Q6.15 Roundabout (K)

A boy of mass 40 kg stands at the edge of a stationary turntable of radius 1.5 m and moment of inertia 200 kgm2. He throws a ball of mass 0.6 kg tangentially to the turntable with a speed of 8 ms-1. What conservation laws are applicable here? Explain carefully. Calculate

  1. the angular momentum of the boy and turntable
  2. the linear speed of the boy relative to the ground

immediately after the ball is thrown.

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Hint

Don’t forget the angular momentum of the ball after it is thrown. (Key Point 5.16).
 

Solution

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Solution

  1. The only external horizontal force acting on the system consisting of boy, ball and turntable is the force that acts through the axle of the turntable.

    Graphic - No title - angmomcons2

    The existence of this force means that the linear momentum of this system is not conserved. However, since this force acts through the point O it exerts no torque about the point O so that the angular momentum about O is conserved.

    The angular momentum of the system, L, is zero before the ball is thrown, and must therefore be zero afterwards.

    Equating angular momentum before and after the ball is thrown:

    0=L1+L2

    The angular momentum of the ball after throwing is

    L2=-m2rv2=-7.2 kg m2 s-1
    ( a vector pointing into the diagram plane).

    To keep the total angular momentum zero, the throw must give the boy and turntable an opposite angular momentum: L1=7.2 kg m2 s-1, a vector pointing out of the diagram plane).

  2. The total moment of inertia of the boy and turntable about the axis through O is

    Ibt=It+m1r2=200+40×1.52=290 kg m2
    where It is the moment of inertia of the turntable about the axis through O. The angular velocity ω of the boy and turntable multiplied by the moment of inertia Ibt equals the angular momentum of boy and turntable:

    \[L_2 = I_{bt}\omega \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} \omega=2.5\times 10^{-2}{\, rad\,s^{-1}};\]

    Thus, finally

    v1=rω=3.7×10-2 m s-1.