Determine the stopping location of the prize wheel. At this moment it is centered on the number 16. It is spinning with an energy of 58.33 J. It is slowing at a rate of 1.900 rad/s/s.

The disk is made from a wood with a density of 457.0 kg/m^3. The disk is 47.0 mm thick and has a radius of 34.0 cm.

Predict the angular displacement it will go through and predict the number on which the wheel will stop. To be safe, you will also get one number the right and to the left of the number you chose.

The wheel will spin clockwise.

The kinetic energy of a rotating object is equal to half its moment of inertia times the square of its angular velocity. The moment of inertia of a solid disk is half the mass times the square of the radius, and mass is equal to density times volume. Using kinematics, we can use the angular acceleration and angular velocities to find the angular displacement.

First, use the kinetic energy to find the initial angular velocity.

RE = ½ Iω²

Substitute the formula for moment of inertia of a solid disk.

RE = ½ (½ mr²) ω²

RE = ¼ m r² ω²

Next, substitute the formula for mass using density and volume of a cylinder.

RE = ¼ (ρ πr²h) r² ω²

RE = ¼ π ρ r⁴ h ω²

Plug in values and solve for angular velocity.

58.33 J = ¼ π (457.0 kg/m³) (0.340 m)⁴ (0.0470 m) ω²

ω = 16.1 rad/s

Now use kinematics to find the angular displacement. Given:

ω₀ = 16.1 rad/s

ω = 0 rad/s

α = -1.900 rad/s²

Find: Δθ

ω² = ω₀² + 2αΔθ

(0 rad/s)² = (16.1 rad/s)² + 2 (-1.900 rad/s²) Δθ

Δθ = 68.1 rad

The spinner rotates 68.1 radians, or 3901.3 degrees. Dividing by 360 degrees per revolution, the remainder is 0.837 revolutions or 301.3 degrees. Each number on the spinner is 10 degrees, so the spinner will stop 30.1 spaces counterclockwise from number 16. That means it will stop approximately on number 10.