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An object moves in a horizontal circle at constant speed \(v\) (in units of \(\mathrm{m} / \mathrm{s}) .\) It takes the object \(T\) seconds to complete one revolution. Derive an expression that gives the radial acceleration of the ball in terms of \(v\) and \(T,\) but not \(r .\) (a) If the speed doubles, by what factor must the period \(T\) change if \(a_{\text {rad }}\) is to remain unchanged? (b) If the radius doubles, by what factor must the period change to keep \(a_{\text {rad }}\) the same?

Short Answer

Expert verified
To keep radial acceleration constant when speed doubles, the period \(T\) must be doubled. If the radius doubles, the period must also be doubled to maintain the same radial acceleration.

Step by step solution

01

Derive Equation for Radial Acceleration

The radial acceleration in circular motion is given by \(a_{\text {rad }} = \frac{{v^{2}}}{{r}}\). But we want the equation in terms of \(T\) and \(v\). The distance around the circle is the circumference, given by \(C = 2\pi r\). By definition, speed is distance divided by time, so we can write \(v = \frac{{2\pi r}}{T}\). Substituting \(r = \frac{{vT}}{{2\pi}}\) from this equation into the equation for radial acceleration gives \(a_{\text {rad }} = \frac{{v^{2}}}{{\frac{{vT}}{{2\pi}}}} = \frac{{4\pi^{2}}}{{T^{2}}}\cdot v^{2}\).
02

Analyze Change in Speed

We now have the radial acceleration in terms of \(T\) and \(v\), given by \(a_{\text {rad }} = \frac{{4\pi^{2}}}{{T^{2}}}\cdot v^{2}\). If the speed doubles, we substitute \(2v\) into the equation, which gives \(a_{\text {new}} = \frac{{4\pi^{2}}}{{T^{2}}}\cdot (2v)^{2} = 4 \cdot \frac{{4\pi^{2}}}{{T^{2}}}\cdot v^{2}\). For \(a_{\text {new}}\) to remain the same as \(a_{\text {rad }}\), it's necessary to increase \(T\) by a factor of 2.
03

Analyze Change in Radius

If the radius doubles but the radial acceleration \(a_{\text {rad }}\) is to remain the same, it means that the period \(T\) must also be doubled.

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Key Concepts

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

Radial Acceleration
In circular motion, radial acceleration is also known as centripetal acceleration. It acts towards the center of the circle, keeping the object moving along its circular path. The formula to calculate radial acceleration is \( a_{\text{rad}} = \frac{v^2}{r} \), where \( v \) is the speed of the object and \( r \) is the radius of the circle.

However, in specific situations, such as when the question requires radial acceleration in terms of speed \( v \) and period \( T \), we use an alternate approach. We know that an object completing a full circle travels a distance equal to the circumference \( 2\pi r \). Thus, speed \( v \) can be expressed as \( v = \frac{2\pi r}{T} \). Substituting this into the radial acceleration formula, and expressing \( r \) in terms of \( v \) and \( T \), gives \( a_{\text{rad}} = \frac{4\pi^2}{T^2} \cdot v^2 \).

This expression shows how radial acceleration depends on the square of the speed and inversely on the square of the period.
Period of Motion
The period of motion, denoted as \( T \), is the time it takes for an object to complete one full revolution around a circular path. It is an essential aspect of understanding circular motion and directly influences the radial acceleration of an object.

In the equation \( a_{\text{rad}} = \frac{4\pi^2}{T^2} \cdot v^2 \), we see that the radial acceleration varies inversely with the square of the period. This means if the period \( T \) increases, the radial acceleration decreases, and vice versa. This relation is critical to solving problems where you want to keep the radial acceleration constant but change other parameters.

For instance, if the speed is doubled, to keep \( a_{\text{rad}} \) unchanged, \( T \) must also be doubled to counteract the effect of the increased speed on the formula.
Constant Speed
When an object moves in a circle at constant speed, it travels equal distances in equal intervals of time around the circular path. This concept is essential because it simplifies the understanding and calculation of quantities like radial acceleration and period of motion.

The term "constant speed" means that while the direction of the object may continuously change as it moves along the circular path, the magnitude of its velocity remains unchanged. This characteristic makes it feasible to use the formula \( v = \frac{2\pi r}{T} \), expressing the speed directly in terms of the radius \( r \) and period \( T \).

Maintaining constant speed in circular motion ensures a consistent relationship between the speed, period, and radius, integral in deriving expressions for radial acceleration.
Circular Path
A circular path refers to the trajectory an object follows when it moves in a circle. This trajectory is defined by various characteristics, such as having a constant radius and being equidistant from a central point.

Understanding the properties of a circular path is vital in applying the concepts of circular motion. The path's radius influences both the speed and radial acceleration of an object moving along it. The circular nature also ensures that, despite the constant speed, the object's velocity changes because its direction constantly changes.

Consequently, when you change the radius of a circular path—say by doubling it—you affect the period needed to maintain the same radial acceleration. As derived, to keep the radial acceleration constant when the radius is doubled, the period \( T \) must also be doubled, ensuring that the speed remains consistent with the radial acceleration formula.

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Most popular questions from this chapter

A man stands on the roof of a 15.0 -m-tall building and throws a rock with a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

On level ground a shell is fired with an initial velocity of \(40.0 \mathrm{~m} / \mathrm{s}\) at \(60.0^{\circ}\) above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell's initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.

The earth has a radius of \(6380 \mathrm{~km}\) and turns around once on its axis in \(24 \mathrm{~h}\). (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\mathrm{rad}}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We'll see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

At \(t=0\) a rock is projected from ground level with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(53.0^{\circ}\) above the horizontal. Neglect air resistance. At what two times \(t\) is the rock \(5.00 \mathrm{~m}\) above the ground? At each of these two times, what are the horizontal and vertical components of the velocity of the rock? Let \(v_{0 x}\) and \(v_{0 y}\) be in the positive \(x\) - and \(y\) -directions, respectively.

An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at \(150 \mathrm{~m}\) above the level ground when the plane is flying at \(75 \mathrm{~m} / \mathrm{s}\) in a direction \(55^{\circ}\) above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?

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