# Basic Kinetics Question: Calculate Velocity on a Circular Path - Homework Help

• Femme_physics
In summary, the conversation discusses a calculation for the velocity of a rider on a circular path, with given values for weight, initial velocity, height, and radius. The person has attempted a solution using different equations and is confused about the correct answer from a previous year's notebook. Another person helps clarify the mistake in the equation and explains the concept of a point mass.
Femme_physics
Gold Member

## Homework Statement

http://img39.imageshack.us/img39/3778/funkyrider.jpg

So I got this funky rider that starts its movement from the top of the hill A at initial speed (Vo). From that point, the bicycles role on freely (without using pedals) at the circular path.

Calculate the velocity of the rider at the bottom point B.

W= (weight of the rider+bicycle) = 650 [N]
Vo (initial velocity) = 27 km/h
H = 14m
R= 10m

Comment: Ignore friction
C is point mass

## The Attempt at a Solution

HERE is my problem. Last year in my notebook I wrote this:

http://img204.imageshack.us/img204/6220/vfold.jpg

I tried resolving it but I didn't get the same result. This result is the same one in the manual so I reckon it's correct.

I tried

a = -9.81
a = 9.81
a = v2/R = 7.52/10
a = g2/R = 9.812/2

Nothing works. The third one makes slight sense to me because according to the formula of circular motion that's what "a" ought to be, v2/R...problem is that using this I'm not considering "g" and that's impossible, therefor the last a makes most sense to me. None of them gives me my old answer!

Last edited by a moderator:
The equation in your old book is incorrect(the answer is right, somehow ). You seem to have interchanged the addition and multiplication signs

From energy conservation it comes out to be,

$mgh + \frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2$

cutting out m, you get...

$\frac{1}{2}v_f^2 = \frac{1}{2}v_i^2 + gh$

And of course, g is...?

Hmm, how can you just cut out m? dividing everything by m?

I'm a bit confused as to why we shouldn't take m into consideration...but maybe it's only for the first clause, since after all it's being defined to us as point mass...

which is another confusing factor. If the thing is "point mass"-- it shouldn't really have any weight

Yes, I divided everything by m. It's non zero, thus legal to divide.

Mass is taken into consideration for the energy equation, but luckily, it gets divided out. If, somehow, there were some mass joining the original body at an intermediate stage of motion(not effecting the other values), then the final mass would be changed, and you couldn't just 'divide mass out' then.

If the thing is "point mass"-- it shouldn't really have any weight

A 'point mass' means it has mass(and may have weight) but its dimensions are infinitely small.

Last edited:
True, m isn't changed through. Thanks, I'll keep it all in check. Much appreciated Infinitum :)

## What is basic kinetics?

Basic kinetics is the study of the rates at which chemical reactions occur. It involves understanding the factors that influence reaction rates and how to measure and calculate these rates.

## What are the factors that affect reaction rates?

The factors that affect reaction rates include temperature, concentration of reactants, presence of catalysts, surface area of reactants, and the nature of the reactants and products.

## How do you measure reaction rates?

Reaction rates can be measured by monitoring changes in concentration of reactants or products over time. This can be done using techniques such as spectroscopy, chromatography, or titration.

## What is the rate law?

The rate law is an equation that relates the reaction rate to the concentrations of the reactants. It is determined experimentally and often includes the rate constant, which is a measure of how fast the reaction occurs.

## How do you calculate reaction rates?

Reaction rates can be calculated using the rate law equation, which takes into account the concentrations of the reactants and the rate constant. The units of the rate constant will depend on the overall order of the reaction.

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