Solve Centripetal Motion: Find Fnet | Physics Calculations

Click For Summary
SUMMARY

The discussion focuses on calculating the net force (Fnet) acting on a cyclist completing a circular race. The cyclist accelerates uniformly from rest, covering a circular course of radius r in time Δt, with mass m. The initial approach incorrectly used average speed to calculate velocity, leading to an incorrect expression for Fnet. The correct method involves recognizing the need to calculate both radial and tangential forces due to non-uniform circular motion, ultimately applying the Pythagorean Theorem to find the resultant force.

PREREQUISITES
  • Understanding of centripetal force and its formula (Fcentripetal = mv²/r)
  • Knowledge of kinematic equations for uniformly accelerated motion
  • Familiarity with the Pythagorean Theorem in physics contexts
  • Concept of non-uniform circular motion and its implications
NEXT STEPS
  • Review the derivation of centripetal force in non-uniform circular motion
  • Study kinematic equations for objects accelerating from rest
  • Learn how to apply the Pythagorean Theorem in force calculations
  • Explore examples of net force calculations in circular motion scenarios
USEFUL FOR

Physics students, educators, and anyone interested in understanding dynamics of circular motion and force calculations in non-uniform scenarios.

SA32
Messages
31
Reaction score
0
Hi there, I was wondering if anyone could help me with a centripetal motion question.

"A cyclist competes in a one-lap race around a flat, circular course of radius r. Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in a time Δt. The mass of the bicycle (including the rider) is m. What is the magnitude of the net force Fnet acting on the bicycle as it crosses the finish line?

Find Fnet, the magnitude of the net force acting on the cyclist at the finish line.
Express the net force in terms of r, Δt, m, and π."

My thinking was as follows:

In circular motion, Fnet = Fcentripetal = mv2/Δt

I also know that v=2πr/Δt

Substituting the latter equation into the former, I get (m*4π2r)/Δt2

However, I am being told that my answer is "off by a multiplicative factor". I worked through a series of hints for this problem with a friend, only to come up with the same answer. Any help is appreciated!
 
Physics news on Phys.org
SA32 said:
Hi there, I was wondering if anyone could help me with a centripetal motion question.

"A cyclist competes in a one-lap race around a flat, circular course of radius r. Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in a time Δt. The mass of the bicycle (including the rider) is m. What is the magnitude of the net force Fnet acting on the bicycle as it crosses the finish line?

Find Fnet, the magnitude of the net force acting on the cyclist at the finish line.
Express the net force in terms of r, Δt, m, and π."

My thinking was as follows:

In circular motion, Fnet = Fcentripetal = mv2/Δt

I also know that v=2πr/Δt

Substituting the latter equation into the former, I get (m*4π2r)/Δt2

However, I am being told that my answer is "off by a multiplicative factor". I worked through a series of hints for this problem with a friend, only to come up with the same answer. Any help is appreciated!
First you have a typo in the F_c equation, it's mv^2/r, which you have correctly applied though. Your error comes in your equation for v. You have calculated his average speed using that formula. Since his initial speed is 0, what's his final speed at the finish line?
 
Oops! Now I feel stupid.

This is non-uniform circular motion... so I need to find the net radial force and the net tangential force than use the Pythagorean Theorem to solve for Fnet.

Thanks for pointing that out!
 

Similar threads

Replies
6
Views
4K
Replies
55
Views
3K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
17
Views
2K
Replies
4
Views
28K