Problem 14 - College Physics

Question

\(\bullet\) A thin, light string is wrapped around the rim of a 4.00 kg solid uniform disk that is 30.0 \(\mathrm{cm}\) in diameter. A person pulls on the string with a constant force of 100.0 \(\mathrm{N}\) tangent to the disk, as shown in Figure \(10.49 .\) The disk is not attached to anything and is free to move and tum. (a) Find the angular acceleration of the disk about its center of mass and the linear acceleration of its center of mass. (b) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the accelerations in part (a)?

Step-by-Step Solution

Answer

(a) Solid disk: \( \alpha = 333.3 \text{ rad/s}^2 \), \( a = 50.0 \text{ m/s}^2 \). (b) Hollow cylinder: \( \alpha = 166.7 \text{ rad/s}^2 \), \( a = 25.0 \text{ m/s}^2 \).

Step 1: Calculate the Moment of Inertia of the Solid Disk

For a solid disk, the moment of inertia about its center of mass is given by \( I = \frac{1}{2} m r^2 \). Here, \( m = 4.00 \text{ kg} \) and the radius \( r = 0.15 \text{ m} \) (since diameter is 30.0 cm). So, \[ I = \frac{1}{2} \times 4.00 \times (0.15)^2 = 0.045 \text{ kg} \cdot \text{m}^2. \]

Step 2: Calculate Angular Acceleration for the Solid Disk

The angular acceleration \( \alpha \) can be found using Newton's second law for rotation: \( \tau = I \alpha \), where \( \tau \) is the torque. The torque is given by \( \tau = F r = 100.0 \times 0.15 = 15.0 \text{ N} \cdot \text{m} \). Solving for \( \alpha \), we get: \[ \alpha = \frac{\tau}{I} = \frac{15.0}{0.045} = 333.3 \text{ rad/s}^2. \]

Step 3: Calculate Linear Acceleration for the Solid Disk's Center of Mass

The linear acceleration \( a \) of the center of mass of the disk is related to the angular acceleration by \( a = r \alpha \). Thus: \[ a = 0.15 \times 333.3 = 50.0 \text{ m/s}^2. \]

Step 4: Calculate the Moment of Inertia of the Hollow Cylinder

For a hollow thin-walled cylinder, the moment of inertia is \( I = m r^2 \). Using the same mass (4.00 kg) and radius (0.15 m), \[ I = 4.00 \times (0.15)^2 = 0.09 \text{ kg} \cdot \text{m}^2. \]

Step 5: Calculate Angular Acceleration for the Hollow Cylinder

Using the torque \( \tau = 15.0 \text{ N} \cdot \text{m} \) and the moment of inertia from Step 4, apply \( \tau = I \alpha \) to find \( \alpha \): \[ \alpha = \frac{15.0}{0.09} = 166.7 \text{ rad/s}^2. \]

Step 6: Calculate Linear Acceleration for the Hollow Cylinder's Center of Mass

The linear acceleration \( a \) is given by \( a = r \alpha \): \[ a = 0.15 \times 166.7 = 25.0 \text{ m/s}^2. \]

Key Concepts

Moment of Inertia

In the realm of physics, the moment of inertia plays a crucial role when analyzing rotational objects. It measures an object's resistance to changes in its rotation. For extended objects, like disks and cylinders, it depends heavily on how mass is distributed relative to the axis of rotation.

For a solid disk, the moment of inertia (I) can be expressed as:

• \( I = \frac{1}{2} m r^2 \)
• "m" is the mass of the disk
• "r" is the radius

If you have a hollow cylinder, then the formula adjusts to:

• \( I = m r^2 \)

The significant difference here is that a hollow cylinder's mass distribution causes it to have a higher moment of inertia compared to a solid disk of the same mass and radius. This essentially means the hollow cylinder offers more resistance to changes in rotational motion.

Torque

Torque is pivotal in understanding how forces cause objects to rotate. In simple terms, torque is the rotational counterpart of force. If you apply a force to an object at a certain distance from its pivot point or axis of rotation, you're essentially creating torque.

Mathematically, it's calculated as:

• \( \tau = F r \)
• "F" is the force applied
• "r" is the lever arm or distance from the axis of rotation

In the example exercise, a force is applied tangent to the disk, resulting in torque being exerted on the disk. This torque is responsible for changing the angular motion of the disk or cylinder.

Linear Acceleration

When dealing with rotational objects like disks and cylinders, linear acceleration comes into play as it relates the rate of change of velocity of the object to its rotational speed through the concept of angular acceleration.

Linear and angular acceleration are connected by the radius "r":

• \( a = r \alpha \)
• "a" is the linear acceleration
• "\alpha" is the angular acceleration

This formula links how fast an object speeds up in a straight path to how fast it rotates. In other words, as angular acceleration increases, so does linear acceleration, and vice versa, provided the radius remains constant.

Solid Disk

A solid disk, when rotating, exhibits unique characteristics influenced by its shape and mass distribution. Usually, calculations involving a disk refer to its regular, uniform mass distribution with a significant amount of mass concentrated near its center. This trait makes solid disks somewhat easier to rotate compared to hollow cylinders.

Angle-wise, a disk's moment of inertia, \( I = \frac{1}{2} m r^2 \), characterizes how it will respond to applied torques, making it clear that it requires less torque to achieve the same angular acceleration as a hollow cylinder would. The calculation of angular and linear accelerations in exercises often start with determining this moment of inertia.

Hollow Cylinder

Hollow cylinders present an interesting case in rotational dynamics because their mass is predominantly situated away from the axis of rotation. This structural detail greatly impacts their inertia characteristics.

The moment of inertia for a hollow cylinder is characterized by:

• \( I = m r^2 \)

This higher moment of inertia stems from the mass being distributed farther from the rotational axis, making the hollow cylinder more resistant to rotational changes than a solid disk of the same dimensions. Consequently, it requires greater torque to achieve the same angular acceleration, causing any linear accelerations observed to be less, given the same initial conditions as a solid disk.