Physics - Rate of Rotation?

I need help solving this problem, any help would be appreciated.

A student stands on a platform that is free to rotate and holds two dumbbells, each at a distance of 65 cm from his central axis. Another student gives him a push and starts the system of student, dumbbells, and platform rotating at 0.75 rev/s. The student on the platform then pulls the dumbbells in close to his chest so that they are each 22 cm from his central axis. Each dumbbell has a mass of 1.00 kg and the rotational inertia of the student, platform, and dumbbells is initially 7.40 kg · m2. Model each arm as a uniform rod of mass 3.00 kg with one end at the central axis; the length of the arm is initially 65 cm and then is reduced to 22 cm. What is his new rate of rotation?

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  • Ossi G
    Lv 7
    9 years ago
    Favorite Answer

    find the difference in rotation inertia of the arms + dumbbells:

    initial Inertia of both arms alone: I = 1/12*m*l^2

    I i = 1/12*6*1.3^2 = 0.845 kgm^2

    inertia of both dumbbells:

    I = 2*mR^2 = 2*1*0.65^2 = 0.845 kgm^2

    together: 1.69 kgm^2

    -------

    final inertia of both arms:

    I = 1/12*6*0.44^2 = 0.0968 kgm^2

    inertia of both dumbells:

    I = 2*1*0.22^2 = 0.0968 kgm^2

    together 0.1936 kgm^2

    the difference I is 1.69-0.1936 = 1.4964 kgm^2

    So the new inertia of the system is 7.4 - 1.4964 = 5.9036 kgm^2

    the angular momentum is conserved:

    L = I1w1 = I2w2 where w1 = 2pi*f = 2pi*0.75 = 4.712 rad/s (f = frequency)

    7.4*4.712 = 5.9036*w2

    w2 = 5.906 rad/s

    f = w/(2pi) = 5.906/(2pi) = 0.939 rev/s <--- ans. (new rate of rotation)

    O.G.

  • dery
    Lv 4
    4 years ago

    if there aren't any exterior forces, angular momentum is conserved, hence I(in the previous) w(in the previous) = I (after) w (after) the prompt of inertia of a uniform sphere is two/5 MR^2 if the mass continues to be consistent, the only change is in radius, so we've w(after)=w(in the previous)[(I(in the previous)/I(after)... w(after)=w(in the previous)[2/5MR(in the previous)^2/2/5... w(after)=w(in the previous)(R(in the previous)/(R(after)... w(after)=w(in the previous)(10^-5)^2 w(after)=w(in the previous)x10^10 so the angular speed will be 10^10 swifter than a million rev in line with 25 days, or approx a million rev in line with 0.00022s

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