/*! 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 70 The motor of a table saw is rota... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 70

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by moans of a V-belt. A circular saw blade of diameter 0.208 \(\mathrm{m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't to its teeth.

Short Answer

Expert verified
The tangential speed is approximately 706.857 m/s. The radial acceleration is about 4.81 脳 10^7 m/s虏.

Step by step solution

01

Find the rotational speed of the second pulley

The motor pulley drives the second pulley that is half its diameter. Therefore, the rotational speed of the second pulley is double the speed of the motor pulley. The motor speed is 3450 rev/min, so the speed of the second pulley is \(2 \times 3450 = 6900 \text{ rev/min}\). Convert this to radians per second: \[6900 \frac{2\pi}{60} = 690 \pi \text{ rad/s}.\]
02

Calculate the tangential speed of the rim of the blade

The linear or tangential speed (v) of the rim of the blade is related to the angular speed (\(\omega\)) by the formula \(v = r\omega\), where \(r\) is the radius of the blade. The diameter of the blade is 0.208 m, so \(r = \frac{0.208}{2} = 0.104\) m. The rotational speed of the blade is the same as the second pulley, which is \(690\pi\) rad/s. Therefore, \[v = 0.104 \times 690\pi = 224.64\pi \text{ m/s} \approx 706.857 \text{ m/s}.\]
03

Calculate the radial (centripetal) acceleration

The radial or centripetal acceleration \(a_r\) is given by \(a_r = r\omega^2\). Using \(r = 0.104\) m and \(\omega = 690\pi\), we find: \[a_r = 0.104 \times (690\pi)^2 = 48763.4\pi^2 \text{ m/s}^2 \approx 4.81 \times 10^7 \text{ m/s}^2.\]
04

Interpret the significance of radial acceleration

The radial acceleration of \(4.81 \times 10^7 \text{ m/s}^2\) is extremely large. This indicates that the points on the rim of the blade experience a significant force directed towards the center of the circular path, resulting in sawdust being ejected due to the inability of the weak forces (like friction) to hold it against such acceleration.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast something rotates. It's like the speedometer for rotation. When we talk about angular velocity, we are typically referring to how many rotations or angles an object spins through over time. In mathematical terms, angular velocity, represented by the Greek letter \( \omega \), is expressed in radians per second (rad/s).

In the context of a rotating object like a circular saw blade, angular velocity helps in understanding how fast the blade is spinning around its center. The angular velocity is crucial because it directly influences the blade's tangential speed, which is the speed at which points on the outer edge of the blade move. In the original problem, we saw that the motor's speed translates to the angular velocity of connected components. The motor's rotation at 3450 rev/min results in the second pulley's increased angular speed due to the difference in pulley sizes, calculated as \( 690 \pi \) rad/s.

Understanding angular velocity allows us to determine other properties of rotating systems, like how fast sawdust might fly off from the blade or how quickly a piece of wood is thrown by the saw.
Centripetal Acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. This isn't about speeding up or slowing down but rather about keeping the object moving in that circular path. Without centripetal acceleration, the object would move off in a straight line.

For this reason, centripetal acceleration is crucial in any rotational setting. Its formula is given by \( a_r = r\omega^2 \), where \( r \) is the radius of the circle, and \( \omega \) is the angular velocity. It measures how fast something must "pull" toward the center to keep moving in a curve.

In the exercise provided, the radial acceleration of the circular saw blade's rim is extremely high, highlighting why sawdust or other materials can't easily stick to the blade. The massive acceleration value \( 4.81 \times 10^7 \text{ m/s}^2 \) suggests that any particles entering the rotational path of the blade are immediately pushed outward. This is an excellent demonstration of how centripetal acceleration functions in everyday tools.
Tangential Speed
Tangential speed is the speed at which a point on the edge of a rotating object moves along its path. Imagine a spinning record; the outer edge moves further than the center in the same time period, meaning it moves faster. So, tangential speed is crucial for understanding how different parts of a rotating object behave.

In a rotating disk or blade, tangential speed \( v \) is calculated using the formula \( v = r\omega \), where \( r \) is the radius, and \( \omega \) is the angular velocity. This speed represents how fast points on the outer edge of the blade are moving.

In the problem, the tangential speed of the saw blade is 706.857 m/s, meaning any particle on the blade's edge is moving at this enormous speed. This speed accounts for why a small piece of wood is thrown swiftly when caught in the blade; it gains this tangential velocity due to the sharp edge's rotation. Understanding tangential speed aids in visualizing the quick motion and power of rotating systems.

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Most popular questions from this chapter

When a toy car is rapidly scooted across the floor, it stores energy in a flywheel. The car has mass \(0.180 \mathrm{kg},\) and its flywheel has moment of inertia \(4.00 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The car is 15.0 \(\mathrm{cm}\) long. An advertisement claims that the car can travel at a scale speed of up to 700 \(\mathrm{km} / \mathrm{h}(440 \mathrm{mi} / \mathrm{h}) .\) The scale speed is the speed of the toy car multiplied by the ratio of the length of an actual car. to the length of the toy. Assume a length of 3.0 \(\mathrm{m}\) for a real car. (a) For a scale speed of 700 \(\mathrm{km} / \mathrm{h}\) , what is the actual translational speed of the car? (b) If all the kinetic energy that is initially in the flywheel is converted to the translational kinetic energy of the toy, how much energy is originally stored in the flywheel? (c) What initial angular velocity of the flywheel was needed to store the amount of energy calculated in part (b)?

A sphere with radius \(R=0.200 \mathrm{m}\) has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(\rho=3.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}-\left(9.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{4}\right) r .\) (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 \(\mathrm{cm} .\) If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 \(\mathrm{m} / \mathrm{s} ?\) The rear wheel has radius 0.330 \(\mathrm{m} .\)

A disk of radius 25.0 \(\mathrm{cm}\) cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.65). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Find \(A .\) (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .\) .

While redesigning a rocket engine, you want to reduce its weight by replacing a solid spherical part with a hollow spherical shell of the same size. The parts rotate about an axis through their center. You need to make sure that the new part always has the same rotational kinetic energy as the original part had at any given rate of rotation. If the original part had mass \(M,\) what must be the mass of the new part?
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