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Chapter 7: Problem 17

(a) What is the tangential acceleration of a bug on the rim of a 10 -in.-diameter disk if the disk moves from rest to an angular speed of 78 rev/min in \(3.0\) s? (b) When the disk is at its final speed, what is the tangential velocity of the bug? (c) One second after the bug starts from rest, what are its tangential acceleration, centripetal acceleration, and total acceleration?

Short Answer

Expert verified
The tangential acceleration of the bug on the disk is \(6.52 \, \text{m/s}^2\). When the disk is at its final speed, the tangential velocity of the bug is \(0.68 \, \text{m/s}\). After one second from rest, the bug's tangential acceleration is \(20.4 \, \text{m/s}^2\), the centripetal acceleration is \(0.021 \, \text{m/s}^2\), and the total acceleration is \(20.4 \, \text{m/s}^2\).

Step by step solution

01

Conversion of Units for Angular Speed

The given angular speed is 78 revolutions per minute, we need to convert this unit to radian per second, which is the SI unit for angular speed. Since 1 revolution is equal to \(2\pi\) radian and 1 minute is equal to 60 seconds, the angular speed in radian per second is \(78 \times \frac{2\pi}{60}\).
02

Calculate the Tangential Accelerations

Tangential acceleration can be calculated using the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\). Substituting the values we get \(\alpha = \frac{78 \times \frac{2\pi}{60}}{3}\). For part c, since it only took 1 second to start from rest, the tangential acceleration is \(\alpha = \frac{78 \times \frac{2\pi}{60}}{1}\).
03

Calculate Tangential Velocity

Tangential velocity can be calculated using \(v_t = r\omega\). Substituting the given values (Remember the diameter is 10 inches, so the radius is 5 inches), we get \(v_t = \frac{5 \times 78 \times 2\pi}{60}\).
04

Calculate the Centripetal Accelerations

Centripetal acceleration is calculated by \(\frac{v_t^2}{r}\). Substituting the obtained tangential velocity from Step 3 and the radius, we can find the centripetal acceleration.
05

Calculate the Total Accelerations

Since the total acceleration is the vector sum of the tangential and centripetal accelerations, we use the formula \(a_t = \sqrt{a^2 + a_c^2}\) where \(a\) is the tangential acceleration and \(a_c\) is the centripetal acceleration. Calculating the total acceleration according to the formula will give the required answer for part c.

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Key Concepts

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

Angular Speed Conversion
When solving problems involving objects in rotational motion, it's often necessary to convert angular speed from revolutions per minute (rev/min) to radians per second (rad/s). This conversion is crucial because radians per second is the standard unit used in physics for angular speed. Here's a simple conversion process:
  • A full revolution is equivalent to an angle of \(2\pi\) radians.
  • There are 60 seconds in a minute.
Thus, to convert a speed of 78 rev/min to rad/s, we multiply: \[ 78 \times \frac{2\pi}{60} \approx 8.168 \text{ rad/s} \] This value is essential for computing further characteristics of rotational motion like tangential velocity and acceleration.
Tangential Velocity
Tangential velocity, denoted as \(v_t\), is the linear speed of an object moving along a circular path. It is calculated using the formula \(v_t = r\omega\), where \(r\) is the radius of the circular path and \(\omega\) is the angular speed in rad/s.
In our example, the radius of the disk is half its diameter, so \(r = 5\) inches. Using the previously converted angular speed, we calculate the tangential velocity:
  • Radius \(r = 5 \text{ in}\)
  • Angular speed \(\omega = 8.168 \text{ rad/s}\)
Plug these into the formula: \[ v_t = 5 \times 8.168 \approx 40.84 \text{ in/s} \] This shows how fast the bug travels along the edge of the disk.
Centripetal Acceleration
Centripetal acceleration is the acceleration that keeps an object moving along a circular path. It is directed towards the center of the circle and is given by the formula \(a_c = \frac{v_t^2}{r}\), where \(v_t\) is the tangential velocity and \(r\) is the radius of the circle.
In our problem, after computing the tangential velocity as \(40.84\) in/s, the centripetal acceleration is calculated as follows:
  • \(v_t = 40.84 \text{ in/s}\)
  • \(r = 5 \text{ in}\)
Thus: \[ a_c = \frac{40.84^2}{5} \approx 333.43 \text{ in/s}^2 \] This acceleration acts to change the direction of the bug's velocity, keeping it moving in a circle.
Total Acceleration
Total acceleration in rotational dynamics combines both tangential acceleration \(a\) and centripetal acceleration \(a_c\). It represents the net effect of these acceleration components on the object.
The total acceleration can be found using the formula: \[ a_t = \sqrt{a^2 + a_c^2} \] Here, \(a\) is the tangential acceleration, calculated previously when the tangential velocity was increasing at a steady rate.
  • Calculated tangential acceleration \(a\) and centripetal acceleration \(a_c\) combine to provide a comprehensive measure of the total effect on motion.
For our bug, using values for both \(a\) and \(a_c\), we find: \[ a_t = \sqrt{a^2 + 333.43^2} \] By plugging in the values for \(a\), you can solve for \(a_t\), providing insight into the overall acceleration experienced by the bug moving along the disk.

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Most popular questions from this chapter

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?

A \(40.0\) -kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of \(18.0 \mathrm{~m}\). (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?

In a popular amusement park ride, a rotating cylinder of radius \(3.00 \mathrm{~m}\) is set in rotation at an angular speed of \(5.00 \mathrm{rad} / \mathrm{s}\), as in Figure \(\mathrm{P} 7.75 .\) The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? (Hint: Recall that the magnitude of the maximum force of static friction is equal to \(\mu n\), where \(n\) is the normal force-in this case, the force causing the centripetal acceleration.)

It has been suggested that rotating cylinders about \(10 \mathrm{mi}\) long and \(5.0 \mathrm{mi}\) in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?

Part of a roller-coaster ride involves coasting down an incline and entering a loop \(8.00 \mathrm{~m}\) in diameter. For safety considerations, the roller coaster's speed at the top of the loop must be such that the force of the seat on a rider is equal in magnitude to the rider's weight. From what height above the bottom of the loop must the roller coaster descend to satisfy this requirement?
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