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Electric Drill. According to the shop manual, when drilling a \(12.7-\mathrm{mm}\) -diameter hole in wood, plastic, or aluminum, a drill should have a speed of 1250 rev/min. For a 12.7 -mm-diameter drill bit turning at a constant 1250 rev \(/ \mathrm{mm}\) , find (a) the maximum linear speed of any part of the hit and (b) the maximum radial acceleration of any part of the bit.

Short Answer

Expert verified
(a) The maximum linear speed is \(0.831\, \text{m/s}\). (b) The maximum radial acceleration is \(108.9\, \text{m/s}^2\).

Step by step solution

01

Identify the Given Values

The diameter of the drill bit is given as \(12.7\, \mathrm{mm}\). The rotational speed is \(1250\, \text{rev/min}\).
02

Convert Units

Convert the diameter to meters for easier calculation: \(12.7\, \mathrm{mm} = 0.0127\, \mathrm{m}\). Also convert the revolutions per minute to radians per second: \[1250\, \text{rev/min} = \frac{1250 \times 2\pi}{60}\, \text{rad/s} \approx 130.9\, \text{rad/s}.\]
03

Calculate the Maximum Linear Speed

The maximum linear speed \(v\) is at the edge of the drill bit and is calculated using the formula \(v = \omega r\), where \(\omega\) is the angular speed and \(r\) is the radius of the bit. The radius \(r = \frac{0.0127}{2}\, \mathrm{m} = 0.00635\, \mathrm{m}\). Therefore, \[v = 130.9\, \text{rad/s} \times 0.00635\, \mathrm{m} \approx 0.831 \, \text{m/s}.\]
04

Calculate the Maximum Radial Acceleration

Radial (centripetal) acceleration \(a_c\) is given by the formula \(a_c = \omega^2 r\). Substitute \(\omega = 130.9\, \text{rad/s}\) and \(r = 0.00635\, \mathrm{m}\) to find \[a_c = (130.9)^2 \times 0.00635 \approx 108.9\, \text{m/s}^2.\]

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Key Concepts

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

Linear Speed
Linear speed refers to how fast an object moves along the path of its motion. When dealing with a rotating object, like a drill bit, the maximum linear speed typically occurs at the perimeter or edge of the object. Imagine you're on a merry-go-round: the farther you are from the center, the faster you're moving in relation to the ground. A similar concept applies to rotating objects.
To calculate the linear speed of a point on the edge of a drill bit, you can use the formula:
  • \( v = \omega r \)
Here, \( v \) is the linear speed, \( \omega \) is the angular speed, and \( r \) is the radius of the circle traced out by the edge of the bit. In our case, the drill bit has a diameter of \(12.7\, \mathrm{mm}\), so its radius is half of that: \(0.00635\, \mathrm{m}\).
Given the angular speed \(\omega = 130.9\, \text{rad/s}\), the linear speed \( v \) comes out to about \(0.831\, \text{m/s}\). This means the edge of the bit moves at a speed of \(0.831\, \text{m/s}\) relative to the center of the circle it makes as it spins.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, is the rate of change of velocity directed towards the center of the circular path. This is crucial because without this inward force, the rotating object would fly off in a straight line due to inertia. Imagine twirling a stone tied to a string; the tension in the string provides the centripetal force pulling the stone towards your hand.
For a rotating drill bit, the maximum radial acceleration can be calculated using:
  • \( a_c = \omega^2 r \)
In this formula, \( \omega \) is the angular speed, and \( r \) is the radius of the circular path followed by the drill bit's edge. By substituting \( \omega = 130.9\, \text{rad/s} \) and \( r = 0.00635\, \mathrm{m} \), the radial acceleration \( a_c \) is approximately \( 108.9\, \text{m/s}^2 \).
This high value shows how much acceleration is needed to keep the edge of the drill bit rotating in its circular path, preventing it from moving off in a tangent.
Angular Speed
Angular speed describes how quickly an object rotates or revolves relative to another point, usually the center of rotation. It's like the speedometer for rotational motion, measuring how fast an object goes around a circle.
In rotational motion, angular speed is denoted by \( \omega \) and is measured in radians per second \(\text{rad/s}\). This differs from linear speed, which measures how fast an object moves along a path. The relationship between the two in circular motion is given by:
  • \( \omega = \frac{v}{r} \)
  • \( v = \omega r \) (for linear speed)
For our drill bit, the angular speed was initially given as \( 1250\, \text{rev/min} \). After converting revolutions per minute to radians per second, we find \( \omega \approx 130.9 \text{rad/s} \). This conversion is crucial because radians are the standard unit in physics for measuring rotation angles.
Understanding angular speed is vital, as it helps determine other important quantities like linear speed and radial acceleration, ultimately allowing us to better grasp the dynamics of rotating objects.

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Most popular questions from this chapter

A turntable rotates with a constant 2.25 \(\mathrm{rad} / \mathrm{s}^{2}\) angular acceleration. After 4.00 \(\mathrm{s}\) it has rotated through an angle of 60.0 \(\mathrm{rad}\) . What was the angular velocity of the wheel at the beginning of the \(4.00-\mathrm{s}\) interval?

Utracentrifuge. Find the required angular speed (in rev/min) of an ultracentrifuge for the radial acceleration of a point 2.50 \(\mathrm{cm}\) from the axis to equal \(400,000 \mathrm{g}\) (that is, \(400,000\) times the acceleration due to gravity).

A thin, uniform rod is bent into a square of side length \(a\) . If the total mass is \(M\) , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

You are to design a rotating cylindrical axle to lift \(800-\mathrm{N}\) buckets of cement from the ground to a rooftop 78.0 \(\mathrm{m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turms, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 \(\mathrm{cm} / \mathrm{s}\) when it is turning at 7.5 \(\mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

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