is. It probably is really a trivial calculation, but I'm just not seeing it. :(
     
  7. Nov 25, 2003 #6

    Doc Al

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    What makes you think θ = 90?

    r is a vector from the pivot point to the point where the force is applied. Where is the force applied? (Think gravity.)

    So... What's the direction of r? What's the direction of the force?
    What is the angle θ between these two vectors? (That's the meaning of θ in your torque definition: Torque = rF sinθ )
     
  8. Nov 25, 2003 #7
    Ok. The force of gravity is pointing downwards, thus r vector is also pointing downwards. Which means my theta is 0.

    Like Doc said, r is a vector from the pivot point to the point where the force is applied. So is the r zero in this case?
     
  9. Nov 25, 2003 #8

    NateTG

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    Yes. τ=0.

    I didn't read the problem well, and thought that the question was about the torque when the stick was horizontal.
     
  10. Nov 25, 2003 #9

    Doc Al

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    No, r is not zero. (You can treat the weight as acting at the center of mass of the stick; r goes from the pivot point to the center of mass.) But who cares, since theta is 0!
     
  11. Nov 25, 2003 #10
    Lol, duh. I kept on thinking that sin 0 equaled 1. Man, I need some sleep. :p

    part d)Ok, the only forces working is gravity and the centripetal force.

    I know that centripetal force is F=m(v^2/r)
    and the force of gravity is going to be m*g

    I can get angular velocity by V=omega*r

    so if I used conservation of energy, it goes from potential energy to kinetic energy
    so mgh=.5*I*omega^2
    or mgh=.5(1/12*m*L^2)*omega^2

    so m=7.7kg
    g=9.8m/s/s
    L=.60m
    but what is h equal to?

    Basically all I got to do is find omega, convert it to radial acceleration, and use that to find the centripetal force, and add that force to the force of gravity.

    I'm prolly going about this wrong, aren't I?
     
  12. Nov 25, 2003 #11

    NateTG

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    Well, it's not the total height, but the change in height you care about, but otherwise you look like you're right on the money.
     
  13. Nov 25, 2003 #12

    Doc Al

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    Careful with the lingo here. "Centripetal Force" is not a separate kind of force like gravity or tension in a rope. The forces on the stick are: gravity (down) and the contact force exerted by the pivot (pulling up). These force add up to be a centripetal force.
    As NateTG points out, h should be the change in height (of the center of mass).
     
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