A potter’s wheel with rotational inertia 24 kg m2 is spinning freely with no friction at 50 rpm. A piece of clay is dropped from above the wheel and sticks 0.5 m from the center of the wheel. Afterward, the rotational speed of the wheel and clay has dropped to 48 rpm. The mass of the piece of clay is closest to

A) 2 kg B) 4 kg C) 100 kg D) 190 kg

To solve this problem, we can use the principle of conservation of angular momentum. The initial angular momentum of the system (potter's wheel + clay) should be equal to the final angular momentum.

Angular momentum is given by the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Given data:
Initial moment of inertia (Ii) = 24 kg m^2
Initial angular velocity (ωi) = 50 rpm = (50/60) × 2π radians/s
Final angular velocity (ωf) = 48 rpm = (48/60) × 2π radians/s
Radius (r) = 0.5 m

Now, let's calculate the initial angular momentum (Li):
Li = Ii * ωi

Next, let's calculate the final angular momentum (Lf):
Lf = (Ii + m * r^2) * ωf

We can equate Li and Lf:
Ii * ωi = (Ii + m * r^2) * ωf

Simplifying further:
Ii * ωi = Ii * ωf + m * r^2 * ωf
Ii * ωi - Ii * ωf = m * r^2 * ωf
Ii * (ωi - ωf) = m * r^2 * ωf

Now, substitute the given values and solve for the mass (m):
(24 kg m^2) * [(50/60) × 2π - (48/60) × 2π] = m * (0.5 m)^2 * (48/60) × 2π

Simplifying further:
(24 kg m^2) * (2π/60) * (2/60) = m * (0.5 m)^2 * (48/60) × 2π
2 * (2π/60) * (2/60) = m * (0.5^2) * (48/60)
(2π/60) * (2/60) = m * (0.25) * (48/60)
1 = m * (0.25) * (4/5)
1 = m * (4/20)
1 = m * (1/5)

Solving further:
m = 5 kg

Therefore, the mass of the piece of clay is closest to 5 kg, making option A) 2 kg the closest answer.