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Physics: Rotation Motion

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform disk 0.3m in diameter and having a mass of 2 kg isfree to rotate about its horizontal axis on frictionless bearings.An object with a mass of 0.05 kg is attached to a string wound around the rim of the disk. The object is released from rest and descends with constant linear acceleration. Calculate: (a) the moment of inertia of the disk; (b) the linear acceleration of the descending object; (c) the angular acceleration of the disk; and(d) the tension in the string.

    Diameter of Disk : 0.3 m
    mass of disk = 2 Kg
    mass of attached object = 0.05 kg

    2. Relevant equations
    I = 1/2*mass*radius^2
    Linear acceleration =Force/Mass
    Angular Acceleration=linear acceleration/radius
    3. The attempt at a solution
    A. I = 1/2(2 kg)(0.15 m)^2 = 0.0225 kg.m^2
    B. LA= [(0.05 kg)(9.81 m/s^2)] / 2 kg = 0.24525 m/s^2
    C. AA= (0.2452 m/s^2) / 0.15 m = 1.635 rad/s^2
    D. Not sure where to begin

    I have attempted using known equations to me. I am not sure if I am using right equations to begin with.
    Help will be greatly appreciated.
  2. jcsd
  3. Nov 2, 2014 #2
    for d), what is the force applied to the wheel by the hanging mass?
  4. Nov 2, 2014 #3
    0.05 kg* 9.81 m/s^2 = 0.4905 Newtons
  5. Nov 2, 2014 #4

    Simon Bridge

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    ... it is best to use physics rather than memorized equations.

    Given: disk, diameter d and mass M, turned by a falling mass m connected to the rim.

    a) moment of inertia of the disk .. ##I=\frac{1}{8}Md^2## you used ##I=\frac{1}{2}Mr^2:r=d/2## which is fine.
    Presumably you looked up the formula.

    b) acceleration of falling mass ... you did: ##a=mg/M##, why?
    Did you try drawing free body diagrams for the falling mass and the disk?

    c) angular acceleration of disk, you did ##\alpha = a/r## which is fair enough, but do you know why?

    d) ... start from the free body diagram you needed in part (b).

    Note: the force of the hanging mass on the string (or the rim of the disk) is only going to be ##mg## if the mass is not accelerating.
  6. Nov 2, 2014 #5
    A. For a uniform disk, I = (1/2)MR2. (But the Moment of Inertia can vary depending on the shape of the rigid object.)

    B. Use Newton's 2nd law to find the net force acting on the object. Then isolate the linear acceleration. There are 2 forces involved. You won't have a numerical solution.

    C. Use the linear acceleration and radius to determine the angular acceleration. You won't have a numerical solution.

    D. Use "Net Torque = I*α" (Newton's 2nd Law of Motion for torque) and "Torque = Net Force*r". Substitute the Part C for α and then solve for T. This part is tricky, but I followed through with the isolating the tension and it worked out. Just keep at it if you get stuck.

    Once you solve for the (EDITED) tension, you can go back and solve for the numerical values of linear and angular acceleration.

    Remember to identify the disc's mass and the object's mass with different symbols.
    Last edited: Nov 2, 2014
  7. Nov 2, 2014 #6

    Simon Bridge

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    ... the resulting system of equations can be solved numerically.
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