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Math Help - Rotations and moments

  1. #1
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    Rotations and moments

    I'm studing rotationary stuff at the moment. Things like angular velocity, ang. momentum, ang. acceleration and moment of inertia. But, my concepts aren't very clear. Especially in relating rotation with linear motion.

    ------------------


    Q. A load of mass 20kg is being raised on the end of a light, inextensible rope by means of a winch. The pully of the winch has a radius of 0.2m and moment of inertia 2.5kg.m.m (normal to the plane, through center of pulley). The pulley's bearings are assumed to be smooth. The body starts from rest and accelerates upwards at 0.4m/s.s. Find:-
    (a) the torque driving the pulley
    (b) the Kinetic Energy of the system after 3 seconds.
    ------------------
    My conventions:-
    Right = +ive
    Up = +ive
    Clockwise = +ive
    Variables are considered only in magnitudes, i.e. they are positive.

    I had some severe problems with answering this question. Only when I learnt to ignore the tension (T) at the bottom and the mass did I get it right. But my mind isn't clear. I can't just ignore two forces like that for too long.

    I can't seem to express my problem. But basically, I don't know how the mass, and the two tensions relate to each other and how the linear motion/torque relates to the rotational. How they fit in to the picture.

    So first, I'll just solve the question correctly:
    (Net) Torque = I x (alpha)
    (alpha = angular acceleration)

    (a)
    linear acceleration = radius x alpha
    ==> alpha = 0.4 / 0.2 = 2

    T - 20 x 9.8 = 20 x 0.4 ==> T = 204N

    D - 0.2 x T = 2.5 x 2 ==> D = 25.8N.m.

    (b)
    KE = 0.5 x I x w.w = 0.5 x 2.5 x 6 x 6 = 45J

    Where does the load, and the tension right above the load, fit into the question, WITHOUT CALCULATING THE TENSION AT THE TOP? I mean, what if I didn't calculate the tension in part (a) and instead used the individual torques of the load and the bottom-tension??? Can their torques even apply into the question?

    And why don't the two tensions just nullify each other???

    And what's with the moment of inertia and angular acceleration? How can their multiplication give the resultant torque when I've been simply using (force x perpendicular distance from pivot) all the time before this?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fez_329 View Post
    I'm studing rotationary stuff at the moment. Things like angular velocity, ang. momentum, ang. acceleration and moment of inertia. But, my concepts aren't very clear. Especially in relating rotation with linear motion.

    ------------------


    Q. A load of mass 20kg is being raised on the end of a light, inextensible rope by means of a winch. The pully of the winch has a radius of 0.2m and moment of inertia 2.5kg.m.m (normal to the plane, through center of pulley). The pulley's bearings are assumed to be smooth. The body starts from rest and accelerates upwards at 0.4m/s.s. Find:-
    (a) the torque driving the pulley
    (b) the Kinetic Energy of the system after 3 seconds.
    ------------------
    My conventions:-
    Right = +ive
    Up = +ive
    Clockwise = +ive
    Variables are considered only in magnitudes, i.e. they are positive.

    I had some severe problems with answering this question. Only when I learnt to ignore the tension (T) at the bottom and the mass did I get it right. But my mind isn't clear. I can't just ignore two forces like that for too long.

    I can't seem to express my problem. But basically, I don't know how the mass, and the two tensions relate to each other and how the linear motion/torque relates to the rotational. How they fit in to the picture.

    So first, I'll just solve the question correctly:
    (Net) Torque = I x (alpha)
    (alpha = angular acceleration)

    (a)
    linear acceleration = radius x alpha
    ==> alpha = 0.4 / 0.2 = 2

    T - 20 x 9.8 = 20 x 0.4 ==> T = 204N

    D - 0.2 x T = 2.5 x 2 ==> D = 25.8N.m.

    (b)
    KE = 0.5 x I x w.w = 0.5 x 2.5 x 6 x 6 = 45J

    Where does the load, and the tension right above the load, fit into the question, WITHOUT CALCULATING THE TENSION AT THE TOP? I mean, what if I didn't calculate the tension in part (a) and instead used the individual torques of the load and the bottom-tension??? Can their torques even apply into the question?

    And why don't the two tensions just nullify each other???

    And what's with the moment of inertia and angular acceleration? How can their multiplication give the resultant torque when I've been simply using (force x perpendicular distance from pivot) all the time before this?
    The tension is the same magnitude at both ends, it is just acting in
    opposite directions, so it does not matter which end you consider.

    This is because of Newton's third law, and the fact that you are assuming
    a light rope.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by fez_329 View Post
    And what's with the moment of inertia and angular acceleration? How can their multiplication give the resultant torque when I've been simply using (force x perpendicular distance from pivot) all the time before this?
    I don't know what you have been doing before but the angular acceleration
    satisfies:

    Torque = Moment of Inertia x angular acceteration

    it is a result of integrating Newton's equations over the body.

    RonL
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