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Chapter 10: Problem 7

A certain type of propeller blade can be modeled as a thin uniform bar 2.50 m long and of mass 24.0 \(\mathrm{kg}\) that is free to rotate about a frictionless axle perpendicular to the bar at its midpoint. If a technician strikes this blade with a mallet 1.15 \(\mathrm{m}\) from the center with a 35.0 \(\mathrm{N}\) force perpendicular to the blade, find the maximum angular acceleration the blade could achieve.

Short Answer

Expert verified
The maximum angular acceleration is \( 3.22\, \text{rad/s}^2 \).

Step by step solution

01

Identify the Moment of Inertia Formula

The moment of inertia (I) for a thin rod rotating about its center is given by the formula \( I = \frac{1}{12} m L^2 \), where \( m \) is the mass and \( L \) is the length of the rod.
02

Substitute Values into Moment of Inertia Formula

Substitute the given mass \( m = 24.0\, \text{kg} \) and length \( L = 2.50\, \text{m} \) into the moment of inertia formula: \( I = \frac{1}{12} (24.0\, \text{kg})(2.50\, \text{m})^2 \).
03

Calculate the Moment of Inertia

Calculate the value: \( I = \frac{1}{12} \times 24 \times 6.25 = 12.5\, \text{kg} \cdot \text{m}^2 \).
04

Identify the Torque Formula

The torque (\( \tau \)) applied by the force is calculated using the formula \( \tau = r \times F \), where \( r \) is the distance from the pivot and \( F \) is the perpendicular force applied.
05

Substitute Values into Torque Formula

Substitute the distance \( r = 1.15\, \text{m} \) and force \( F = 35.0\, \text{N} \) into the torque formula: \( \tau = 1.15\, \text{m} \times 35.0\, \text{N} \).
06

Calculate the Torque

Calculate the value: \( \tau = 40.25\, \text{Nm} \).
07

Identify the Angular Acceleration Formula

Angular acceleration (\( \alpha \)) is given by \( \alpha = \frac{\tau}{I} \), where \( \tau \) is the torque and \( I \) is the moment of inertia.
08

Substitute Values into Angular Acceleration Formula

Substitute the calculated values: \( \alpha = \frac{40.25\, \text{Nm}}{12.5\, \text{kg} \cdot \text{m}^2} \).
09

Calculate the Angular Acceleration

Calculate the value: \( \alpha = 3.22\, \text{rad/s}^2 \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a key concept in rotational dynamics, akin to mass in linear motion. It describes how difficult it is to change the rotational state of an object. For a thin rod rotating about its center, the moment of inertia is calculated using the formula: \( I = \frac{1}{12} m L^2 \). Here, \( m \) represents the mass of the rod, and \( L \) stands for its length.
  • In our propeller blade example, the mass \( m \) is 24.0 kg and the length \( L \) is 2.50 m.
  • So the moment of inertia \( I \) becomes: \( \frac{1}{12} \times 24.0 \, \text{kg} \times (2.50 \text{m}^2) \), which simplifies to 12.5 kg\cdot m^2.

Understanding the moment of inertia is crucial for solving rotational motion problems, as it directly influences how much torque is required to achieve a desired angular acceleration.
Rotational Motion
Rotational motion refers to the movement of an object around a central point or axis. In the case of the propeller blade, it rotates around its midpoint due to the force applied by the technician. Just like linear motion depends on force, rotational motion depends on torque.
  • Torque (\( \tau \)) can be thought of as a rotational equivalent of force. It's calculated using \( \tau = r \times F \), where \( r \) is the distance from the axis of rotation and \( F \) is the force applied perpendicularly.
  • For our rotating blade, \( r \) is 1.15 m, and the force \( F \) is 35.0 N, resulting in a torque of 40.25 Nm.

Rotational motion is common in real-world applications and understanding how torque causes rotation is vital for careers in physics and engineering.
Physics Problem-Solving
Solving physics problems often involves understanding and applying basic principles to new situations. It requires breaking down the problem into manageable steps and systematically working through calculations.
  • Start by identifying the known quantities, such as mass, length, force, and distance in the case of rotational motion.
  • Apply relevant formulas to those quantities. For example, use the moment of inertia formula to find the rotational resistance and the torque formula to find the rotational force.
  • Keep track of units and consistently substitute them into formulas to ensure dimensional accuracy.
  • Combine your findings to calculate desired quantities like angular acceleration using \( \alpha = \frac{\tau}{I} \), ensuring each step logically follows the last.

Mastering the skill of solving physics problems involves practice and a clear understanding of fundamental concepts. With time, the process of identifying, substituting, and calculating will become intuitive.

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Most popular questions from this chapter

A solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

\(\bullet\) 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)?

A solid disk of radius 8.50 \(\mathrm{cm}\) and mass \(1.25 \mathrm{kg},\) which is rolling at a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) , begins rolling without slipping up a \(10.0^{\circ}\) slope. How long will it take for the disk to come to a stop?

\bullet A small block on a frictionless horizontal surface has a mass of 0.0250 \(\mathrm{kg}\) . It is attached to a massless cord passing through a hole in the surface. (See Figure \(10.54 .\) The block is originally revolving at a distance of 0.300 \(\mathrm{m}\) from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 \(\mathrm{m} .\) You may treat the block as a particle. (a) Is angular momentum conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?
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