Question

A thin uniform cylindrical turntable of radius 2.3 m and mass 22 kg rotates in a horizontal plane with an initial angular speed of 7.3 rad/s. The turntable bearing is frictionless. A clump of clay of mass 8.7 kg is dropped onto the turntable and sticks at a point 1.5 m from the point of rotation. Treat the clay as a point mass. Find the angular speed of the clay and turntable. Answer in units of rad/s.

Answer 1

Answer:

ω = 5.5 rad/s

Explanation:

• Assuming no external torques present during the instant that the clump of clay is dropped on the turntable, total angular momentum must be conserved.
• The angular momentum of a rotating rigid body, can be expressed as follows:

       $L = I * \omega (1)$

       where I = moment of inertia regarding the axis of rotation, and ω =

       angular speed of the rotating body.

• Since the angular momentum must keep constant, this means that it must be satisfied the following equality:

       $L_{o} = L_{f} (2)$

       where L₀ = I₀ * ω₀,  Lf = If * ωf.

       I₀ is the moment of inertia of a solid disk rotating around an axis

       passing through its center, as follows:

      $I_{o} =\frac{1}{2} * m* r^{2} = \frac{1}{2} * 22 kg*(2.3m)^{2} = 58.2 kgm2 (3)$

      If, is the moment of inertia after dropping the clump of clay, which adds

      its own moment of inertia as a point mass, as follows:

     $I_{f} =\frac{1}{2} * m* r^{2} + m_{cl} * r_{cl}^{2} =58.2 kgm2 + (8.7kg)*(1.5m)^{2} \\ = 58.2 kgm + 19.6 kgm2 = 77.8 kgm2 (4)$

• Replacing I₀, If and ω₀ in (2), we can solve for ωf, as follows:

        $\omega_{f} = \frac{I_{o} *\omega_{o} }{I_{f}} = \frac{58.2kgm2*7.3rad/s}{77.8kgm2} = 5.5 rad/s (5)$