Problem 10 - University Physics with Modern Physics

Question

A uniform disk with mass 40.0 \(\mathrm{kg}\) and radius 0.200 \(\mathrm{m}\) is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force \(F=30.0 \mathrm{N}\) is applied tangent to the rim of the disk. (a) What is the magnitude \(v\) of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution? (b) What is the magnitude \(a\) of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

Step-by-Step Solution

Answer

(a) The tangential velocity is 0.868 m/s. (b) The resultant acceleration is 4.05 m/s².

Step 1: Calculate Moment of Inertia

The moment of inertia for a uniform disk pivoted through its center is given by \( I = \frac{1}{2} m r^2 \).Given: Mass \( m = 40.0 \, \mathrm{kg} \) and Radius \( r = 0.200 \, \mathrm{m} \).Substitute to find \( I \):\[I = \frac{1}{2} \times 40.0 \, \mathrm{kg} \times (0.200 \, \mathrm{m})^2 = 0.800 \, \mathrm{kg \, m^2}\]

Step 2: Calculate Angular Acceleration

Torque \( \tau \) is produced by the force applied tangent to the rim and is given by \( \tau = F \times r \).Using \( F = 30.0 \, \mathrm{N} \) and \( r = 0.200 \, \mathrm{m} \):\[\tau = 30.0 \, \mathrm{N} \times 0.200 \, \mathrm{m} = 6.0 \, \mathrm{Nm}\]Angular acceleration \( \alpha \) can be calculated with \( \alpha = \frac{\tau}{I} \):\[\alpha = \frac{6.0 \, \mathrm{Nm}}{0.800 \, \mathrm{kg \, m^2}} = 7.5 \, \mathrm{rad/s^2}\]

Step 3: Calculate Final Angular Velocity

The disk starts from rest and turns through an angle of \( \theta = 0.200 \) revolutions. Convert revolutions to radians: \[\theta = 0.200 \times 2 \pi = 1.257 \, \mathrm{rad}\]The relation \( \omega^2 = \omega_0^2 + 2\alpha\theta \) gives the final angular velocity \( \omega \) (initial angular velocity \( \omega_0 = 0 \)):\[\omega = \sqrt{2 \times 7.5 \, \mathrm{rad/s^2} \times 1.257 \, \mathrm{rad}} = \sqrt{18.855} \approx 4.34 \, \mathrm{rad/s}\]

Step 4: Calculate Tangential Velocity

The tangential velocity \( v \) of a point on the rim of the disk is related to its angular velocity by \( v = r \omega \).Substitute \( r = 0.200 \, \mathrm{m} \) and \( \omega = 4.34 \, \mathrm{rad/s} \) to find \( v \):\[v = 0.200 \, \mathrm{m} \times 4.34 \, \mathrm{rad/s} = 0.868 \, \mathrm{m/s}\]

Step 5: Calculate Resultant Acceleration

Resultant acceleration is the vector sum of tangential and centripetal acceleration.Tangential acceleration \( a_t \) is given by \( a_t = r \alpha \).\[a_t = 0.200 \, \mathrm{m} \times 7.5 \, \mathrm{rad/s^2} = 1.5 \, \mathrm{m/s^2}\]Centripetal acceleration \( a_c \) is given by \( a_c = r \omega^2 \):\[a_c = 0.200 \, \mathrm{m} \times (4.34 \, \mathrm{rad/s})^2 = 3.76 \, \mathrm{m/s^2}\]Resultant acceleration \( a = \sqrt{a_t^2 + a_c^2} \):\[a = \sqrt{(1.5 \, \mathrm{m/s^2})^2 + (3.76 \, \mathrm{m/s^2})^2} = \sqrt{2.25 + 14.14} = \sqrt{16.39} \approx 4.05 \, \mathrm{m/s^2}\]

Key Concepts

Moment of Inertia

Moment of inertia, often denoted as \( I \), is a fundamental concept in rotational motion. It measures an object's resistance to changes in its rotational motion. Think of it like mass in linear motion—just as heavier objects are harder to push, objects with larger moments of inertia are harder to spin.
The formula for the moment of inertia depends on the shape of the object and the axis of rotation. For a uniform disk rotating about its center, the moment of inertia is given by:\[ I = \frac{1}{2} m r^2 \]where \( m \) is the mass of the disk and \( r \) is its radius.
In the given exercise, we calculated the moment of inertia of the disk by substituting the provided values (mass = 40 kg, radius = 0.200 m) into the formula to obtain:\[ I = \frac{1}{2} \times 40 \, \mathrm{kg} \times (0.200 \, \mathrm{m})^2 = 0.800 \, \mathrm{kg \, m^2} \]Understanding this concept helps us predict how the disk will respond when a force is applied.

Angular Acceleration

Angular acceleration, denoted by \( \alpha \), is a crucial aspect of rotational motion. It represents how quickly the rotational speed of an object is changing. It’s the rotational equivalent of linear acceleration.
To find angular acceleration, we use the relationship between torque \( \tau \) and moment of inertia \( I \):\[ \tau = I \alpha \]In simpler terms, the torque, or twisting force, applied on an object, causes it to accelerate rotationally. For our disk, torque is calculated from the force applied tangentially at the edge:\[ \tau = F \times r = 30.0 \, \mathrm{N} \times 0.200 \, \mathrm{m} = 6.0 \, \mathrm{Nm} \]The angular acceleration is then:\[ \alpha = \frac{\tau}{I} = \frac{6.0 \, \mathrm{Nm}}{0.800 \, \mathrm{kg \, m^2}} = 7.5 \, \mathrm{rad/s^2} \]Knowing the angular acceleration allows us to determine how quickly the disk's rotation speed increases.

Tangential Velocity

Tangential velocity relates to the speed of an object moving along the edge of a rotating disk. Imagine a point on the edge of the disk that gives you an idea of how fast it moves around the circle.
Tangential velocity \( v \) can be calculated by the formula:\[ v = r \omega \]where \( r \) is the radius and \( \omega \) is the angular velocity. The relation between angular and tangential velocities is crucial to understanding how rotational motion translates into linear motion. The angular velocity \( \omega \) is found as the disk turns through a specific angle using:\[ \omega = \sqrt{2 \alpha \theta} \]After computing the angular velocity, the tangential velocity for our exercise was:\[ v = 0.200 \, \mathrm{m} \times 4.34 \, \mathrm{rad/s} = 0.868 \, \mathrm{m/s} \]This value shows how fast a point on the rim of the disk moves tangentially as the disk rotates.

Resultant Acceleration

Resultant acceleration on the rim of a rotating object combines tangential and centripetal accelerations. It's a vector sum indicating the total acceleration a point on the edge experiences.

• **Tangential Acceleration** \( a_t \): This is due to angular acceleration and is calculated as:

\[ a_t = r \alpha = 0.200 \, \mathrm{m} \times 7.5 \, \mathrm{rad/s^2} = 1.5 \, \mathrm{m/s^2} \]

• **Centripetal Acceleration** \( a_c \): This results from the change in direction as the disk rotates and is given by:

\[ a_c = r \omega^2 = 0.200 \, \mathrm{m} \times (4.34 \, \mathrm{rad/s})^2 = 3.76 \, \mathrm{m/s^2} \]To find the resultant acceleration \( a \), we use the Pythagorean theorem:\[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{(1.5 \, \mathrm{m/s^2})^2 + (3.76 \, \mathrm{m/s^2})^2} = 4.05 \, \mathrm{m/s^2} \]This resultant acceleration combines both types of motion, giving us a comprehensive view of the forces acting on the particle.