B Centripetal acceleration

Recently, I was looking into centripetal acceleration and there's something I don't understand.

According to my book, during uniform circular motion, the acceleration is

a= v^2/r

where v is the speed at which the object is moving and r is the radius of the circle.

However, this formula is not treating acceleration as a vector because the result of this formula is a scalar while acceleration is a vector. I think that this formula refers to the modulus of the acceleration vector. I'd like to know an explanation about how calculations could be made while treating acceleration and velocity as vectors. i.e. calculating the acceleration vector in a moment t. As acceleration is changing, is there jerk? Were there any, what would it be?
 

Orodruin

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The acceleration is just the second time derivative of the position vector. Start by writing down the position vector for uniform circular motion and differentiate it twice and you will have a vector expression for the acceleration. The jerk is just the time derivative of the acceleration so you can compute that too with exactly the same method, just take another time derivative.
 

Doc Al

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However, this formula is not treating acceleration as a vector because the result of this formula is a scalar while acceleration is a vector. I think that this formula refers to the modulus of the acceleration vector.
Sure, that formula just gives the magnitude of the acceleration. A complete description would also include the direction of the acceleration vector. That direction, of course, is toward the center of the circle. (Thus the name centripetal, which just means 'toward the center'.)
 
The acceleration is just the second time derivative of the position vector. Start by writing down the position vector for uniform circular motion and differentiate it twice and you will have a vector expression for the acceleration. The jerk is just the time derivative of the acceleration so you can compute that too with exactly the same method, just take another time derivative.
The acceleration is just the second time derivative of the position vector. Start by writing down the position vector for uniform circular motion and differentiate it twice and you will have a vector expression for the acceleration. The jerk is just the time derivative of the acceleration so you can compute that too with exactly the same method, just take another time derivative.
Thanks. I didn't know the position vector for uniform circular motion, but I looked it up and did the derivatives. Your advice was useful,
 
Sure, that formula just gives the magnitude of the acceleration. A complete description would also include the direction of the acceleration vector. That direction, of course, is toward the center of the circle. (Thus the name centripetal, which just means 'toward the center'.)
Yeah, I knew it was towards the center but I wanted to know the formula.
 
By the way, in case it wasn't clear, my question was already answered. How do I close the thread?
 
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Sure, that formula just gives the magnitude of the acceleration. A complete description would also include the direction of the acceleration vector. That direction, of course, is toward the center of the circle. (Thus the name centripetal, which just means 'toward the center'.)
Hi Doc Al ,

Could you please clarify one doubt - If an object moves with uniform speed around a circle (fixed radius R) , the magnitude of centripetal acceleration is constant . The direction is towards the circle . But since the direction is always towards the center , is it correct to say that acceleration is constant ??

If I look things from the perspective of polar coordinates , then acceleration is always directed radially inwards . But from the perspective of cartesian coordinates , the direction continuously changes .

Please give your views .

Thanks
 

vanhees71

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The most simple way is to do the calculation. For the motion with constant angular velocity around the ##z##"=axis of a Cartesian coordinate system you have
$$\vec{x}=R [\vec{e}_x \cos(\omega t)+\vec{e}_y \sin(\omega t)].$$
Now you take the time derivatives to get the velocity
$$\vec{v}=\dot{\vec{x}}=R \omega [-\vec{e}_x \sin(\omega t)+\vec{e}_y \cos(\omega t)].$$
The magnitude is
$$v=|\vec{v}|=R \omega, \quad \text{because} \quad \sin^2 (\omega t)+\cos^2(\omega t)=1.$$
Another derivative gives the acceleration
$$\vec{a}=\dot{\vec{v}}=-R \omega^2 [\vec{e}_x \cos(\omega t)+\vec{e}_y \sin(\omega t)].$$
The magnitude is
$$a=\vec{a}=R \omega^2=\frac{v^2}{R}.$$
 

Doc Al

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Hi Doc Al ,

Could you please clarify one doubt - If an object moves with uniform speed around a circle (fixed radius R) , the magnitude of centripetal acceleration is constant . The direction is towards the circle . But since the direction is always towards the center , is it correct to say that acceleration is constant ??

If I look things from the perspective of polar coordinates , then acceleration is always directed radially inwards . But from the perspective of cartesian coordinates , the direction continuously changes .

Please give your views .

Thanks
I would say that the magnitude is constant, but the acceleration vector continually changes. Sure, using polar coordinates it's easy to write the direction of the acceleration as ##\hat r##, but viewed from an inertial frame the direction that ##\hat r## indicates is always changing.
 
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The direction of acceleration changes in polar coordinates as well. The unit vectors in polar coordinates have variable directions so even though the acceleration is always radial and has constant magnitude, its derivative is not zero.
 
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so even though the acceleration is always radial and has constant magnitude, its derivative is not zero.
Nice !
 

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