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Chapter 9: Problem 64

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 means of a V-belt. A circular saw blade of diameter 0.208 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鈥檛 stick to its teeth.

Short Answer

Expert verified
(a) The tangential speed of the blade is approximately 75.1 m/s. (b) The radial acceleration is approximately 54217 m/s虏.

Step by step solution

01

Calculate the Angular Speed of the Motor

The motor rotates at 3450 revolutions per minute (rev/min). To convert this to radians per second (rad/s), we use the formula: \[ \omega_{1} = \frac{3450 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60 \text{ s/min}} \]. Calculating this gives \( \omega_{1} \approx 361.1 \text{ rad/s} \).
02

Determine the Angular Speed of the Second Pulley

The second pulley has half the diameter of the first pulley, which means it will rotate at double the angular speed of the motor because the linear speeds at the pulleys' rims are equal: \( \omega_{2} = 2 \omega_{1} = 2 \times 361.1 \text{ rad/s} = 722.2 \text{ rad/s} \).
03

Find the Tangential Speed of the Blade

The blade shares the shaft with the second pulley, so it has the same angular speed \( \omega_{2} = 722.2 \text{ rad/s} \). The tangential speed \( v \) of the blade's rim is given by \( v = \omega_{2} \times r \), where \( r \) is the radius of the blade, \( r = \frac{0.208}{2} \) m. Calculating \[ v = 722.2 \times 0.104 \approx 75.1 \text{ m/s} \].
04

Calculate the Radial (Centripetal) Acceleration

The radial acceleration \( a_{r} \) of a point on a rotating object is given by \( a_{r} = \omega^{2} \times r \). Using \( \omega_{2} = 722.2 \text{ rad/s} \) and \( r = 0.104 \text{ m} \), we calculate \[ a_{r} = 722.2^{2} \times 0.104 \approx 54217 \text{ m/s}^2 \].

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Key Concepts

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

Angular Speed
Angular speed refers to how quickly an object rotates around a central point. This concept is crucial when dealing with circular motion in systems like a table saw.
Angular speed is measured in radians per second (rad/s), which quantifies how much rotation takes place per unit time. For example, in the exercise, the motor starts at 3450 revolutions per minute. To convert this into angular speed in radians per second, we use the conversion formula:
  • Multiply the revolutions per minute by \(2\pi\) to account for the full circle in radians.
  • Divide by 60 to convert minutes into seconds.
This converts the motor's speed to approximately 361.1 rad/s.
Since the second pulley's diameter is halved, and its circumference鈥攖he crucial factor for linear speed鈥攔elies on the radius, the pulley rotates at twice the motor鈥檚 angular speed, thus having 722.2 rad/s as its angular speed.
Radial (Centripetal) Acceleration
Radial acceleration, or centripetal acceleration, is what keeps an object moving in a circular path. This force acts towards the center of the circle, ensuring the continued path of rotation.
In the context of the saw blade, the radial acceleration is a critical factor as it explains why sawdust and other debris do not cling to the blade. The formula to find radial acceleration in a rotating object is:
  • \( a_{r} = \omega^2 \times r \)
where \(\omega\) is the angular speed, and \(r\) is the radius of the rotation.
For the blade, with an angular speed of 722.2 rad/s and a radius of 0.104 m, the radial acceleration is calculated to be an astounding 54217 m/s虏. This high acceleration rapidly throws off any small particles from the blade, making sure it remains clean and efficient during operation.
Tangential Speed
Tangential speed refers to the linear speed of any point on a rotating object that moves along the tangent to its circular path.
For an object rotating around a circle, like the blade of a saw, the tangential speed can be found using the formula:
  • \( v = \omega \times r \)
where \(v\) is the tangential speed, \(\omega\) is the angular speed, and \(r\) is the radius of the circle.
In the problem context, since the blade rotates at an angular speed of 722.2 rad/s and has a radius of 0.104 m, the tangential speed at the rim of the blade is about 75.1 m/s. Importantly, this is the speed at which the piece of wood is "thrown" by the blade, explaining its rapid ejection from the saw.

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Most popular questions from this chapter

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius 20.0 cm. A 1.50-kg stone is attached to a very light wire that is wrapped around the rim of the pulley (\(\textbf{Fig. E9.43}\)), and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

A uniform disk with radius \(R =\) 0.400 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t) =\) (1.10 rad/s)\(t +\) (6.30 rad/s\(^2)t^2\). What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

A light, flexible rope is wrapped several times around a \(hollow\) cylinder, with a weight of 40.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of 5.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is \(P\)?

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t) = \gamma t^2 - \beta t^3\), where \(\gamma =\) 3.20 rad/s\(^2\) and \(\beta =\) 0.500 rad/s\(^3\). (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of t does it occur?

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.45. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 m and moment of inertia for rotation about the axle of \(I =\) 0.480 kg \(\cdot\) m\(^2\). A small block with mass 0.340 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-\)9.00 J of work as the block descends 3.00 m. What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 m?
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