Problem 77 - University Physics with Modern Physics

Question

Two metal disks, one with radius \(R_1 =\) 2.50 cm and mass \(M_1 =\) 0.80 kg and the other with radius \(R_2 =\) 5.00 cm and mass \(M_2 =\) 1.60 kg, are welded together and mounted on a frictionless axis through their common center (\(\textbf{Fig. P9.77}\)). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block is suspended from the free end of the string. If the block is released from rest at a distance of 2.00 m above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

Step-by-Step Solution

Answer

(a) Total moment of inertia is \(2.25 \times 10^{-3} \text{ kg}\cdot \text{m}^2\). (b) Speed before striking floor is \(\approx 4.04 \text{ m/s}\) (c) Speed is \(\approx 8.08 \text{ m/s}\) with the larger disk; larger disk gives greater speed.

Step 1: Calculate Moment of Inertia of Each Disk

The moment of inertia for a solid disk about its center is calculated using the formula \( I = \frac{1}{2} M R^2 \). Let's compute this for both disks. For the first disk with \( M_1 = 0.80 \text{ kg} \) and \( R_1 = 2.50 \text{ cm} = 0.025 \text{ m} \):\[ I_1 = \frac{1}{2} \times 0.80 \times (0.025)^2 = 2.5 \times 10^{-4} \text{ kg}\cdot \text{m}^2 \]For the second disk with \( M_2 = 1.60 \text{ kg} \) and \( R_2 = 5.00 \text{ cm} = 0.05 \text{ m} \):\[ I_2 = \frac{1}{2} \times 1.60 \times (0.05)^2 = 2.0 \times 10^{-3} \text{ kg}\cdot \text{m}^2 \]

Step 2: Calculate Total Moment of Inertia

The total moment of inertia for the system is the sum of the individual moments of inertia of the disks:\[ I = I_1 + I_2 = 2.5 \times 10^{-4} + 2.0 \times 10^{-3} = 2.25 \times 10^{-3} \text{ kg}\cdot \text{m}^2 \]

Step 3: Analyze Motion with String on Smaller Disk

When the string is wrapped around the smaller disk (radius \( R_1 = 0.025 \text{ m} \)), the torque \( \tau \) due to the block is \( \tau = m g R_1 \). Using Newton's second law for rotation, \( \tau = I \alpha \), we solve for angular acceleration \( \alpha \):\[ \alpha = \frac{m g R_1}{I} \]where \( m = 1.50 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \).Calculate:\[ \alpha = \frac{1.50 \times 9.81 \times 0.025}{2.25 \times 10^{-3}} \approx 163.33 \text{ rad/s}^2 \]The linear acceleration \( a \) of the block is related to \( \alpha \) by \( a = \alpha R_1 \):\[ a = 163.33 \times 0.025 = 4.08 \text{ m/s}^2 \]Using \( v^2 = u^2 + 2a s \) where \( u = 0 \) and \( s = 2.0 \text{ m} \):\[ v^2 = 0 + 2 \times 4.08 \times 2.0 = 16.32 \]\[ v = \sqrt{16.32} \approx 4.04 \text{ m/s} \]

Step 4: Analyze Motion with String on Larger Disk

Now consider the string wrapped around the larger disk (radius \( R_2 = 0.05 \text{ m} \)). Use the same process:\[ \alpha = \frac{m g R_2}{I} = \frac{1.50 \times 9.81 \times 0.05}{2.25 \times 10^{-3}} \approx 326.67 \text{ rad/s}^2 \]Linear acceleration \( a = \alpha R_2 \):\[ a = 326.67 \times 0.05 = 16.33 \text{ m/s}^2 \]Using \( v^2 = u^2 + 2a s \) where \( u = 0 \) and \( s = 2.0 \text{ m} \):\[ v^2 = 0 + 2 \times 16.33 \times 2.0 = 65.32 \]\[ v = \sqrt{65.32} \approx 8.08 \text{ m/s} \]

Step 5: Compare Final Speeds

The final speed of the block is greater when the string is wrapped around the larger disk (speed \( \approx 8.08 \text{ m/s} \)) than when it is wrapped around the smaller disk (speed \( \approx 4.04 \text{ m/s} \)). This is because a larger radius provides greater torque for the same tension force, leading to larger angular and thus linear acceleration.

Key Concepts

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate. It's analogous to force in linear motion, but just as force is about moving things, torque is about spinning things.
Torque (\( \tau \)) depends on three factors: the force applied (\( F \)), the distance from the point of rotation (lever arm radius \( r \)), and the angle at which the force is applied. The formula for torque is:

• \( \tau = r \times F \times \sin(\theta) \)

In this example, the tension in the string creates a force, and the radius of the disk acts as the lever arm. The angle is 90 degrees, meaning \( \sin(90^\circ) = 1 \), maximizing the torque's effectiveness. That's why,

• Torque affects how quickly an object can start spinning.
• Greater torque results in faster spin initiation given the same mass moment of inertia.

When dealing with multiple disks as described in our problem, understanding torque helps determine how motion transfers from one rotational body to another.

Angular Acceleration

Angular acceleration (\( \alpha \)) refers to how quickly an object's rotational speed changes. It's the rotational equivalent of linear acceleration and is a key player in determining rotational dynamics.

• Angular acceleration is a vector, meaning it has both magnitude and direction.
• The formula connecting it with torque is \( \alpha = \frac{\tau}{I} \).

In this exercise, we calculated angular acceleration for situations where the string is wrapped around different disks:

• For the smaller disk, angular acceleration was found to be 163.33 rad/s².
• For the larger disk, it was 326.67 rad/s².

Angular acceleration shows us differences in how rapidly each configuration can spin up depending on the disk radius. A larger radius offers more torque, therefore a greater angular acceleration, meaning the system spins up faster. This demonstrates how gaining control over spinning can be achieved by altering the physical parameters of a system.

Linear Acceleration

Linear acceleration (\( a \)) tells us how quickly an object speeds up or slows down along a straight line. It's related to angular acceleration when rotating objects are involved, especially where we wish to determine how this rotational effect translates to linear motion.

• Linear acceleration is measured in m/s², unlike angular acceleration, which is in rad/s².
• It is linked to angular acceleration using the disk's radius: \( a = \alpha \times r \).

In the given problem:

• With the string on the smaller disk, linear acceleration was calculated as 4.08 m/s².
• On the larger disk, it increased to 16.33 m/s².

Notice, linear acceleration depends directly on both the angular acceleration and the radius of the path. This correlation means the block falls faster when leveraging the larger disk because of its increased linear acceleration, proving a greater radius can accelerate linear movement faster, influencing the final speed of the moving block.