Planet and Car rotational movement

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SUMMARY

The discussion focuses on calculating the period of rotation for a newly discovered planet with a mean radius of 1230 km. A vehicle on the planet's surface, moving in the same direction as the planet's rotation at a speed of 130 km/h, has an angular velocity 6.18 times that of the planet. The calculated period of the planet's rotation is approximately 307.944 hours. Additionally, when the vehicle reverses direction, it must achieve an angular velocity equal and opposite to that of the planet, which requires further clarification on the speedometer reading.

PREREQUISITES
  • Understanding of angular velocity and its relationship to linear velocity
  • Familiarity with the equations v = rω and T = 2π/ω
  • Basic knowledge of rotational dynamics
  • Concept of relative motion in physics
NEXT STEPS
  • Explore the implications of angular velocity in different reference frames
  • Study the effects of reversing direction on angular and linear velocities
  • Investigate the concept of relative velocity in rotating systems
  • Learn more about planetary rotation and its effects on surface vehicles
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Students studying physics, particularly those focusing on rotational dynamics, as well as educators seeking to clarify concepts related to angular velocity and relative motion.

qqchan
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Homework Statement



A newly discovered planet has a mean radius of 1230 km. A vehicle on the planet's surface is moving in the same direction as the planet's rotation, and its speedometer reads 130 km/h. If the angular velocity of the vehicle about the planet's center is 6.18 times as large as the angular velocity of the planet, what is the period of the planet's rotation?

If the vehicle reverses direction, how fast must it travel (as measured by the speedometer) to have an angular velocity that is equal and opposite to the planet's?



Homework Equations



v=rω
T = 2∏/ω


The Attempt at a Solution



Since the value from the speedometer is relative to the ground, the following equation is developed:
v = 6.18rωplanet
v = 130 + rωplanet

v - rωplanet = 130
6.18rωplanet - rωplanet = 130
5.18rωplanet= 130

ωplanet = 130/(5.18 *1230)

T = 2∏/ωplanet
≈307.944

However when the vehicle reverses its direction to have an angular velocity similar to that of the planet, wouldn't the result of the speedometer be zero? However, the teacher said it isn't zero.
planet - rωplanet = Vspeedometer

I would like to know what is wrong in my thought process when attempting to solve this equation
 
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qqchan said:

Homework Statement



A newly discovered planet has a mean radius of 1230 km. A vehicle on the planet's surface is moving in the same direction as the planet's rotation, and its speedometer reads 130 km/h. If the angular velocity of the vehicle about the planet's center is 6.18 times as large as the angular velocity of the planet, what is the period of the planet's rotation?

If the vehicle reverses direction, how fast must it travel (as measured by the speedometer) to have an angular velocity that is equal and opposite to the planet's?



Homework Equations



v=rω
T = 2∏/ω


The Attempt at a Solution



Since the value from the speedometer is relative to the ground, the following equation is developed:
v = 6.18rωplanet
v = 130 + rωplanet

v - rωplanet = 130
6.18rωplanet - rωplanet = 130
5.18rωplanet= 130

ωplanet = 130/(5.18 *1230)

T = 2∏/ωplanet
≈307.944

However when the vehicle reverses its direction to have an angular velocity similar to that of the planet, wouldn't the result of the speedometer be zero? However, the teacher said it isn't zero.
planet - rωplanet = Vspeedometer

I would like to know what is wrong in my thought process when attempting to solve this equation

Well, yes. If you want the car to have the same angular velocity as the planet, ##r \omega_v=r \omega_{planet}##, then the velocity is 0. But you don't want that, you want ##r \omega_v= -r \omega_{planet}##.
 

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