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Rotation Newton's second law

  1. Apr 1, 2004 #1
    A string wrapped around a cylinder, and is held by a hand (to the up right of the cylinder so the cylinder is rotating clockwise) that is accelerated upward so that the centre of mass of the cylinder does not move. a) Find the tension in the string. b) Find the angular acceleration of the cylinder. c) Find the acceleration of the hand.

    a)The only force that exerts the a torque on the cylinder is the tension:
    Segma torque = I *alpha
    T*R = I * alpha

    The hand force is: Sigma Fy=ma,y
    Mg – T = ma,t (I am not sure if its minus or plus here)

    at=R(alpha)

    (mg-T)/m=R(TR/I)
    so T = mg/(1+(mR^2/I) this is the tension of the string

    b) how to get the angular acceleration here?


    c) mg-(mg)/(1+(mR^2/I)) = m a,t
    so a,t =(1/(1+I/mR^2)g this is the acceleration of the hand.

    Is my reasoning correct?
     
    Last edited: Apr 1, 2004
  2. jcsd
  3. Apr 1, 2004 #2

    NateTG

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    a) You know that the center of mass is not accelerated so
    [tex]F_{net}=0[/tex]
    Since there are only two forces, this should be fairily easy to apply

    b) Once you have the torque, you can use:
    [tex]\tau=I\alpha[/tex]

    c) How about:
    [tex]\theta = \frac{1}{2} \alpha t^2[/tex]
    so
    [tex] 1/2 a t^2 = x = r \theta = \frac{1}{2} r \alphta t^2[/tex]
    so
    [tex] a = r \alpha [/tex]
     
  4. Apr 1, 2004 #3
    Do you find it wrong in a) T = mg/(1+(mR^2/I) ?

    I concidere I = 1/2MR^2 here

    so torque = 1/2M(R^2) * alpha

    but you said F,net = 0?

    Was my reasoning correct with the calculation I gave?

    I am confused.
     
    Last edited: Apr 1, 2004
  5. Apr 1, 2004 #4

    NateTG

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    There are two forces acting on the cylinder:
    The tension of the string
    Gravity.

    Since the cylinder is stationary.
    [tex]F_{net}=0=mg+T \rightarrow T=-mg[/tex]

    The force of gravity is acting on the center of rotation so it exerts zero torque. That leaves the torque due to tension which is:
    [tex]\tau= rT = -rmg[/tex]
    Now we can use
    [tex]\tau=I\alpha \rightarrow \frac{\tau}{I}=\alpha=\frac{-mgr}{\frac{1}{2}mr^2}=\frac{-2g}{r}[/tex]
     
  6. Apr 2, 2004 #5
    But the hand is accelrating upward?!
    I am really lost with this one :(
     
    Last edited: Apr 2, 2004
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