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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 cm from the axis of spin. (The distance varies from person to person.) What is the radial acceleration (in m/s\(^2\) and in \(g\)'s) of the endolymph fluid?

Short Answer

Expert verified
The radial acceleration is approximately 109.82 m/s\(^2\) or 11.20 g's.

Step by step solution

01

Convert Revolutions per Second to Radians per Second

Since 1 revolution is equal to \(2\pi\) radians, we need to convert the skater's spin rate into radians per second. The skater is spinning at 3.0 revolutions per second: \(3.0 \times 2\pi = 6\pi\) radians per second.
02

Determine Radial Acceleration Formula

The radial acceleration \(a_r\) for an object moving in a circle is given by \(a_r = \omega^2 r\), where \(\omega\) is the angular velocity in radians per second and \(r\) is the radius in meters.
03

Substitute Values into Radial Acceleration Formula

First, convert 7.0 cm to meters: \(7.0 \text{ cm} = 0.07 \text{ m}\). Then substitute the known values into the formula: \(a_r = (6\pi)^2 \times 0.07 = 11.09 \pi^2 \) m/s\(^2\).
04

Calculate Radial Acceleration in m/s²

Compute the value of \(11.09 \pi^2\) to get the radial acceleration in meters per second squared: \(a_r \approx 109.82\) m/s\(^2\).
05

Convert Radial Acceleration to g's

To express the radial acceleration in terms of g's (where \(1 g = 9.81\) m/s\(^2\)), divide the radial acceleration by 9.81: \(\frac{109.82}{9.81} \approx 11.20 \text{ g's}\).

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Key Concepts

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

Radial Acceleration
Radial acceleration is a key concept in circular motion, particularly when understanding the forces acting on a rotating body. Radial acceleration, also known as centripetal acceleration, points towards the center of the circle along which an object is moving. This means it is always directed inward, regardless of the direction of the object’s actual travel. It was calculated as 109.82 m/s² in the scenario with the skater.

The formula for radial acceleration is:
  • \(a_r = \omega^2 r\)
Where:
  • \(\omega\) is the angular velocity in radians per second.
  • \(r\) is the radius (distance from the center of the path) in meters.
Computing it gives insights into how fast the object changes direction while moving along the curved path.
Conversion of Units
Converting units is essential in physics because it ensures all measurements align with each other, making calculations accurate. In this problem, the conversion is from revolutions per second to radians per second. Since one revolution is equivalent to \(2\pi\) radians, the skater's spin is converted by multiplying by \(2\pi\) to get \(6\pi\) radians per second. This step is crucial to use the radial acceleration formula effectively.

Moreover, the unit conversion from centimeters to meters (7.0 cm to 0.07 m) ensures our calculations are consistent with the standard units used in physics equations. Always remember to:
  • Multiply revolutions by \(2\pi\) to convert to radians.
  • Convert centimeters to meters by dividing by 100.
These equivalent conversions enable better communication and understanding of scientific data.
Circular Motion
Circular motion describes the movement of an object along the circumference of a circle or rotation along a circular path. This concept is visible in a spinning skater's situation. The endolymph fluid in the inner ear moves in a circle around the skater’s head, causing him to experience dizziness while spinning.

Important Characteristics of Circular Motion:
  • Constant speed but changing velocity due to direction changes.
  • Presence of centripetal force which maintains the circular path.
  • Radial acceleration directed towards the center of the motion path.
Understanding these principles helps explicate the nature of motion not just in sports but in many real-world phenomena like the operation of engines and celestial orbits.
Angular Velocity
Angular velocity is a measure of the rate of rotation, showing how fast an object spins around a specific axis. It's a vector quantity, which includes both the magnitude and the direction (in this case, around the vertical axis). For our spinning skater, the angular velocity was calculated as \(6\pi\) radians per second.

Formulaically expressed as:
  • \(\omega = \frac{\Delta \theta}{\Delta t}\)
Where:
  • \(\Delta \theta\) is the angle of rotation in radians.
  • \(\Delta t\) is the time over which the rotation occurs.
In practice, this helps determine the speed of rotating parts of machines or the Earth’s rotation. It’s crucial in linking circular motion to linear motion for complete physical analysis.

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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 m/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.

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 m/s to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 m/s relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

A rocket is fired at an angle from the top of a tower of height \(h_0\) = 50.0 m. Because of the design of the engines, its position coordinates are of the form \(x(t) = A + Bt^2 \)and \(y(t) = C + Dt^3\), where \(A, B, C,\) and \(D\) are constants. The acceleration of the rocket 1.00 s after firing is \(\vec{a} = (4.00 \hat{i}+ 3.00\hat{j}) m/s^2\). Take the origin of coordinates to be at the base of the tower. (a) Find the constants \(A, B, C,\) and \(D\), including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the \(x\)- and \(y\)-components of the rocket's velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

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

If \(\vec{r} = bt^2\hat{\imath} + ct^3\hat{\jmath}\), where \(b\) and \(c\) are positive constants, when does the velocity vector make an angle of 45.0\(^\circ\) with the \(x\)- and \(y\)-axes?

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