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Math Help - Can someone please help me with this rotational mechanics question please?

  1. #1
    s3a
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    Can someone please help me with this rotational mechanics question please?

    The question is: http://i.imgur.com/pB7pm.png

    I get (a) 66.1175 rad/s^2 but am stuck right away on (b). It seems to me that (a) and (b) should have the same answer however that is not the case. As for (c) and (d), I haven't been able to get to them yet but it would appreciated if someone could show me their work for (b) => (d).

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    The question is: http://i.imgur.com/pB7pm.png

    I get (a) 66.1175 rad/s^2 but am stuck right away on (b). It seems to me that (a) and (b) should have the same answer however that is not the case. As for (c) and (d), I haven't been able to get to them yet but it would appreciated if someone could show me their work for (b) => (d).

    Thanks in advance!

    \tau=r\times F

    \tau=I\alpha

    Angular acceleration depends on the torque, and the torque depends on the angle between F and r. Therefore angular acceleration depends on the angle between F and r. The moment you release the block, F is perpendicular to r. However as the block falls the angle between F and r becomes more obtuse, making the torque smaller and the angular acceleration smaller.
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  3. #3
    s3a
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    It sounds like I need derivatives and rates of change but this course does not involve calculus...so how would I go about this mathematically?
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  4. #4
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    Quote Originally Posted by s3a View Post
    It sounds like I need derivatives and rates of change but this course does not involve calculus...so how would I go about this mathematically?
    set up a system of two equations.

    let m = block mass

    I = cylinder's rotational inertia

    T = tension in the string

    R = cylinder radius

    a = linear acceleration of the block

    \alpha = angular acceleration of the cylinder


    net force acting on the block ...

    mg - T = ma

    net torque acting on the cylinder ...

    TR = I\alpha

    the "tie" between the two equations is \alpha = \frac{a}{R}

    solve the system for the linear acceleration of the block and the rest should be rather easy to do.
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