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Newton's 2nd Law for Rotation

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data
    In the figure http://i.imgur.com/Y5Vc7.gif shows a uniform disk (M = 2.5kg, R = 0.20 m) mounted on a fixed horizontal axle. A block (m = 1.2kg) hangs from a massless cord that is wrapped around the rim of the disk. The cord does not slip and there is no friction.

    Find the acceleration of the block.

    2. Relevant equations
    [tex]\tau_{net} = I\alpha[/tex]
    [tex]a_t = \alpha r[/tex]

    3. The attempt at a solution
    1. [tex]F_{net} = -ma = T - mg[/tex]
    2. [tex]a = -\frac{T}{m} + g[/tex]

    3. [tex]\tau_{net} = I\alpha = \frac{1}{2}MR^2\alpha = -TR[/tex]
    4. [tex]\alpha = \frac{a_t}{r} = \frac{a}{r}[/tex] the tangential acceleration is also the linear acceleration.

    5. [tex]-TR = \frac{1}{2}MR^2(\frac{a}{r})[/tex]
    6. [tex]T = -\frac{1}{2}Ma[/tex]

    Plugging equation 6 into 2 yields:
    [tex]a = \frac{Ma}{2m}+g[/tex]
    Some manipulation

    [tex]a = \frac{g}{1-\frac{M}{2m}}[/tex]
    [tex]a = \frac{9.8}{1-\frac{2.5}{2*1.2}} = -235.2[/tex]

    This is obviously wrong.

    5. Solution
    7. [tex]F_{net} = ma = T - mg[/tex]
    8. [tex]a = \frac{T}{m} - g[/tex]

    plugging equation 6 into equation 8 yields:
    [tex]a = -\frac{g}{1+\frac{M}{2m}}[/tex]
    [tex]a = -\frac{g}{1+\frac{2.5}{2*1.2}} = -4.8 ms[/tex]

    6. My question
    Obviously my model for the net force is wrong. I used -ma instead of ma.

    Why is it okay for me to consider the direction of the net torque to be -TR instead of TR for equation 3, but it isn't okay for me to consider the direction of the net force to be -ma instead of ma for equation 1? The net force is pointing downward isn't it?

    Also vice-versa
    Why is it okay for me to consider the direction of the net force to be ma instead of -ma for equation 1, but it isn't okay to consider the direction of the net torque TR instead of -TR for equation 3? shouldn't i be consistent?

    Thanks in advance. This stuff always gets me in physics.
     
    Last edited: Apr 8, 2012
  2. jcsd
  3. Apr 8, 2012 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    F is always equal to ma. They are vectors, which means they have both magnitude and direction. In this 1D problem, this translates to them being either negative or positive, depending on whether they point downward or upward. But they must both have *same* sign, regardless of what it is. If you write F = -ma, then what you are saying is that the net force points in the opposite direction of the acceleration! I.e. if the net force is upward the acceleration is downward, and vice versa. This is clearly not the case.

    What's tricky here is that the sign here is intrinsic i.e. it is not indicated explicitly. F and a are numbers that could either be positive or negative e.g. F = +10 N, a = +4 m/s^2, or F = -10 N, a = -4m/s^2. So if the net force is pointing downward, then that will be reflected in your numerical answer. But the reason you don't put the negative sign in explicitly is because F and a are assumed to be signed quantities that carry that sign already.

    EDIT: the reason why it was okay for you to write -TR was because you defined T as the force *on the hanging mass* due to the tension in the rope. The force *on the disc* due to the tension in the rope will necessarily be -T then. It's a different force.
     
    Last edited: Apr 8, 2012
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