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Chapter 3: Problem 91

For no apparent reason, a poodle is running at a constant speed of \(v=5.00 \mathrm{m} / \mathrm{s}\) in a circle with radius \(R=2.50 \mathrm{m} .\) Let \(\vec{v}_{1}\) be the velocity vector at time \(t_{1},\) and let \(\vec{v}_{2}\) be the velocity vector at time \(t_{2}\) . Consider \(\Delta \vec{v}=\vec{v}_{2}-\vec{v}_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Recall that \(\vec{a}_{\mathrm{av}}=\Delta \vec{v} / \Delta t .\) For \(\Delta t=0.5 \mathrm{s}, 0.1 \mathrm{s},\) and 0.05 \(\mathrm{s}\) , calculate the magnitude (to four significant figures) and direction (relative to \(\vec{v}_{1} )\) of the average acceleration \(\vec{d}_{\mathrm{ev}}\) . Compare your results to the general expression for the instantaneous acceleration \(\vec{a}\) for uniform circular motion that is derived in the text.

Short Answer

Expert verified
As \( \Delta t \to 0 \), the average acceleration \( \vec{a}_{\text{av}} \) approaches the centripetal acceleration of \( 10 \text{ m/s}^2 \), perpendicular to \( \vec{v}_1 \).

Step by step solution

01

Understand the Problem

We need to determine the average acceleration for a poodle running in a circle at a constant speed, for given time intervals. Additionally, we will compare this result with instantaneous acceleration in circular motion.
02

Recall Circular Motion Concepts

In uniform circular motion, the speed is constant and the velocity vector changes direction. The acceleration is centripetal, pointing towards the center of the circle, with magnitude given by the formula \( a = \frac{v^2}{R} \).
03

Calculate Angular Displacement

The angular displacement \( \Delta \theta \) is found using \( \Delta \theta = \omega \Delta t \), where \( \omega = \frac{v}{R} \) is the angular velocity. We can use \( \Delta \theta \) to determine the change in direction of velocity.
04

Find Change in Velocity Vectors

The change in velocity \( \Delta \vec{v} = \vec{v}_2 - \vec{v}_1 \) is perpendicular to \( \vec{v}_1 \) and \( \vec{v}_2 \) and its magnitude can be calculated using trigonometry: \( \Delta v = 2v \sin(\frac{\Delta \theta}{2}) \).
05

Compute Average Acceleration

The average acceleration \( \vec{a}_{\text{av}} = \frac{\Delta \vec{v}}{\Delta t} \). Substitute \( \Delta v \) obtained from the previous step and calculate it for \( \Delta t = 0.5 \text{s}, 0.1 \text{s}, 0.05 \text{s} \).
06

Compare with Instantaneous Acceleration

The instantaneous centripetal acceleration has a constant magnitude \( a = \frac{v^2}{R} = \frac{5^2}{2.5} = 10 \text{ m/s}^2 \). Comparing with the average acceleration values calculated, observe how \( \vec{a}_{\text{av}} \) approaches \( 10 \text{ m/s}^2 \) as \( \Delta t \) decreases.

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Key Concepts

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

Uniform circular motion
Uniform circular motion refers to the movement of an object at a constant speed along a circular path. In this type of motion, while the speed remains unchanged, the direction of the velocity vector continuously changes. This constant change in direction is what distinguishes uniform circular motion from linear motion.
- The velocity vector in circular motion is tangent to the circle at any point. - Since the speed is constant, any change in velocity is only due to the change in direction. - This change in direction necessitates a force, known as the centripetal force, to keep the object on its circular path.
The presence of centripetal force results in centripetal acceleration, pointing towards the center of the circle. This is a central concept in understanding circular motion, especially when discussing the average and instantaneous acceleration of objects, such as a poodle, moving along a circular track.
Centripetal acceleration
Centripetal acceleration is a fundamental concept in circular motion. It is the acceleration experienced by an object moving in a circle at a constant speed. Despite the speed remaining constant, the continuous change in direction of the velocity vector causes acceleration.
- The formula for centripetal acceleration is given by \[ a = \frac{v^2}{R} \]where \( v \) is the speed of the object and \( R \) is the radius of the circle.- Centripetal acceleration always points towards the center of the circular path.
This inward acceleration is necessary for keeping the object in circular motion. Without it, the object would move in a straight line, as per Newton's first law of motion. In this context, the problem asked to calculate average acceleration over different time intervals, which helps in understanding how closely the average acceleration approaches the instantaneous centripetal acceleration as the time interval shrinks.
Instantaneous acceleration
Instantaneous acceleration in circular motion is the acceleration an object experiences at any given point along its path. This is different from average acceleration, which is determined over a time interval.
- As the time interval \( \Delta t \) becomes very small, the average acceleration approaches the instantaneous acceleration value.- In uniform circular motion, instantaneous acceleration is equivalent to the centripetal acceleration.- For our example, the instantaneous centripetal acceleration is calculated using the formula \[ a = \frac{v^2}{R} = \frac{5^2}{2.5} = 10 \text{ m/s}^2 \]
This value remains constant for any point along the circle, provided that the object maintains the same speed and does not depart from the circular path. Understanding instantaneous acceleration is key to analyzing motion at a precise moment, as demonstrated by your calculations comparing average with instantaneous acceleration as \( \Delta t \) varies.

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Most popular questions from this chapter

A railroad flatear is traveling to the right at a speed of 13.0 \(\mathrm{m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. 3.43\()\) . What is the velocity (magnitude and direction) of the motor scooter relative to the flatcar if its velocity relative to the observer on the ground is (a) 18.0 \(\mathrm{m} / \mathrm{s}\) to the right? (b) 3.0 \(\mathrm{m} / \mathrm{s}\) to the left? (c) zero?

Cycloid. A particle moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R(\omega t-\sin \omega t) \quad y(t)=R(1-\cos \omega t) $$ where \(R\) and \(\omega\) are constants. (a) Sketch the trajectory of the particle. (This is the trajectory of a point on the rim of a wheel that is rolling at a constant speed on a horizontal surface. The curve traced out by such a point as it moves through space is called a cycloid.) (b) Determine the velocity components and the acceleration components of the particle at any time \(t\) . (c) At which times is the particle momentarily at rest? What are the coordinates of the particle at these times? What are the magnitude and direction of the acceleration at these times? (d) Does the magnitude of the acceleration depend on time? Compare to uniform circular motion.

An elevator is moving upward at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . A bolt in the elevator ceiling 3.00 \(\mathrm{m}\) above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings ofm the building? (d) According to the observer in part(c), what distance

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Kicking a Field Goal. In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 \(\mathrm{ft}\) above the ground, and the ball is kicked from ground level, 36.0 \(\mathrm{ft}\) horzontally from the bar (Fig. 3.48\() .\) Football regulations are stated in English units, but convert to SI units for this problem. (a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\) .
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