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

b. 29.2 rev/min

Explanation:

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

       

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

       

        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:⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/minThe final angular momentum can be written as follows:

       

       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:

   

       ⇒ Lo =Lf = If*ωf

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

       

Answer form flower;

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j