Dynamics question

A car starts from rest and travels along a circular path of radius 200 m. The magnitude of its velocity increases at a rate of 2 m/s2. After 10 s, what is the magnitude of the acceleration of the car (in m/s2)?

?

Reply 1

Papkin

Magnitude of acceleration in circular motion is:

a = \frac {v^{2}}{r}

Reply 2

Brainss

Papkin

Magnitude of acceleration in circular motion is:

a = \frac {v^{2}}{r}

I've never learned this, so I'd appreciate further explanation (I can't remember from A-level physics, admittedly, and we've only done fluid dynamics not general dynamics in our first year of Civil Engineering). A friend of mine doesn't know how to do it and asked me. :o:

Reply 3

Brainss

Physics Enemy

Isn't the answer already given? i.e.) 2 ms^-2? :confused:

That's what I thought at first, but apparently acceleration is always changing in circular motion because it is a vector.

Reply 4

Brainss

Physics Enemy

But it's asking for magnitude anyway, not whether it's pointing up or down or left to right at that magnitude 10 secs later.

...and the magnitude is always changing in circular motion. Otherwise it would simply be just 2ms^-2. You're obviously supposed to use an arbitrary point as the start as 2ms^-2, but it's going to constantly change as you go around the circle. Are you able to do this?

2ms^-2 is in one direction. A circle has an infinite number of directions.

(edited 13 years ago)

Reply 5

Brainss

Physics Enemy

If you're going round and round in a circle then I don't see why the magnitude of acceleration would change because of that. If the question is saying what it says, to me I interpet that as |a| = 2 ms^-2.

Unfortunately I can't add any further input.

So by your logic a

90^{\circ}

rotation from the starting point, in a circle, would still have an absolute acceleration of

2ms^{-2}

when its non-absolute value is simply zero?

Reply 6

Robob

Brainss

So by your logic a

90^{\circ}

rotation from the starting point, in a circle, would still have an absolute acceleration of

2ms^{-2}

when its non-absolute value is simply zero?

The magnitude is constant, the direction changes.

Reply 7

Papkin

Brainss

Note that body moving around an object at a constant speed has in fact non-zero acceleration, because velocity is a vector, not a scalar, and therefore change of its direction is caused by an accelaration pointing towards the centre of the circle. (So that the velocity will always be tangent to the circle)

Reply 8

Brainss

Robob

The magnitude is constant, the direction changes.

So it's a trick question and the answer is in fact simply

2ms^{-2}

Reply 9

Papkin

Brainss

So it's a trick question and the answer is in fact simply

2ms^{-2}

Isn't the magnitude

2 \sqrt{2} \frac {m}{s^{2}}

?
2 inwards, and 2 tangent?

(edited 13 years ago)

Reply 10

Robob

Brainss

So it's a trick question and the answer is in fact simply

2ms^{-2}

Sorry, didn't even read the question. The component of acceleration along the direction of motion is, there is also a centripetal component which you need to work out and then add the two using pythagoras.

Reply 11

Stonebridge

Yes. After 10 s the linear component of velocity will have increased to 20m/s (10s x 2 m/s/s)
The resultant acceleration at this time will be the sum of the linear acceleration (2 m/s/s given) and the centripetal acceleration (v² /r) where v is now 20m/s
This will be the sum of two vectors at right angles and can easily be found from Pythagoras. The direction isn't required.