- #1
DaalChawal
- 85
- 0
Calculating rates of two points moving along a circle
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SUMMARY
The discussion focuses on calculating the speeds of two particles moving along a circular path, with speeds denoted as $v_1$ and $v_2$. When moving in opposite directions, their rate of closure is $(v_1 + v_2)$, while moving in the same direction results in a rate of opening of $(v_1 - v_2)$. The solution yields $v_1 = 18 \, ft/sec$ and $v_2 = 12 \, ft/sec$, emphasizing that the problem pertains to speeds rather than velocities due to the circular motion involved.
PREREQUISITES- Understanding of circular motion dynamics
- Basic algebra for solving equations
- Knowledge of speed versus velocity concepts
- Familiarity with the concept of circumference in relation to distance
- Study the principles of circular motion in physics
- Learn how to set up and solve systems of equations
- Explore the differences between speed and velocity in various contexts
- Investigate real-world applications of relative motion in circular paths
Students studying physics, educators teaching circular motion concepts, and anyone interested in understanding relative motion in circular paths.
- #2
skeeter
- 1,103
- 1
Let $v_1 > v_2$ be the respective speeds (rates) in feet per second of the two particles.
Moving in opposite directions, their rate of closure is $(v_1+v_2)$
Moving in the same direction, their rate of opening is $(v_1-v_2)$
Set up a system of two equations and solve for both speeds.
Moving in opposite directions, their rate of closure is $(v_1+v_2)$
Moving in the same direction, their rate of opening is $(v_1-v_2)$
Set up a system of two equations and solve for both speeds.
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- #3
DaalChawal
- 85
- 0
You have added them like they are moving on straight line...In circle velocity changes as direction is changing.
- #4
skeeter
- 1,103
- 1
DaalChawal said:
If the two particles move in opposite directions, the sum of their respective distances traveled in 5 seconds will be one full circumference length when they meet again.
If the two particles move in the same direction, the faster particle will move ahead of the slower particle, hence the difference between their respective distances traveled in 25 seconds will be one full circumference length when they meet again.
Using this method, I get $v_1 = 18 \, ft/sec$ and $v_2 = 12 \, ft/sec$. You can check the results yourself.
- #5
HOI
- 921
- 2
This problem has nothing to do "velocity". The problem asks for their speeds, not their velocities.
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