In a tire-throwing competition, a man holding a \(23.5-\mathrm{kg}\) car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. The tire starts from rest and is then accelerated in a circular path. The orbital radius \(r\) for the tire's center of mass is \(1.10 \mathrm{~m},\) and the path is horizontal to the ground. The figure shows a top view of the tire's circular path, and the dot at the center marks the rotation axis. The man applies a constant torque of \(20.0 \mathrm{~N} \mathrm{~m}\) to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius \(R=0.35 \mathrm{~m}\) from its center. a) What is the time, \(t_{\text {throw }}\) required for the tire to complete three full revolutions? b) What is the final linear speed of the tire's center of mass (after three full revolutions)? c) If, instead of assuming that all of the mass of the tire is at a distance \(0.35 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.30 \mathrm{~m}\) and outer radius \(0.40 \mathrm{~m}\), how does this change your answers to parts (a) and (b)?

Step by step solution

01

1. Calculate the moment of inertia

Using the formula for the moment of inertia for a point mass (I=MR²), where M is the mass and R is the radius: I = (23.5 kg)(0.35 m)² = 2.87 kg m²

02

2. Calculate the angular acceleration

Using the torque formula (τ = Iα), we can solve for the angular acceleration (α): 20.0 Nm = (2.87 kg m²)α α = 6.97 rad/s²

03

3. Calculate the angular displacement

The tire undergoes three full revolutions, so we can find the total angular displacement (θ) using the following formula: θ = 3 × 2π = 18.85 rad

04

4. Calculate the time for three full revolutions

Using the angular kinematic equation (θ = ω₀t + 0.5αt²) and substituting the known values (ω₀ = 0, θ = 18.85 rad, α = 6.97 rad/s²): 18.85 rad = 0.5 (6.97 rad/s²) t² Solving for t (time): t_throw = 2.33 s

05

5. Calculate the final angular velocity

Using the angular kinematic equation (ω_f = ω₀ + αt) with the calculated time and known initial angular velocity (ω₀ = 0, α = 6.97 rad/s², t = 2.33 s): ω_f = 6.97 (2.33) = 16.25 rad/s

06

6. Calculate the final linear speed

To find the final linear speed of the tire's center of mass (v_f), multiply the final angular velocity (ω_f) by the orbital radius (r=1.10 m): v_f = ω_f × r = (16.25 rad/s)(1.10 m) = 17.88 m/s For parts (a) and (b), we now have: a) t_throw = 2.33 s b) v_f = 17.88 m/s Now, let's address part (c) by calculating the moment of inertia using the hollow disk model.

07

7. Calculate the new moment of inertia

Using the formula for the moment of inertia for a hollow disk I_hollow_disk = 0.5 × M × (r_outer² + r_inner²) where M is the mass and the inner and outer radii I_hollow_disk = 0.5 × (23.5 kg)(0.30 m² + 0.40 m²) = 4.245 kg m²

08

8. Calculate the new angular acceleration

Using the new moment of inertia and the torque formula, we can find the new angular acceleration: 20.0 Nm = (4.245 kg m²)α α_new = 4.71 rad/s²

09

9. Calculate the new time for three full revolutions

Using the same angular kinematic equation as before, but with the new angular acceleration: 18.85 rad = 0.5 (4.71 rad/s²) t² Solving for the new time: t_new_throw = 2.84 s

10

10. Calculate the new final angular velocity

Using the angular kinematic equation with the new time and angular acceleration: ω_new_f = 4.71 (2.84) = 13.37 rad/s

11

11. Calculate the new final linear speed

To find the updated final linear speed, multiply the new final angular velocity by the orbital radius: v_new_f = ω_new_f × r = (13.37 rad/s)(1.10 m) = 14.70 m/s For part (c), the new time to complete three full revolutions is 2.84 s and the new final linear speed of the center of mass is 14.70 m/s.

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