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A dentist's drill starts from rest. After \(3.20 \mathrm{~s}\) of constant angular acceleration, it turns at a rate of \(2.51 \times 10^{4} \mathrm{rev} / \mathrm{min}\). (a) Find the drill's angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

Short Answer

Expert verified
(a) The angular acceleration of the drill is \(822 rad/s\(^2\) \) (b) The drill rotates about \( 4190 \) radians.

Step by step solution

01

Converting units

First convert the given final speed from rev/min to rad/s for easier use in the formulas. Note that \( 1 \text{ rev} = 2 \pi \text{ rad} \), and there are 60 seconds in one minute. So the final angular speed is \( \omega = 2.51 \times 10^{4} \text{ rev/min} \times \frac{2 \pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 2630 \text{ rad/s} \).
02

Computing Angular Acceleration

Use the first kinematic equation to solve for \( \alpha \). Start by setting \( \omega_i = 0 \), because the drill starts from rest. Then, input the known values into the equation \( \alpha = (\omega - \omega_i) / t = (2630 rad/s - 0 rad/s) / 3.20 s = 821.875 rad/s\(^2\) \). Thus the angular acceleration is \( \alpha = 822 rad/s\(^2\) \)
03

Computing the Angle

Use the second kinematic equation to find \( \theta \). (We know that the drill starts from rest, so \( \omega_i = 0 \).) By plugging in the values we get, \( \theta = 0.5 * 822 rad/s\(^2\) * (3.20 s)\(^2\) = 4191 rad \) Thus, the drill rotates through about 4190 radians.

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Key Concepts

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

Angular Acceleration
Angular acceleration refers to how quickly the angular velocity of an object changes with time. It is an extremely useful measure, especially when talking about things that rotate or spin since it helps us describe how they start, stop, or change their speeds. In mathematical terms, angular acceleration (\( \alpha \)) can be expressed as the change in angular velocity (\( \Delta \omega \)) over the change in time (\( \Delta t \)).The formula used is:\[\alpha = \frac{\Delta \omega}{\Delta t}\]### Simple Case- If an object starts from rest, its initial angular velocity (\( \omega_i \)) is zero.- The change in angular velocity becomes simply the final angular velocity (\( \omega \)) since \( \omega_i = 0 \).- Therefore, the angular acceleration becomes \( \alpha = \frac{\omega}{t} \). In the case of our dentist's drill, we see that over \(3.20i seconds\), it achieved a final speed of \(2630 \text{ rad/s}\). Using the above formula, the angular acceleration was calculated to be \(822 \text{ rad/s}^2\). This means every second, the drill's angular speed increased by \(822 \text{ rad/s}\), helping it reach its final speed quickly.
Angular Velocity
Angular velocity describes how fast something rotates. Unlike linear velocity, which deals with straight-moving objects, angular velocity is concerned with objects that turn around a specific axis—like a rotating drill. Angular velocity (\( \omega \)) is often measured in radians per second or degrees per second, though radians per second are most common in scientific calculations.### Calculating Angular Velocity- To convert angular velocity from revolutions per minute (rev/min) to radians per second (rad/s), a couple of conversions are necessary: - One complete revolution is \(2\pi \text{ radians}\). - One minute is \(60 \text{ seconds}\). - If given \(2.51 \times 10^4 \text{ rev/min}\), it changes to \(2630 \text{ rad/s}\) after conversion.Essentially, angular velocity tells us how fast an object spins and is crucial for understanding rotational motion. For example, the drill's angular velocity indicates how rapidly it spins as it performs operations. Understanding this rate helps ensure efficiency during use, like in dental procedures where precision and speed are needed.
Angular Displacement
Angular displacement measures how far an object has rotated or turned around a particular point or axis. It is usually measured in radians or degrees, with radians being preferred in most calculations due to their ability to simplify mathematical expressions.### Calculating Angular Displacement- When starting from rest (\( \omega_i = 0 \)), the basic kinematic equation for angular motion is: \[\theta = \frac{1}{2} \alpha t^2\]- In this formula, \( \theta \) represents angular displacement, \( \alpha \) is the angular acceleration, and \( t \) is the time.- From our problem, using an angular acceleration of \(822 \text{ rad/s}^2\) over \(3.20 \text{ seconds}\), the drill covers an angle of about \(4191\) radians. This tells us how far or through what angle the drill has turned, providing insight into the extent of the work done or the area covered during rotation. Understanding angular displacement helps in knowing how much rotation is needed for various real-life applications, enhancing accuracy in tasks that require specific rotational movements.

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Most popular questions from this chapter

ecp The pilot of an airplane executes a constant-speed loop-the-loop maneuver in a vertical circle as in Figure 7.15b. The speed of the airplane is \(2.00 \times 10^{2} \mathrm{~m} / \mathrm{s}\), andthe radius of the circle is \(3.20 \times 10^{3} \mathrm{~m}\). (a) What is the pilot's apparent weight at the lowest point of the circle if his true weight is \(712 \mathrm{~N}\) ? (b) What is his apparent weight at the highest point of the circle? (c) Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body. Under what conditions does this occur? (d) What speed would have resulted in the pilot experiencing weightlessness at the top of the loop?

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A \(50.0\) -kg child stands at the rim of a merry-go-round of radius \(2.00 \mathrm{~m}\), rotating with an angular speed of \(3.00 \mathrm{rad} / \mathrm{s} .\) (a) What is the child's centripetal acceleration? (b) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path? (c) What minimum coefficient of static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the merry-goround?

Casting of molten metal is important in many industrial processes. Centrifugal casting is used for manufacturing pipes, bearings, and many other structures. A cylindrical enclosure is rotated rapidly and steadily about a horizontal axis, as in Figure \(\mathrm{P} 7.64 .\) Molten metal is poured into the rotating cylinder and then cooled, forming the finished product. Turning the cylinder at a high rotation rate forces the solidifying metal strongly to the outside. Any bubbles are displaced toward the axis so that unwanted voids will not be present in the casting. Suppose a copper sleeve of inner radius \(2.10 \mathrm{~cm}\) and outer radius \(2.20 \mathrm{~cm}\) is to be cast. To eliminate bubbles and give high structural integrity, the centripetal acceleration of each bit of metal should be \(100 \mathrm{~g}\). What rate of rotation is required? State the answer in revolutions per minute.

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