# (Circ Motion) Why does using a vector diagram and the centripetal acceleration formula give different values? (1 Viewer)

#### erucibon

##### Member
Why does using a vector diagram and the centripetal acceleration formula give different values? (using exact values in both so it isn't rounding errors)

#### fan96

##### 617 pages
Why does using a vector diagram and the centripetal acceleration formula give different values? (using exact values in both so it isn't rounding errors)
Can you give an example?

One possible reason may be that the formula you're using has implicit assumptions (e.g. uniform circular motion).

#### Drdusk

##### π
Moderator
Can you give an example?

One possible reason may be that the formula you're using has implicit assumptions (e.g. uniform circular motion).
I think it might be because your doing the operations incorrectly. For example you can't just add or subtract two vectors the same way as numbers like 2+3. You need to consider what direction they're pointing and work out the resultant vectors magnitude appropriately using Pythagorus and what not.

Regardless an example would suffice for us to clear this up.

#### erucibon

##### Member
A mass lies on a perfectly smooth horizontal surface and is connected to a 1.2m string. The mass completes 5 rotations per second. Position B differs from Position A by 60 degrees.

Calculating the acceleration using the centripetal acceleration formula with 5 rotations per second as the speed gives 1184ms-2. By calculating the time it takes to move to B (0.033s, exact value is used in acceleration calculation), and then drawing a vector diagram to find change in velocity and then finding acceleration through change in velocity over time, I get 1131ms-2 (using exact values).

$\bg_white a_c = \frac{(5\times1.2\times2\pi)^2}{1.2}$

$\bg_white a=\frac{5\times1.2\times2\pi}{\frac{\frac{60}{360}\times2\pi\times1.2}{5\times1.2\times2\pi}}$

Shouldn't the magnitude of centripetal acceleration be the same regardless of whether it is moving from one point to another?

#### fan96

##### 617 pages
Your second acceleration formula doesn't actually give the correct instantaneous acceleration - rather, an approximation of it.

Consider a more extreme case:

After the mass leaves point $\bg_white A$, it will again return to $\bg_white A$ (after $\bg_white 0.2$ seconds have passed).
The vector for change in velocity here is, of course, the zero vector (with magnitude zero).
This implies the acceleration is zero.

Now this is completely true (depending on what is meant by acceleration), but it's not the answer we want.

Why? Because this method only considers two points - the start and the end. The path of the mass was never taken into account.
Between $\bg_white A$ and $\bg_white B$, the mass may as well have gone to the moon and back - as long as its initial and final velocities didn't change, we'd have no way of telling.

That's not to say that the second method is useless. With a bit of basic calculus, you can add a restriction of (uniform) circular motion and derive the centripetal force formula from it.

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#### Sydney Learning Hub

##### New Member
Hi,

vector diagram: averages out the acceleration of the object between 2 points while
centripetal acceleration formula: finds the instantaneous acceleration.

The difference lies in using differentiation (the formula) versus the small change in values (the diagram).
Then again, when drawing the diagram you are not drawing the acceleration but the final velocity and the initial velocity to find the change in values.

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Sydney Learning Hub
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