A Bicycle Momentum Problem...?

A bicycle has wheels of radius 0.29 m. Each wheel has a rotational inertia of 0.079 kgm2 about its axle. The total mass of the bicycle including the wheels and the rider is 78 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?

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  • 1 decade ago
    Favorite Answer

    K.E 1/2 m v^2 = 1/2 x 78 x v^2 = 39 v^2

    R.K.E = 2[1/2 x moment of inertia x angular speed^2] as there are two wheels.

    but as you know that v = r x angular speed

    angular speed squared = v^2 / r^2

    substituting this in the second equation we get

    R.K.E = 2 x1/2 x 0.079 x v^2/0.29^2 = 0.9394 v^2

    T.KE = KE + RKE = 39v^2 + 0.9394v^2

    Now dividing 0.9394 v^2 with 39.9394 v^2 we get

    0.0235 , the ratio required.

    Hope my steps are clear enough...

  • 1 decade ago

    Find the rotational kinetic energy of the wheel and the translational kinetic energy of the bicycle. Add them together to find total kinetic energy. Simply find the ratio of rotational kinetic energy to total kinetic energy.

    RKE/(RKE+TKE)

  • Anonymous
    1 decade ago

    Total K.E = translation KE + Rotation K.E of wheels

    = 0.5*m*v^2 + Iw^2 + Iw^2

    w = angular velocity

    I = polar momemt of inertia of one wheel

    Also w = v * r

    Assuming there is no slipping of the tires, the horizontal velocity is same as the rotational velocity of the tires

    So, the ratio = rotational K.E/ Total K.E

    = (I*w*2/2 + I*w^2)/(I*w*2/2 + I*w^2 + 0.5*m*v^2)

    = (I*v^2/r^2)/(I*v^2/r^2 + 0.5*mv^2)

    = I/(I+0.5*m*r^2)

    substituting values,

    answer is 0.0235

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