Uniform circular motion -- How can radial acceleration have a calculated value?

[Gurasees]

In uniform circular motion, direction of particle is changing at every moment but its speed remains the same. If the magnitude of velocity or speed remains the same, change in magnitude of velocity is zero. Then how come radial acceleration can have a calculated value since acceleration = change in magnitude of velocity/ change in time. Yes, the object has an acceleration due to change in direction but how can we possibly obtain a value?

 

[scottdave]

This video may shed some insight for you. I was helpful for me, explaining derivatives of vectors.

 

[A.T.]

acceleration = change in magnitude of velocity/ change in time

Where did you get that definition of acceleration from?

 

[Drakkith]

In uniform circular motion, direction of particle is changing at every moment but its speed remains the same. If the magnitude of velocity or speed remains the same, change in magnitude of velocity is zero. Then how come radial acceleration can have a calculated value since acceleration = change in magnitude of velocity/ change in time. Yes, the object has an acceleration due to change in direction but how can we possibly obtain a value?

Acceleration is defined as the change in the velocity divided by the change in time, of: $\vec a = \frac{dV}{dt}$
Since velocity is changing, this requires that there be an acceleration. Note that acceleration is not defined as the change in the magnitude of the velocity, but simply the change in the velocity.

You can find a simple calculus derivation here: https://en.wikipedia.org/wiki/Centripetal_force#Calculus_derivation

 

[scottdave]

Acceleration is defined as the change in the velocity divided by the change in time, of: $\vec a = \frac{dV}{dt}$ Note that acceleration is not defined as the change in the magnitude of the velocity, but simply the change in the velocity.

Yes. Note that acceleration is a vector quantity and is often in a different direction than velocity. In the case of uniform circular motion, it is always at 90° to the velocity direction (towards center of the circle). I think the video does a nice job of explaining this.

 

[Gurasees]

Yes. Note that acceleration is a vector quantity and is often in a different direction than velocity. In the case of uniform circular motion, it is always at 90° to the velocity direction (towards center of the circle). I think the video does a nice job of explaining this.

Actually yeah definition that i wrote for acceleration is wrong. Here acceleration is associated with change in direction. But what i am asking is how can we obtain a numerical value of acceleration if there is no change in numerical value of velocity?

 

[A.T.]

... if there is no change in numerical value of velocity

Velocity is a vector which does change.

 

[Windadct]

A vector is a "numerical" value. Just not a scalar. Speed is a Scalar, Velocity is a vector. Since the Force (vector) and the Velocity of the particle are always at 90 degrees - no work is done, but there is action/reaction.

F=ma where F and a are vectors is critical to proper analysis,

MANY- heck if not all, cases involving normal vectors are counter-intuitive, and are worth special attention. If you master the vector math behind the precession of a gyroscope, for example - you will know more physics than 99.9% of the population.

It may seem like a simple concept - but complete comprehension is very valuable.

 

[Drakkith]

Actually yeah definition that i wrote for acceleration is wrong. Here acceleration is associated with change in direction. But what i am asking is how can we obtain a numerical value of acceleration if there is no change in numerical value of velocity?

The numerical value of the velocity's vector components is continuously changing. This is true in both Cartesian and polar/spherical coordinates.

 

[vanhees71]

Instead of unclear words formulae can only help the understanding ;-)).

Take the example of a particle running around on a circle of radius $R$ around the origin in the $xy$ plane with constant angular velocity $\omega$. The position vector is given by
$$\vec{x}(t)=R \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix}.$$
The velocity is
$$\vec{v}=\dot{\vec{x}} = R \omega \begin{pmatrix} -\sin(\omega t) \\ \cos (\omega t) \\ 0 \end{pmatrix}$$
and the acceleration
$$\vec{a}=\dot{\vec{v}}=R \omega^2 \begin{pmatrix} -\cos(\omega t) \\ -\sin(\omega t) \\ 0 \end{pmatrix}=-\omega^2 \vec{x},$$
i.e., the acceleration is radially towards the center with the magnitude $a=|\vec{a}|=\omega^2 R$. To keep the particle on the circle you need the corresponding force, called the centripetal force, $\vec{F}=m \vec{a}=-m \omega^2 \vec{x}$.

 

[jtbell]

Acceleration is defined as the change in the velocity divided by the change in time, of: $\vec a = \frac {dV}{dt}$

To clarify: both velocity and acceleration are vectors, so this should be written as $$\vec a = \frac {d \vec v}{dt}$$ or, component by component: $$a_x = \frac{dv_x}{dt} \\ a_y = \frac{dv_y}{dt} \\ a_z = \frac{dv_z}{dt}$$ or, in the matrix notation that vanhees71 used: $$\vec a = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} = \frac {d}{dt} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix}$$

 

[Arjan82]

Without formula's: acceleration in one direction is independent of the acceleration in any other direction (as long as the directions are normal to each other). Thus if you look at your object, it's velocity in horizontal direction does change all the time, and therefore it has an acceleration in horizontal direction. This is also independently true for the vertical direction.

 

[CWatters]

what i am asking is how can we obtain a numerical value of acceleration if there is no change in numerical value of velocity?

Why should it be a problem to calculate a numerical value of acceleration in such a case?

Lets say you have a ball going North at 3m/s and later it's found to be going South at 3m/s. The numerical value of the velocity (aka speed) hasn't changed, only the direction has changed. To calculate the acceleration you first need to calculate the change in velocity which in this case is 6m/s South.