/*! 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 26 Our balance is maintained, at le... [FREE SOLUTION] | 91Ó°ÊÓ

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Our balance is maintained, at least in part, by the endolymph fluid in the inner ear. Spinning displaces this fluid, causing dizziness. Suppose that a skater is spinning very fast at 3.0 revolutions per second about a vertical axis through the center of his head. Take the inner ear to be approximately \(7.0 \mathrm{~cm}\) from the axis of spin. (The distance varies from person to person.) What is the radial acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) of the endolymph fluid?

Short Answer

Expert verified
The radial acceleration of the endolymph fluid is 414 \(\mathrm{m} / \mathrm{s}^{2}\) or 42.2 g's.

Step by step solution

01

Convert angular velocity to radian per second

The skater is spinning at a rate of 3.0 revolutions per second. We have to convert this to radian per second as the formula for radial acceleration requires this unit. There are \(2\pi\) radians in one revolution. Therefore, the angular velocity in radian per second is \(3.0 \times 2\pi\) rad/s.
02

Substitute the values into the radial acceleration formula

The radial acceleration of the fluid can be calculated using the formula \(\mathrm{a} =\mathrm{r}\omega^{2}\). The radius \(r\) is given as 7.0 cm which needs to be converted to metres. 7.0 cm equals 0.07 m. The angular velocity \(\omega\) is the value we calculated in the first step, which is \(3.0 \times 2\pi\) rad/s. Now, you can substitute these values into the formula and solve for \(a\), the radial acceleration.
03

Convert acceleration from m/s^2 to g's

To convert the radial acceleration from \(\mathrm{m} / \mathrm{s}^{2}\) to \(g\) 's, divide the acceleration by 9.8 \(\mathrm{m} / \mathrm{s}^{2}\), which is the acceleration due to gravity here on Earth.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves around a fixed point or axis. It is essentially the rotational equivalent to linear velocity, which most people are familiar with from their everyday experiences, such as driving a car. The angular velocity tells us the angle through which an object turns in a specific amount of time.

In the context of our skater, we're considering the rate at which the skater spins around the vertical axis through the center of his head. As mentioned in the solution, the skater's angular velocity is initially given in revolutions per second which is a common unit for measuring the speed of turntables, wheels, and, in our case, figure skaters!
Revolutions Per Second
When we talk about revolutions per second, we're referring to the number of complete rotations an object makes in one second. It's an intuitive way to express rotational speeds because it tells us how many times an object spins around in one second. In everyday life, you might not come across objects spinning incredibly fast, but in machinery and certain sports like figure skating or gymnastics, high rotational speeds are common.

Our skater, for example, is rotating at a high speed of 3.0 revolutions per second. This information is essential but needs to be converted to a more standardized unit of radians per second to use in our radial acceleration calculations.
Radian Conversion
Radians are the standard unit of angular measurements used in physics and mathematics. Unlike degrees, which are based on dividing a circle into 360 arbitrary units, radians provide a measure that directly relates the arc length of a circle to its radius. There are exactly \(2\pi\) radians in one revolution, because the circumference of a circle is \(2\pi r\), and the arc length for one full circle is the circumference itself.

To convert from revolutions per second to radians per second, you multiply by \(2\pi\). If a skater spins at 3.0 revolutions per second, using radian conversion, the angular velocity in radians per second would be \(3.0 \times 2\pi\) rad/s, which is essential for calculating radial acceleration.
Acceleration Due to Gravity
Gravity is a force that pulls objects towards each other, and on the surface of the Earth, it gives objects an acceleration of \(9.8 \mathrm{m} / \mathrm{s}^{2}\), downwards towards the center of the planet. This acceleration is often represented by the symbol \(g\) and is a universal value that helps in understanding motion and forces. It's crucial when converting the radial acceleration from the standard meters per second squared into 'g's, a unit which compares an acceleration to the standard acceleration due to gravity on Earth.

After calculating the radial acceleration in \( \mathrm{m} / \mathrm{s}^{2}\), as done in the skater exercise, we divide that value by \(9.8 \mathrm{m} / \mathrm{s}^{2}\) to find how many times greater the radial acceleration is compared to Earth's gravity. This comparison is often used in fields that require an understanding of forces exerted on bodies, such as in designing roller coasters or understanding the physical stresses on astronauts during liftoff.

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Most popular questions from this chapter

Firefighters use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) as it leaves the end of the hose and then exhibits projectile motion. The firefighters adjust the angle of elevation \(\alpha\) of the hose until the water takes \(3.00 \mathrm{~s}\) to reach a building \(45.0 \mathrm{~m}\) away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. \(\mathbf{P 3 . 6 3 )}\). The takeoff ramp was inclined at \(53.0^{\circ},\) the river was \(40.0 \mathrm{~m}\) wide, and the far bank was \(15.0 \mathrm{~m}\) lower than the top of the ramp. The river itself was \(100 \mathrm{~m}\) below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 60 minutes, reaching a point \(2.0 \mathrm{~km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) \(5.0 \mathrm{~km}\) downstream from the turnaround point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

In the middle of the night you are standing a horizontal distance of \(14.0 \mathrm{~m}\) from the high fence that surrounds the estate of your rich uncle. The top of the fence is \(5.00 \mathrm{~m}\) above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of \(1.60 \mathrm{~m}\) above the ground and at an angle of \(56.0^{\circ}\) above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance beyond the fence will the rock land on the ground?

A major leaguer hits a baseball so that it leaves the bat at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(36.9^{8}\) above the horizontal. Ignore air resistance. (a) At what \(t w o\) times is the baseball at a height of \(10.0 \mathrm{~m}\) above the point at which it left the bat? (b) Calculate the horizonta and vertical components of the baseball's velocity at each of the two times calculated in part (a). (c) What are the magnitude and directior of the baseball's velocity when it returns to the level at which it lef the bat?

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