/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 (III) A particle revolves in a h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 27

(III) A particle revolves in a horizontal circle of radius 1.95 m. At a particular instant, its acceleration is 1.05 m/s\(^2\), in a direction that makes an angle of 25.0\(^\circ\) to its direction of motion. Determine its speed (\(a\)) at this moment, and (\(b\)) 2.00 s later, assuming constant tangential acceleration.

Short Answer

Expert verified
The speed at the moment is approximately 2.05 m/s. Two seconds later, the speed is approximately 2.76 m/s.

Step by step solution

01

Understand Components of Acceleration

The total acceleration of the particle has two components: the radial (or centripetal) acceleration and the tangential acceleration. Given that the total acceleration is 1.05 m/s虏 at a 25-degree angle to the direction of motion, the radial acceleration can be found as follows: \( a_r = 1.05 \cos(25^\circ) \). The tangential acceleration is given by \( a_t = 1.05 \sin(25^\circ) \).
02

Calculate Radial Component (Centripetal Acceleration)

The radial acceleration (centripetal acceleration) is responsible for keeping the particle in its circular path. Using the formula \( a_r = v^2 / r \), where \( v \) is the speed and \( r \) is the radius of the circle (1.95 m), we set the calculated radial acceleration from Step 1 equal to this expression to solve for the speed \( v \).
03

Solve for Initial Speed

Substitute the expression for radial acceleration from Step 1 into the equation from Step 2 to solve for the speed \( v \). This gives us \( v = \sqrt{1.95 \times 1.05 \cos(25^\circ)} \). Calculate \( v \) to find the initial speed at the given instant.
04

Solve for Speed 2 Seconds Later

Assuming constant tangential acceleration, we use the kinematic equation \( v_f = v_i + a_t t \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity found in the previous step, \( a_t \) is the tangential acceleration calculated, and \( t = 2.00 \) seconds. Substitute these values to find the speed 2 seconds later.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is the type of acceleration that acts on an object moving along a circular path, directing it towards the center of the circle. It is crucial for maintaining the circular motion of an object, preventing it from flying off in a straight line due to inertia.
When a particle moves in a circular path, as our exercise suggests, the direction of centripetal acceleration is always perpendicular to the direction of motion. This makes it distinct yet equally important compared to other components of acceleration. To compute this acceleration, we use the formula:
  • \( a_r = \frac{v^2}{r} \)
Where \( v \) is the speed of the particle and \( r \) is the radius of the circle. In practical terms, this means that an increase in speed or a smaller radius results in greater centripetal acceleration. By understanding these relationships, you can discern how different elements of circular motion interact with each other.
Tangential Acceleration
Tangential acceleration is the acceleration that occurs when there is a change in the speed of an object moving along a circular path. This is different from centripetal acceleration, which only changes the direction and not the speed of the object.
In our problem, tangential acceleration can be calculated using the component of total acceleration aligned with the direction of motion. Cosine and sine functions help us decompose the total acceleration into radial and tangential components, respectively:
  • \( a_t = 1.05 \sin(25^\circ) \)
This formula implies that the tangential acceleration is proportional to the sine of the angle between the acceleration vector and the direction of motion. This is key to understanding how an object can speed up or slow down while still following a circular path.
Kinematic Equations
Kinematic equations are the core tools used to describe and predict the motion of objects. In the context of our exercise, these equations help us evaluate changes in velocity over time, especially when acceleration is constant.
The kinematic equation we use for circular motion with tangential acceleration is:
  • \( v_f = v_i + a_t t \)
Where:
  • \( v_f \) is the final velocity.
  • \( v_i \) is the initial velocity.
  • \( a_t \) is the tangential acceleration.
  • \( t \) is the time elapsed.
This equation essentially bridges the initial conditions of motion with its subsequent changes, based on the tangential acceleration experienced. By employing this, you can seamlessly calculate how an object's speed evolves over time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a train that rounds a curve with a radius of 570 m at a speed of 160 km/h (approximately 100 mi/h ). (\(a\)) Calculate the friction force needed on a train passenger of mass 55 kg if the track is not banked and the train does not tilt. (\(b\)) Calculate the friction force on the passenger if the train tilts at an angle of 8.0\(^\circ\) toward the center of the curve.

(II) A coin is placed 13.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 38.0 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

Two equal-mass stars maintain a constant distance apart of 8.0 \(\times\) 10\(^{11}\) m and revolve about a point midway between them at a rate of one revolution every 12.6 yr. (\(a\)) Why don't the two stars crash into one another due to the gravitational force between them? (\(b\)) What must be the mass of each star?

(II) What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves (\(a\)) upward with constant speed 5.0 m/s, (\(b\)) downward with constant speed 5.0 m/s, (\(c\)) with an upward acceleration 0.23 \(g\), (\(d\)) with a downward acceleration 0.23 \(g\), and (\(e\)) in free fall?

(II) Determine the time it takes for a satellite to orbit the Earth in a circular \(\textbf{near-Earth orbit}\). A "near-Earth" orbit is at a height above the surface of the Earth that is very small compared to the radius of the Earth. [\(Hint\): You may take the acceleration due to gravity as essentially the same as that on the surface.] Does your result depend on the mass of the satellite?
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Translational Dynamics

Read Explanation

Solid State Physics

Read Explanation

Mechanics and Materials

Read Explanation

Nuclear Physics

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.