GSU asked in Science & MathematicsPhysics · 1 decade ago

Angular velocity (rads/sec)?

Please solve, Thank You!

A uniform circular disk with mass (20kg and radius = 5cm) rolls (without slipping) down a hill with a vertical drop of 2 meters, what is the disk’s angular velocity (rads/sec) at the bottom of the hill. Find the moment of inertia

2 Answers

  • Edward
    Lv 7
    1 decade ago
    Favorite Answer

    The ph_yo has the right start, however it is a angular velocity that is in question

    So the potential energy at the top equal to the kinetic energy (Ke) on the bottom.

    Pe=Ke (but Ke is rotational energy!)

    mgh = .5 I w^2

    m – mass of the cylinder

    g - acceleration due to gravity

    h – height at the top

    I - moment of inertia of the cylinder

    w – angular velocity

    I= .5m r^2

    r – radius of the cylinder

    since mgh = .5 I w^2

    we have w=sqrt(2mgh/I)


    w= sqrt(2mgh/.5mr^2)

    w=2/r sqrt(gh)

    w=2/ .05 sqrt (9.81 x 2)=

    w= 177 rad /sec

    now moment of inertia

    I=.5m r^2

    I=.5 (2) (.05)^2

    I=0.025 kg m^2

    Have fun

  • Anonymous
    1 decade ago


    potential = kinetic, assuming the disk start from rest

    mgh = 1/2 m v^2

    v^2 = 2gh ==> v = sqrt( 2*9.81*2) = 6.26

    w = v/r = 6.26/.05 = 125.3 rads/sec.

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