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Chapter 8: Problem 52

An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. What is the angle?

Short Answer

Expert verified
The angle is 1 radian.

Step by step solution

01

Understand the Relationship between Accelerations

Centripetal acceleration \(a_c\) is given by \(a_c = r\omega^2\), where \(r\) is the radius and \(\omega\) is the angular velocity. Tangential acceleration \(a_t\) is given by \(a_t = r\alpha\), where \(\alpha\) is the angular acceleration. From the problem, we know \(a_c = 2a_t\).
02

Substitute Angular Acceleration

Using \(a_c = 2a_t\), we get \(r\omega^2 = 2(r\alpha)\). This simplifies to \(\omega^2 = 2\alpha\).
03

Use Kinematic Equation for Angular Motion

The angular motion equation \(\omega^2 = \omega_0^2 + 2\alpha\theta\) can be simplified since the drill starts from rest, leading to \(\omega^2 = 2\alpha\theta\).
04

Solve for the Angle

Substitute \(\omega^2 = 2\alpha\) into the equation \(\omega^2 = 2\alpha\theta\) to find \(2\alpha = 2\alpha\theta\). Simplifying this gives \(\theta = 1\).
05

Conclude with Angle

From the simplification, the angle \(\theta\) through which the drill rotates when the centripetal acceleration equals twice the tangential acceleration is \(\theta = 1\) radian.

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a key concept when discussing objects in circular motion. It refers to the acceleration that keeps an object moving in a circular path. The direction of centripetal acceleration is always towards the center of the circle on which the object is moving. To calculate the centripetal acceleration, we use the formula:
  • \( a_c = r \omega^2 \)
where \( r \) is the radius of the circle and \( \omega \) is the angular velocity.
The importance of centripetal acceleration stems from the need to maintain the circular motion of an object. Without it, the object would move in a straight line due to inertia. When considering a rotating drill, as in our example, the centripetal acceleration at any point ensures that the drill bit continues its circular journey instead of flying off.
Understanding that centripetal acceleration can greatly exceed tangential acceleration is crucial in mastering the balance of forces involved in angular motion.
Tangential Acceleration
Tangential acceleration describes the change in the speed of an object along its circular path. It acts in the direction of the tangent to the circle at the object's position. If an object speeds up or slows down along the curve of the circle, it experiences tangential acceleration.
  • The formula is: \( a_t = r \alpha \)
where \( \alpha \) represents angular acceleration and \( r \) is the radius of the circular path.
In practical terms, imagine a car accelerating on a curved track. As it speeds up, its tangential acceleration changes due to increased speed. Similarly, a drill increasing its speed sees its tangential acceleration change as well.
Tangential acceleration is essential in scenarios where rotational speed is changing, highlighting its role in dynamic rotational motion.
Angular Acceleration
Angular acceleration refers to how quickly an object speeds up or slows down its rotation. It is the rate of change of angular velocity over time:
  • Denoted by \( \alpha \), it measures in radians per second squared (rad/s虏)
Angular acceleration is tied directly to both centripetal and tangential accelerations. In our exercise, the angular acceleration \( \alpha \) is the key linking factor, allowing us to connect the centripetal velocity (\( \omega \)) and the tangential changes in speed.
Consider the formula \( \omega^2 = 2\alpha\theta \), where the drill, starting from rest, sees its angular velocity build up due to constant angular acceleration. This relation helps us find how far the drill turns before certain acceleration conditions are met.
By grasping angular acceleration, one understands how constantly changing forces maintain the motion, whether starting, stopping, or adjusting speeds in rotary systems.

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Most popular questions from this chapter

mmh The drawing shows a chain-saw blade. The rotating sprocket tip at the end of the guide bar has a radius of \(4.0 \times 10^{-2} \mathrm{m} .\) The linear speed of a chain link at point A is 5.6 m/s. Find the angular speed of the sprocket tip in rev/s.

In \(9.5 \mathrm{~s}\) a fisherman winds \(2.6 \mathrm{~m}\) of fishing line onto a reel whose radius is \(3.0 \mathrm{~cm}\) (assumed to be constant as an approximation). The line is being reeled in at a constant speed. Determine the angular speed of the reel.

A baseball pitcher throws a base- ball horizontally at a linear speed of 42.5 \(\mathrm{m} / \mathrm{s}\) (about 95 \(\mathrm{mi} / \mathrm{h} ) .\) Before being caught, the bascball travels a horizontal distance of 16.5 \(\mathrm{m}\) and rotates through an angle of 49.0 \(\mathrm{rad.}\) The baseball has a radius of 3.67 \(\mathrm{cm}\) and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the buseball?

The take-up reel of a cassette tape has an average radius of 1.4 cm. Find the length of tape (in meters) that passes around the reel in 13 s when the reel rotates at an average angular speed of 3.4 rad/s.

A compact disc (CD) contains music on a spiral track. Music is pul onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. Since \(v_{\mathrm{T}}=r \omega, \mathrm{a}\) CD rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music neur the inner part of the disc. For music at the outer edge (r 0.0568 m), the angular speed is 3.50 rev/s. Find (a) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of 0.0249 m from the center of a CD.
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