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Chapter 10: Problem 2

A dentist's drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of \(2.51 \times 10^{4}\) rev/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
The dentist's drill has an angular acceleration of about \( 826 \) rad/\( s^{2} \) and the drill rotates through an angle of about \( 4,211 \) radians.

Step by step solution

01

Convert rev/min to rad/s

Given that the final angular speed of the drill is given as \(2.51 \times 10^{4}\) rev/min. This must be converted to radians per second (rad/s) using the equation \( \omega_f = 2.51 \times 10^{4} \times \frac{2\pi}{60} \).
02

Calculate the angular acceleration

Using the kinematics equation \( \omega_{f} = \omega_{i} + \alpha t\), where \( \alpha\) is the angular acceleration, \( t = 3.20s\) is the time, \( \omega_{f}\) is the final angular speed, and \( \omega_{i}\) is the initial angular speed (which is zero because the drill starts from rest), you can isolate and find \( \alpha = \frac{\omega_{f}}{t}\).
03

Calculate the angle of rotation

Applying the kinematics equation \( \theta = \omega_{i}t + 0.5 \alpha t^{2}\), where \( \theta\) is the angle of rotation, \( \alpha\) is the angular acceleration calculated in Step 2, \( t = 3.20s\) is the time, and \( \omega_{i}\) is the initial angular speed (which is zero because the drill starts from rest), you have \( \theta = 0.5 \alpha t^{2} \).

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Key Concepts

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

Kinematics Equations
In the world of physics, kinematics equations are crucial for describing the motion of objects, without having to consider the forces that cause such motion. They are particularly useful in understanding scenarios involving constant acceleration, such as a dentist's drill speeding up from a stationary position.

When it comes to rotational motion, these kinematic equations can be adapted by replacing linear displacement, velocity, and acceleration with their angular counterparts: angular displacement (theta, \theta), angular velocity (omega, \(\omega\)), and angular acceleration (alpha, \(\alpha\)). An example of such an equation is \(\omega_{f} = \omega_{i} + \alpha t\), which allows us to calculate angular acceleration if the initial and final angular velocities are known, as well as the time taken to change between these velocities. This relationship was used in the step-by-step solution to solve for the drill's angular acceleration.
Angular Velocity
Angular velocity represents how fast an object rotates or revolves relative to another point, which is usually the center of a circle or the axis of rotation. It is denoted by the symbol \(\omega\) and measured in radians per second (rad/s).

Understanding angular velocity is key to interpreting rotational motion. For example, when the dentist's drill reaches an angular velocity of 2.51 x 104 revolutions per minute after 3.20 seconds, we realize the rate at which the drill is spinning. To use this information in kinematics equations, we often need to convert from revolutions per minute to radians per second, because radians provide a standard, unitless measure that can be applied universally, regardless of the size of the circle.
Radians per Second
The radian is the standard unit of angular measure used in many areas of mathematics. To grasp the concept of radians per second, imagine the arc length of a circle that is equal to the radius of that circle 鈥� this defines an angle of one radian. Since there are 2\(\pi\) radians in a full circle, this conversion is essential when dealing with rotational speeds or angular velocities.

For instance, when the solution to the exercise involves converting from rev/min (commonly used in practical scenarios) to rad/s (used in kinematics equations), we multiply the given value by \(\frac{2\pi}{60}\) since there are 2\(\pi\) radians in one revolution and 60 seconds in one minute. This conversion allows us to work with the angle and speed of rotation in a format that's compatible with the standard kinematic formulas. Understanding radians per second as a measure of angular speed is fundamental for solving problems related to rotation in physics.

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Most popular questions from this chapter

The density of the Earth, at any distance \(r\) from its center, is approximately $$ \rho=[14.2-11.6(r / R)] \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} $$ where \(R\) is the radius of the Earth. Show that this density leads to a moment of inertia \(I=0.330 M R^{2}\) about an axis through the center, where \(M\) is the mass of the Earth.

A car accelerates uniformly from rest and reaches a speed of \(22.0 \mathrm{m} / \mathrm{s}\) in \(9.00 \mathrm{s} .\) If the diameter of a tire is \(58.0 \mathrm{cm}\) find (a) the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of \(12.0 \mathrm{rad} / \mathrm{s}\) in 3.00 s. Find (a) the magnitude of the angular acceleration of the wheel and (b) the angle in radians through which it rotates in this time.

Consider a tall building located on the Earth's equator. As the Earth rotates, a person on the top floor of the building moves faster than someone on the ground with respect to an inertial reference frame, because the latter person is closer to the Earth's axis. Consequently, if an object is dropped from the top floor to the ground a distance \(h\) below, it lands east of the point vertically below where it was dropped. (a) How far to the east will the object land? Express your answer in terms of \(h, g,\) and the angular speed \(\omega\) of the Earth. Neglect air resistance, and assume that the free-fall acceleration is constant over this range of heights. (b) Evaluate the eastward displacement for \(h=50.0 \mathrm{m}\) (c) In your judgment, were we justified in ignoring this aspect of the Coriolis effect in our previous study of free fall?

A rotating wheel requires \(3.00 \mathrm{s}\) to rotate through 37.0 revolutions. Its angular speed at the end of the \(3.00-s\) interval is \(98.0 \mathrm{rad} / \mathrm{s} .\) What is the constant angular acceleration of the wheel?
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