Question

Hi can someone please tell me how i can show that f=mv^2/R may be written as F=mw^2R

Answer 1

The arc length of a circle of radius R and angle \( theta \) is
\[ \theta R \]
Hence if a mass moves an angular distance \( \Delta \theta \) in time \( \Delta t \), it has velocity
\[ v = \frac{\Delta \theta}{\Delta t} R = \omega R \] by definition of angular velocity \( \omega \).

Now substitute this expression into your first one to get the second.

Answer 2

im just a little confused bc in my book it says w= theta/time

Answer 3

Yes, \[ \omega = \frac{d\omega}{dt} \] the rate of change of angle. That is, \( \omega \) is the angular velocity.

For example, if the angular velocity \( \omega = 2\pi \), then every second, the circle/disk makes one complete revolution of \( 2\pi \) radians.

Answer 4

In that case, if you the circle had radius r, a point on the circle traveled a distance of
\[ 2\pi r \] in that one second and hence it's average velocity has magnitude
\[ v = \omega r = 2 \pi r \] also.

Answer 5

*average instantaneous velocity v. It's average velocity of course is zero. But at any given moment its instantaneous velocity or speed is \( \omega r \).

Answer 6

can u please show how you got v=dellta theta/change time R=wR step by step bc i am a little stuck

Answer 7

Let's come at this a different way. Suppose an object is in uniform circular motion of radius R and angular velocity \( \omega \). Over a very small time period \( \Delta t \), what is the arc length the object has moved?

Answer 8

To answer that question, first tell what angle the object has moved.

Answer 9

arc length= R(radius)w ( angular velocity)

Answer 10

No. What is the angle the object moves in time \( \Delta t \).

Answer 11

If the object is moving with angular speed w (for omega), then in time Dt (for delta t), the **angle** it has swept out is
theta = w.Dt

make sense?

Answer 12

because that is what angular velocity omega means. By definition, w is the amount of angle moved in 1 second. Hence in Dt seconds, the object must have moved an angle of
\[ \theta = \omega \Delta t \]
Yes?

Answer 13

yah

Answer 14

Hence in time Dt, what **distance** has the object moved?

Answer 15

arc length= Rxw not sure

Answer 16

If it has moved an angle of theta, how far in **distance** has it moved?

Answer 17

what is the arc length for the angle theta, in other words?

Answer 18

other formulas that were given with this problem were v=2piRn/deltat , w=2pin/delta t
and w= Dtheta/ delta t.. this problem is like a warm up to the chapter i was wondering can i use this formulas to derive it or do i have to use your method

Answer 19

In time \( \Delta t \) the angle moved is
\[ \Delta \theta = \omega \Delta t \]
Hence in that time, \( \Delta t \), the distance moved is
\[ \Delta s = \Delta \theta . R \]
I'm taking you back to just basic, basic definition here.

Answer 20

Now what is \( \Delta \theta \), we know that
\[ \Delta \theta = \omega \Delta t \]
Hence
\[ \Delta s = \Delta \theta R = \omega \Delta t . R \]

Answer 21

Divide both sides by delta t and we have
\[ \frac{\Delta s }{\Delta t} = \omega R \]

Answer 22

So far so good?

Answer 23

yah much better

Answer 24

Now what is delta s / delta t equal to?

Answer 25

isnt wR

Answer 26

speed v is
\[ v = \frac{\Delta s}{\Delta t} \] therefore
\[ v = \omega R \]

Answer 27

can i use any of the formulas listed above as well to derive thisquation

Answer 28

v=2piRn/delta t , w=2pin/delta t and w= Delta theta/ delta t..

Answer 29

These have unusual notation. What is n here, frequency?

Answer 30

R radius , n revolutions

Answer 31

(James, how do you do fractions in the equation editor? I'd like to volunteer an explanation.)

Answer 32

Yes, frequency.

In short, yes, you can use these equations to get to the same results. But in an important way, the method I have shown you is more fundamental. As an exercise for yourself, use your equations to find the same result.

Answer 33

@Underhill, go for it. Fraction are written as
\frac{a}{b}
\[ \frac{a}{b} \]

Answer 34

*fractions

Answer 35

Awesome!

Answer 36

can someone please show how i can derive it using the formulas.. having a very difficult time

Answer 37

\[F = m\frac{v^2}{R}\]Since\[v = \omega \times R = \frac{radians}{second}\times\frac{meters}{radian}\]then \[v^2 = \omega^2R^2\]and therefore\[F = m\frac{v^2}{R} = m\frac{\omega^2R^2}{R} = m \omega^2R\]

Answer 38

It's no more complicated than this. Omega is the angular velocity. V is the linear velocity. You can already see how closely related these two equations are.

You can make the connection by realizing that omega is radians/second, and that in any circle that the radius corresponds exactly to one radian. Think of the radius not as a simple distance, but in terms of "meters per radian" because that is exactly what it is.

Multiplying omega (radians/second) by the radius (meters/radian) gives meters/second. Viola! Meters/second = velocity. You're there!