Problem 49 - University Physics with Modern Physics

Question

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

Answer

Question: Considering the given scenario, calculate the following: a) The time it takes for the tire to complete three full revolutions. b) The final linear speed of the tire's center of mass. c) Recalculate the time and final linear speed if we change the moment of inertia to a hollow disk model. Answer: a) t_throw = 2.33 s b) v_f = 17.88 m/s c) New time: 2.84 s, New final linear speed: 14.70 m/s.

Step 1: 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

Step 2: 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

Step 3: 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

Step 4: 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

Step 5: 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

Step 6: 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.

Step 7: 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

Step 8: 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

Step 9: 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

Step 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

Step 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.

Key Concepts

Angular Acceleration

Angular acceleration, denoted by \(\alpha\), is a measure of how quickly the angular velocity of an object changes. It is an essential concept in rotational dynamics, just like linear acceleration is in linear motion. When a torque is applied to an object, it causes the object to rotate, accelerating it at a specific rate. This rate is known as the angular acceleration.

The relationship between torque (\(\tau\)), moment of inertia (\(I\)), and angular acceleration is given by the formula \(\tau = I\alpha\). In our case, the torque applied to the tire was \(20.0\,\text{Nm}\), and using the calculated moment of inertia of \(2.87\,\text{kg m}^2\), we found the angular acceleration to be \(6.97\,\text{rad/s}^2\). This indicates how fast the tire's angular speed is increasing as it spins.

It's worth noting that if the tire's mass distribution changes (like treating it as a hollow disk instead of a solid one), the angular acceleration would change too since the moment of inertia would differ.

Moment of Inertia

The moment of inertia, symbolized by \(I\), is the rotational equivalent of mass in linear motion. It indicates how much torque is needed for a certain angular acceleration given a distribution of mass around an axis. Simply put, it's a measure of an object's resistance to change in its rotation.

For a point mass, the moment of inertia is calculated by \(I = MR^2\), where \(M\) is the mass and \(R\) is the radius from the axis of rotation. In our original setup, the tire was treated as if all its mass was concentrated at \(0.35\,\text{m}\) from its center, leading to a moment of inertia of \(2.87\,\text{kg m}^2\).

However, considering it as a hollow disk with different inner (\(0.30\,\text{m}\)) and outer radii (\(0.40\,\text{m}\)), its moment of inertia becomes \[I = 0.5 \times M \times (r_{\text{outer}}^2 + r_{\text{inner}}^2)\], resulting in \(4.245\,\text{kg m}^2\). This change directly affects the tire's angular acceleration and, subsequently, its rotational dynamics.

Torque

Torque is the force that causes an object to rotate around an axis. It can be thought of as the rotational analog to linear force. In mathematical terms, torque \(\tau\) is the product of the force applied \(F\) and the distance \(r\) from the axis of rotation, given by \(\tau = rF\sin(\theta)\). In this scenario, it represents the man's effort to swing the tire.

When the man applies a torque of \(20.0\,\text{Nm}\), this force contributes to the tire's angular acceleration. The amount of torque needed depends on the moment of inertia. If the moment of inertia is higher, more torque would be required to achieve the same angular acceleration.

• Consistent torque results in constant angular acceleration.
• Torque is directly proportional to both force applied and the radius of application.

Understanding torque allows us to control and predict rotational motion effectively.

Linear Speed

Linear speed, often just called speed, is the distance an object's center of mass travels per unit of time. In the context of rotational dynamics, it becomes relevant when considering the movement along the circumference of a circle.

For an object moving in a circular path, its linear speed \(v\) is related to its angular speed \(\omega\) by the formula \(v = \omega r\), where \(r\) is the radius of the circular path. After calculating the final angular velocity \(\omega_f\) of the tire, its linear speed was determined by this relation.

In the exercise, the final linear speed of the tire after three full turns was \(17.88\,\text{m/s}\). When considering the tire as a hollow disk, it changed to \(14.70\,\text{m/s}\) due to the different moment of inertia affecting the angular acceleration.

• Linear speed affects how quickly an object covers distance.
• It can be directly calculated from angular speed in circular motion scenarios.

Hence, this connection between rotational and linear concepts is crucial in understanding the actual movement in circular paths.