Question

A potter's wheel is a uniform disk of mass of 10.0 kg and radius 20.0 cm. A 2.0-kg lump of clay, roughly cylindrical with radius 3.0 cm, is placed at the center of the wheel. The wheel initially rotates at 30.0 rev/min. The clay then flattens into a disk of radius 8.0 cm. What is the final angular speed of the wheel?
a. 29.6 rev/min
b. 29.2 rev/min
c. 30.8 rev/min
d. 30.4 rev/min
e. 30.0 rev/min

Answer 1

Answer:

b. 29.2 rev/min

Explanation:

• Assuming no external torques acting during the process, total angular momentum must be conserved, as follows:

       $L_{0} = L_{f} (1)$

• The initial angular momentum L₀, can be expressed as follows:

        $L_{0} = I_{0} * \omega_{0} (2)$

        where I₀ = initial moment of inertia = moment of inertia of the disk +

        moment of inertia of the cylinder and ω₀ = initial angular velocity  =

       30.0 rev/min.

• Replacing by the values, we get:$I_{0} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{c} *r_{c} ^{2} = 0.2 kg*m2 +9e-4 kg*m2 = 0.2009 kg*m2 (3)$⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/min
• The final angular momentum can be written as follows:

       $L_{f} = I_{f} * \omega_{f} (4)$

       where If = final moment of inertia = moment of the inertia of the solid

      disk + moment of  inertia of the clay flattened on a disk, and ωf = final

      angular velocity.

• Replacing by the values, we get:

   $I_{f} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{fd} *r_{fd} ^{2} = 0.2 kg*m2 +6.4e-3 kg*m2 = 0.2064 kg*m2 (5)$

       ⇒ Lo =Lf = If*ωf

• Replacing (2) in (1), and solving for ωf, we get:

        $\omega_{f} = \frac{L_{o}}{I_{f} } = \frac{6.027kg*m2*rev/min}{0.2064kg*m2} = 29.2 rev/min (6)$