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The blade of a circular saw has a diarneter of \(20 \mathrm{~cm}\). If this blade fotatee at 7000 revolutions per minute (its maximum safe ipeed), what arc the spced and the centriperal acccicration of a point on the rim?

Short Answer

Expert verified
Speed: \( 73.3 \text{ m/s} \); Centripetal Acceleration: \( 53689 \text{ m/s}^2 \).

Step by step solution

01

Calculate the Circumference of the Blade

To determine the circumference of the circular saw blade, we use the formula for the circumference of a circle, \( C = \pi \times d \), where \( d \) is the diameter:\[C = \pi \times 20 \mathrm{~cm} = 20\pi \mathrm{~cm}\]
02

Determine the Speed of a Point on the Rim

The speed \( v \) of a point on the rim is the product of the number of revolutions per minute and the circumference:\[v = 7000 \text{ revolutions/minute} \times 20\pi \mathrm{~cm/revolution}= 140000\pi \text{ cm/minute}\] We then convert the speed from cm/minute to m/s:\[140000\pi \text{ cm/minute} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{140000\pi}{6000} \text{ m/s} = \frac{700\pi}{30} \text{ m/s} \approx 73.3 \text{ m/s}\]
03

Calculate the Centripetal Acceleration

The formula for centripetal acceleration \( a_c \) is given by \( a_c = \frac{v^2}{r} \), where \( r \) is the radius of the circle. With the diameter being 20 cm, the radius \( r \) is 10 cm or 0.1 meters, and \( v \approx 73.3 \text{ m/s} \):\[a_c = \frac{(73.3)^2}{0.1} \approx 53689 \text{ m/s}^2\]

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Key Concepts

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

Circumference of a Circle
The circumference of a circle is a fundamental concept in geometry that describes the distance around a circular path. It's an important metric because it determines how far a point travels with each complete rotation around the circle. To find the circumference, you multiply the diameter of the circle by pi (\( \pi \approx 3.14159 \)). In the case of the circular saw blade with a diameter of 20 cm, the circumference is calculated as:
  • Formula: \( C = \pi \times d \)
  • Calculation: \( C = \pi \times 20 \text{ cm} = 20\pi \text{ cm} \)
The unit of the circumference here is centimeters, meaning every full rotation of the blade moves its edge through a distance of \(20\pi \) cm. Understanding this distance is crucial for further calculations, like speed and acceleration, especially in determining how fast a point on the rim is moving.
Angular Speed
Angular speed is a measure of how fast an object moves along a circular path. It's expressed in terms of the angle covered per unit of time, often in radians per second or revolutions per minute. In practical terms, it relates to how quickly an object turns around its center point. For the circular saw blade, angular speed is given as 7000 revolutions per minute.
  • This means the blade spins around its center 7000 times in one minute.
  • Each revolution corresponds to moving around the circumference of the circle.
These revolutions can be converted into linear speed (distance traveled per unit of time) by multiplying the angular speed by the circumference of the circle:\[v = \text{angular speed} \times \text{circumference}\]This understanding helps in calculating the speed along the path of the rim, which in our example converts 7000 revolutions per minute into a tangible speed along the edge of the blade.
Circular Motion
Circular motion describes the movement of an object along a circular path. In physics, this type of motion is characterized not just by speed but also by constant changes in direction, which results in acceleration even if the object's speed is stable. This acceleration is known as centripetal acceleration and is always directed toward the center of the circle.
  • The formula for centripetal acceleration is \( a_c = \frac{v^2}{r} \) where \( v \) is the speed and \( r \) is the radius of the circle.
  • For the blade with a radius of 10 cm (or 0.1 m), the centripetal acceleration can be calculated using the previously determined speed on the rim.
The continuous inward acceleration is crucial for maintaining circular motion as it keeps pulling the moving object toward the center, countering the outward inertia that wants to fling the object outside the path. In our example, this concept reveals the immense force required to keep the saw blade operating smoothly at high speeds.

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Most popular questions from this chapter

According to the Gwinness Book of World Recerds, during a catastrophic explosion in Halifax on December 6,1917, William Becher wae thrown throuph the air for nome \(1500 \mathrm{~m}\) and was found, itill alive, in a tree. Aseume that Becluer left the ground and rctumed to the gacund (ignore the height of the trcc) at an anple of \(45^{\circ}\). With what sipeed did he leave the pround? How high did he rise? How long did he stay in flight?

The Earth rotates ahout its axis once in one sidereal day of 23 h 56 min, Calculate the centripetal acceleration of a point located on the equator. Calculate the centripetal mccelcration of a point located at a latitude of \(45^{\circ}\).

A blinp motoring at a constant altirude tus a velocity component of \(15 \mathrm{~km} / \mathrm{h}\) in the north direction and a velocity component of \(15 \mathrm{~km} / \mathrm{h}\) in the eust dircetion. What is the speed of the blimp? What in the dircction of motion of the blimp?

machine on one occavion achicvod a rangc of \(730 \mathrm{~m}\). If this it true, what murt have been the minimum initial speed of the atone as it wis cjccted from the engine? When thrown with this ipeed, bow long would the stone have taken to read its target?

Suppose that the acecleration voctor of a partick moving in the \(x-y\) planc is $$ a=3 i+2 j $$ where the acceleration is measurcd in \(\mathrm{m} / \mathrm{s}^{2}\). The velocity vector and the povition vectur are rero at \(t=0\). (a) What is the velocity vector of this particle as a finction of time? (b) What is the position vector as a function of time?

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