Problem 8 - University Physics with Modern Physics

Question

A uniform disk with mass 40.0 kg and radius 0.200 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 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 approximately 0.868 m/s. (b) The resultant acceleration is approximately 4.06 m/s².

Step 1: Calculate the Moment of Inertia

The moment of inertia \( I \) for a uniform disk pivoted about its center is given by \( I = \frac{1}{2} m r^2 \). Substituting the given values (mass \( m = 40.0 \) kg and radius \( r = 0.200 \) m), we have: \[ I = \frac{1}{2} \cdot 40.0 \text{ kg} \cdot (0.200 \text{ m})^2 = 0.8 \text{ kg m}^2 \]

Step 2: Determine the Angular Acceleration

The torque \( \tau \) exerted by the force is \( \tau = F \cdot r \) where \( F = 30.0 \) N and \( r = 0.200 \) m. Thus, \[ \tau = 30.0 \text{ N} \cdot 0.200 \text{ m} = 6.0 \text{ Nm} \]The angular acceleration \( \alpha \) can be found using \( \tau = I \cdot \alpha \), so:\[ \alpha = \frac{\tau}{I} = \frac{6.0 \text{ Nm}}{0.8 \text{ kg m}^2} = 7.5 \text{ rad/s}^2 \]

Step 3: Calculate Angular Displacement in Radians

The disk turns through 0.200 revolution. Since 1 revolution is \( 2\pi \) radians, the angular displacement \( \theta \) is:\[ \theta = 0.200 \times 2\pi = 0.4\pi \text{ rad} \]

Step 4: Find the Final Angular Velocity

Using the equation \( \omega^2 = \omega_0^2 + 2\alpha\theta \) with initial angular velocity \( \omega_0 = 0 \):\[ \omega^2 = 0 + 2 \times 7.5 \cdot 0.4\pi \]\[ \omega^2 = 6\pi \]\[ \omega = \sqrt{6\pi} \approx 4.34 \text{ rad/s} \]

Step 5: Calculate the Tangential Velocity

The tangential velocity \( v \) is given by \( v = \omega \cdot r \). With \( \omega \approx 4.34 \text{ rad/s} \) and \( r = 0.200 \text{ m} \), we have:\[ v = 4.34 \cdot 0.200 = 0.868 \text{ m/s} \]

Step 6: Determine the Tangential Acceleration

The tangential acceleration \( a_t \) is \( a_t = \alpha \cdot r \). Substituting \( \alpha = 7.5 \text{ rad/s}^2 \) and \( r = 0.200 \text{ m} \):\[ a_t = 7.5 \cdot 0.200 = 1.5 \text{ m/s}^2 \]

Step 7: Calculate the Resultant Acceleration

The resultant acceleration \( a \) is the combination of tangential acceleration \( a_t \) and centripetal acceleration \( a_c = \omega^2 \cdot r \). Calculate \( a_c \):\[ a_c = (4.34)^2 \cdot 0.200 = 3.76 \text{ m/s}^2 \]The resultant acceleration is:\[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{1.5^2 + 3.76^2} \approx 4.06 \text{ m/s}^2 \]

Key Concepts

Moment of Inertia

In the study of rotational motion, the moment of inertia, often represented by the symbol \( I \), is a crucial concept. It can be thought of as the rotational analogue of mass in linear motion. This property quantifies how difficult it is to change the rotational motion of an object around a specific axis. For a disk rotating around its center, the formula for moment of inertia is \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius.
In our exercise with a disk of mass 40.0 kg and radius 0.200 m, we computed the moment of inertia as \( 0.8 \text{ kg m}^2 \). This value informs us about the distribution of the disk’s mass and how it affects rotational acceleration when a force is applied. The larger the moment of inertia, the more torque is required to achieve the same angular acceleration.

Angular Acceleration

Angular acceleration (\( \alpha \)) indicates how quickly an object’s rotational velocity changes. It's similar to how linear acceleration describes changes in speed. The equation for angular acceleration is \( \alpha = \frac{\tau}{I} \), where \( \tau \) is the torque applied, and \( I \) is the moment of inertia. In our context, a constant force applied tangentially to the disk creates a torque \( \tau = F \cdot r \).
For the given disk, this leads to \( \tau = 6.0 \text{ Nm} \) and an angular acceleration of \( 7.5 \text{ rad/s}^2 \).
This value conveys how quickly the disk picks up speed as it starts spinning due to the applied force. It is pivotal for determining other dynamic parameters such as angular velocity and subsequent motion characteristics.

Centripetal Acceleration

Centripetal acceleration is the acceleration required to keep an object moving in a circular path. For any object moving in a circle, this acceleration points towards the center of the circle. It's computed using the formula \( a_c = \omega^2 \cdot r \), where \( \omega \) is the angular velocity.
In the problem, after the disk covers\(~0.2\) revolutions, it achieves an angular velocity of about \( 4.34 \text{ rad/s} \). Consequently, the centripetal acceleration becomes \( 3.76 \text{ m/s}^2 \).
This central-seeking acceleration is essential to maintaining circular motion and is always perpendicular to the object’s tangential velocity.

Tangential Velocity

Tangential velocity refers to the linear speed of a point on the rim of the rotating object. It's essential for understanding how fast a point on the object's edge travels, akin to how fast a car moves along a road. This velocity relates directly to angular velocity via the formula \( v = \omega \cdot r \).
For the given disk problem, after calculating the angular velocity, we found that the tangential velocity is \( 0.868 \text{ m/s} \).
This value informs us how fast a point on the disk's rim moves in a linear path, and it plays a crucial role in explaining the compound motion of rotational systems.