Question

\(\cdot 10.53\) 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 \(R=0.350 \mathrm{~m}\) 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 \(0.350 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.300 \mathrm{~m}\) and outer radius \(0.400 \mathrm{~m},\) how does this change your answers to parts (a) and (b)?

Short answer

Answer: The time needed for the tire to complete three full revolutions is 2.05 seconds, and the final linear speed of the tire's center of mass is 11.61 m/s.

Step-by-step solution

Calculate angular acceleration for the first scenario

The torque applied (τ) can be related to the moment of inertia (I) and angular acceleration (α) as follows: τ = I * α We consider all mass to be concentrated at a distance R from the center. The moment of inertia for a point mass (m) at a distance R is given by: I = m * R^2 In this case, m = 23.5 kg and R = 0.350 m. Now, we can calculate I: I = 23.5 kg * (0.350 m)^2 I = 2.875 kg*m^2 Next, using the given torque of 20 N*m, we can find angular acceleration: τ = I * α 20 N * m = 2.875 kg*m^2 * α α = 6.957 rad/s^2

Calculate time for three revolutions

To find the time it takes for the tire to complete three revolutions, we can use the angular displacement formula (θ), and relate it to the angular acceleration: θ = 1/2 * α * t^2 Since there are three full revolutions, the total angular displacement is: θ = 3 * 2π = 6π rad Now, we can find time (t_throw) by substituting the values: 6π rad = 1/2 * 6.957 rad/s^2 * t^2 t_throw = 1.82 s

Calculate final linear speed

Linear speed (v) can be related to the angular speed (ω) and orbital radius r as follows: v = r * ω We can relate angular acceleration (α) to the final angular speed with the following equation: ω = α * t Now, substituting values: ω = 6.957 rad/s^2 * 1.82 s ω = 12.661 rad/s Now, calculating the linear speed: v = 1.10 m * 12.661 rad/s v = 13.93 m/s Thus, the final linear speed of the tire's center of mass is 13.93 m/s.

Calculate moment of inertia for a hollow disk

Now, we will find a new moment of inertia considering the tire as a hollow disk with inner radius (Ri) of 0.300 m and outer radius (Ro) of 0.400 m. The moment of inertia for a hollow disk is given by: I_hollow = 0.5 * m * (Ri^2 + Ro^2) Substituting given values, we get: I_hollow = 0.5 * 23.5 kg * (0.300 m^2 + 0.400 m^2) I_hollow = 3.883 kg*m^2

Calculate new angular acceleration and time for three revolutions

Using the new moment of inertia, we can calculate the new angular acceleration: 20 N * m = 3.883 kg*m^2 * α α_new = 5.150 rad/s^2 Next, we find the new time for three revolutions: 6π rad = 1/2 * 5.150 rad/s^2 * t^2 t_new = 2.05 s

Calculate new final linear speed

Now, we can find the new final angular speed: ω_new = 5.150 rad/s^2 * 2.05 s ω_new = 10.558 rad/s Finally, calculating the new linear speed: v_new = 1.10 m * 10.558 rad/s v_new = 11.61 m/s In conclusion, considering the tire as a hollow disk changes the time needed for three full revolutions to 2.05 s and the final linear speed of the tire's center of mass to 11.61 m/s.

Key concepts

Torque and Angular Acceleration

Torque is a measure of the force that can cause an object to rotate about an axis. Essentially, it is the rotational equivalent of linear force. In our exercise, the man applies a constant torque to the tire, which in turn produces an angular acceleration.

Angular acceleration, denoted as \( \alpha \), is the rate at which the angular velocity of an object changes with respect to time. It is directly proportional to the torque applied, provided the mass distribution (moment of inertia) of the object does not change. The formula connecting torque \( \tau \) with moment of inertia \( I \) and angular acceleration \( \alpha \) is:
\[ \tau = I \cdot \alpha \
]In practical terms, this means that the more torque applied, the greater the angular acceleration, assuming the moment of inertia is constant. However, in situations where the distribution of mass changes--such as changing from a point mass to a hollow disk--the moment of inertia changes, and thus the angular acceleration changes for the same amount of applied torque.

Moment of Inertia

The moment of inertia is a critical concept when analyzing rotational motion. It is a measure of an object's resistance to changes in its rotational motion and depends on the mass distribution within the object.

For a point mass, the moment of inertia \( I \) is simply the product of the mass \( m \) and the square of its distance \( R \) from the rotation axis, described by the formula \[ I = m \cdot R^2 \
]However, real objects have mass distributed over a volume, not just at a point. In the case of our tire, we initially treated it as if all its mass were concentrated at a specific radius, but a more accurate model is the hollow disk, which requires us to calculate the moment of inertia using the radii of both the inner and outer sides of the tire.

The difference in the moment of inertia due to the mass distribution affects the tire's angular acceleration for a given torque. A higher moment of inertia would mean a lower angular acceleration for the same amount of torque.

Angular Displacement

Angular displacement refers to the angle through which a point or line has been rotated in a specified sense about a specified axis. It gives us a measure of how far an object has rotated compared to its initial position. In our exercise, the tire completes three full revolutions before being released.

Each revolution represents a complete circle, hence an angular displacement of \( 2\pi \) radians. Therefore, for three full revolutions, the total angular displacement is \( 3 \times 2\pi \), or \( 6\pi \) radians. We used this angular displacement to calculate the time taken for the tire to complete the revolutions using the formula \[ \theta = \frac{1}{2} \cdot \alpha \cdot t^2 \
]Once the angular acceleration and the angular displacement are known, this equation lets us solve for the time required to reach that displacement.

Linear Speed

Linear speed is the rate at which an object moves along a path. In the context of rotational motion, the linear speed at the edge of a spinning object is tied to how fast it is spinning - its angular velocity. The formula linking linear speed \( v \) and angular velocity \( \omega \) is:
\[ v = r \cdot \omega \
]where \( r \) is the radius of the circular path. The angular velocity, in turn, can be found if we know the angular acceleration and the time of rotation, using \( \omega = \alpha \cdot t \).

In our exercise, given the angular acceleration and time, we calculate the angular velocity and subsequently the linear speed for the tire. Different moments of inertia due to different representations of the tire's mass distribution (point mass versus hollow disk) result in different times and, therefore, different final linear speeds of the tire's center of mass when completing three full revolutions.