A potter's wheel is a uniform disk of mass 4.50 kg and radius 0.650 m and can spin freely around a vertical axis through its center. With the wheel spinning at an angular speed of 4.70 rad/s, a small piece of clay of mass 0.870 kg is dropped at the outer edge of the wheel and sticks to it. Find the final angular speed of the wheel clay. Treat the piece of clay as a point particle. Group of answer choices

3.39 rad / s.

Explanation:

Given data:

mass of disk = 4.50 Kg

radius of wheel = 0.650 m

mass of the clay = 0.870 kg

The moment of inertial of the wheel = I = 4.5 kg x ( 0.65 m )2 / 2 = 0.95 kg . m2.

Now, applying the principle of angular momentum conservation :

Iω_i = ( I + mr2 )ω_f.

where ω_i = initial angular speed= 4.70 rad/s, ω_f = final angular speed

Hence, ω_f = Iω_i / ( I + mr2 )

= ( 0.95 kg . m2 x 4.7 rad / s ) / [ 0.95 kg . m2 + 0.87 kg x ( 0.65 m )2 ]

= 3.39 rad / s.

Hence, correct answer is  : 3.39 rad / s.

it about 600 mph from what i calculated

i think if it were me i whould divide 300 dide 15

step-by-step explanation: