/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Two metal disks, one with radius... [FREE SOLUTION] | 91Ó°ÊÓ

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Two metal disks, one with radius \(R_{1}=2.50 \mathrm{cm}\) and mass \(M_{1}=0.80 \mathrm{kg}\) and the other with radius \(R_{2}=5.00 \mathrm{cm}\) and mass \(M_{2}=1.60 \mathrm{kg}\) , are welded together and mounted on a frictionless axis through their common center, as in Problem \(9.89 .\) (a) A light string is wrapped around the edge of the smaller disk, and a 1.50 \(\mathrm{kg}\) block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released? (b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater? Does your answer make sense?

Short Answer

Expert verified
The block accelerates faster with the smaller disk: 2.28 m/s² vs. 1.63 m/s² with the larger disk. This is because the smaller radius requires less torque for the same acceleration.

Step by step solution

01

Determine the total moment of inertia

The total moment of inertia, \( I_t \), for two disks welded together is the sum of their individual moments of inertia. The formula for the moment of inertia of a single disk about its central axis is \( I = \frac{1}{2}MR^2 \). Thus:\[ I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} \times 0.80 \, \text{kg} \times (0.025 \, \text{m})^2 = 2.5 \times 10^{-4} \, \text{kg} \, \text{m}^2 \]\[ I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} \times 1.60 \, \text{kg} \times (0.05 \, \text{m})^2 = 2.0 \times 10^{-3} \, \text{kg} \, \text{m}^2 \]\[ I_t = I_1 + I_2 = 2.5 \times 10^{-4} \, \text{kg} \, \text{m}^2 + 2.0 \times 10^{-3} \, \text{kg} \, \text{m}^2 = 2.25 \times 10^{-3} \, \text{kg} \, \text{m}^2 \]
02

Analyze the system with the string around the smaller disk

When the string is wrapped around the smaller disk, the torque \( \tau \) due to the weight of the block is calculated as:\[ \tau = T R_1 \]where \( T \) is the tension in the string and \( R_1 \) is the radius of the disk.The net torque \( \tau = I_t \alpha \), where \( \alpha \) is the angular acceleration.The relation between linear acceleration \( a \) and angular acceleration \( \alpha \) is \( a = R_1 \alpha \).Using the force equation for the block: \( M_3 g - T = M_3 a \).Combining these equations we can solve for \( a \):\[ a = \frac{M_3 g R_1}{I_t + M_3 R_1^2} \]Substituting the values, \( M_3 = 1.50 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \):\[ a = \frac{1.50 \times 9.81 \times 0.025}{2.25 \times 10^{-3} + 1.50 \times 0.025^2} = 2.28 \, \text{m/s}^2 \].
03

Analyze the system with the string around the larger disk

Repeat the previous analysis but using the radius of the larger disk \( R_2 \):\[ a = \frac{M_3 g R_2}{I_t + M_3 R_2^2} \]Substituting \( R_2 = 0.05 \, \text{m} \):\[ a = \frac{1.50 \times 9.81 \times 0.05}{2.25 \times 10^{-3} + 1.50 \times 0.05^2} \approx 1.63 \, \text{m/s}^2 \].
04

Compare accelerations and conclude

In part (a), the acceleration is \( 2.28 \, \text{m/s}^2 \) and in part (b), it is \( 1.63 \, \text{m/s}^2 \). Thus, the acceleration is greater when the string is wrapped around the smaller disk. This makes sense as a smaller radius requires less torque to achieve the same angular acceleration, allowing more of the gravitational force to contribute to the linear acceleration of the block.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Acceleration
Angular acceleration describes how quickly the angular velocity of an object changes. It is a critical concept in rotational dynamics. This parameter can be compared to linear acceleration but applied to rotational motion instead. To find angular acceleration, we use the formula \(\alpha = \frac{\tau}{I_t}\), where \(\tau\) is the torque applied, and \(I_t\) is the moment of inertia.
Central to understanding how forces affect rotating objects, angular acceleration connects with both angular velocity and angular displacement through the kinematic equations of motion. In this exercise, angular acceleration helps us determine how quickly the welded disks accelerate when a torque is applied by the string tied to different-radius disks.
Torque
Torque is the rotational equivalent of linear force and is crucial in determining how an object will rotate. Just as force affects linear acceleration, torque influences angular acceleration. In any system with rotation, such as the combined disk system, torque is defined as the product of the force and the radius at which the force is applied, denoted by \( \tau = F \times R\).
When the string is wrapped around a disk and a block descends, the gravitational force on the block is transferred into torque on the disk. The point where the force is applied on the disk makes a significant difference. A smaller radius, as with the smaller disk, results in a higher magnitude of angular acceleration, given the same force, because less torque is needed relative to a larger radius.
Linear Acceleration
Linear acceleration refers to the rate of change of velocity with respect to time for an object moving along a straight path. It is directly related to the forces acting on the object. In this exercise, linear acceleration is affected by how the disks spin due to the tangential pull of gravity on the block.
The conversion between angular and linear acceleration can be made through the formula \( a = \alpha \times R \), where \( a \) is linear acceleration, \( \alpha \) is angular acceleration, and \( R \) is the radius of the rotation. It illustrates that tying a string around a smaller radius produces a greater linear acceleration, as found in the results of this exercise, due to the efficient use of torque.
Physics Problems
Solving physics problems often involves understanding the relationships between concepts like force, motion, and rotational dynamics. They require a clear approach starting from diagnosing what you know, the mechanics involved, and applying the right equations and principles to find the unknowns.
For this problem, we calculated the effects of radius and mass on the moment of inertia, how torque affects both angular and linear acceleration, and finally how to interpret the results in terms of physical intuition. Such problems test one's ability to switch between linear and rotational thinking, ensuring a thorough grasp of principles such as conservation of energy and Newton's laws, even when they manifest in the form of rotational dynamics.
Solving them helps build a deep understanding of fundamental physics principles, translating complex mechanical systems into simpler, solvable parts.

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