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A certain gyroscope precesses at a rate of 0.50 rad/s when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is 0.165g, what would be its precession rate?

Short Answer

Expert verified
0.0825 rad/s

Step by step solution

01

Understand the Formula for Precession Rate

The precession rate \( \Omega \) of a gyroscope is given by \( \Omega = \frac{\tau}{L} \) where \( \tau \) is the torque and \( L \) is the angular momentum. Torque due to weight on Earth is \( \tau = rmg \). Since torque is also related to gravity as \( \tau = rmg_{moon} \) on the Moon, rewrite the precession rate formula for the moon.
02

Relate Torques on Earth and Moon

On Earth, the precession rate is \( \Omega_{earth} = \frac{rmg_{earth}}{L} \). On the Moon, the precession will be \( \Omega_{moon} = \frac{rmg_{moon}}{L} \). Thus, \( \Omega_{moon} = \Omega_{earth} \times \frac{g_{moon}}{g_{earth}} \).
03

Substitute Known Values

Given \( \Omega_{earth} = 0.50 \text{ rad/s} \), \( g_{moon} = 0.165g_{earth} \), substitute these into the equation: \( \Omega_{moon} = 0.50 \times 0.165 \).
04

Calculate the Precession Rate on the Moon

Perform the calculation: \( \Omega_{moon} = 0.50 \times 0.165 = 0.0825 \text{ rad/s} \). This is the precession rate of the gyroscope on the Moon.

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Key Concepts

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

Torque in Physics
Torque is a measure of the rotational force acting on an object. It is essential to understand this concept when discussing how gyroscopes work. Torque is mathematically defined by the equation \( \tau = rF \), where \( \tau \) is the torque, \( r \) is the distance from the axis of rotation to the point of force application, and \( F \) is the force applied.
For a gyroscope, the force in question is often the gravitational force pulling on it, and this is why torque is related to gravity as \( \tau = rmg \) on Earth and \( \tau = rmg_{moon} \) on the Moon. Here, \( m \) represents mass and \( g \) is the acceleration due to gravity.
  • On Earth, gravity is stronger, so the torque will be higher with the same conditions.
  • On the Moon, \( g_{moon} \approx 0.165g_{earth} \), meaning gravity is weaker, and thus, the torque generated is reduced.
The understanding of torque helps explain how the forces exerted on the gyroscope differ between Earth and the Moon, which directly influences the precession rate.
Angular Momentum
Angular momentum is a key concept when discussing rotational movement. It characterizes the motion of a gyroscope, or any rotating object, by reflecting its rotational inertia and the velocity at which it spins. The formula for angular momentum \( L \) is given by \( L = I\omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
A gyroscope's angular momentum is conserved in the absence of an external torque. This implies that the angular speed and the inertial properties dictate how resistant the gyroscope is to changes in its rotation.
  • A high angular momentum means the gyroscope will maintain its orientation and oppose any changes from external forces.
  • This is crucial for gyroscopes used in navigation and stabilization systems, as it allows them to keep their position or to react predictably to applied forces.
In exercises like this, where torque and precession rate are involved, it's fundamental to grasp that angular momentum provides the gyroscope with stability against perturbations.
Gravity on the Moon
Gravity plays a vital role in determining the forces that act on objects, and consequently, their behavior in different environments. On the Moon, gravity is about 16.5% of Earth's gravity, denoted as \( g_{moon} = 0.165g_{earth} \). This significant difference leads to profound changes in the motion and mechanics of objects like gyroscopes.
In the context of gyroscopes and precession, the reduced gravitational pull on the Moon means:
  • The weight of the gyroscope is less, influencing the torque exerted on it.
  • Gyroscopes will precess more slowly compared to Earth, as seen from the formula relating precession rate \( \Omega \) to the ratio of lunar and Earth gravity \( \frac{g_{moon}}{g_{earth}} \).
  • This change is crucial for systems that rely on gyroscopic devices for precision and stability in lunar environments.
Understanding gravity's influence on the Moon helps explain why the same device behaves differently under distinct gravitational conditions.

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Most popular questions from this chapter

The Hubble Space Telescope is stabilized to within an angle of about 2-millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass 2.0 kg and diameter 5.0 cm, spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of 1.0 \(\times 10^{-6}\) degree during a 5.0-hour exposure of a galaxy?

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \mathrm{~m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s}\). What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m} .\) (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance \(h\) down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0\(^\circ\), and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by \(I = \frac{1}{3} ML^2\), where \(L =\) 6.00 m is the length of the pole and \(M =\) 24.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0\(^\circ\) after the collision?

A block with mass m is revolving with linear speed \(v_1\) in a circle of radius \(r_1\) on a frictionless horizontal surface (see Fig. E10.40). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_2\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_1\) and the radius \(r_1\). (b) Use \(W = \int_{r1}^{r2} \overrightarrow{T} (r) \cdot d\overrightarrow{r}\) to calculate the work done by \(\overrightarrow{T}\) when \(r\) changes from \(r_1\) to \(r_2\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

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