Radial and Tangential Acceleration

[Faiq]

Homework Statement

2 A particle P of mass mkg moves on an arc of a circle with centre O and radius a metres. At time t = 0
the particle is at the point A. At time t seconds, angle POA = sin^2 2t.
Find
(i) the value of t when the transverse component of the acceleration of P is first equal to zero

The answer is $$\frac{d^2\theta}{dt^2} = 0$$
Isn't d^2theta/dt^2 equal to radial acceleration. Since angular velocity is rate of change of theta, thus rate of change of change of theta should be angular acceleration aka radial acceleration?
Then why are the equating it as transverse component?

 

[FactChecker]

If P goes around in a circle, the velocity vector is always at right angles to the radius through P. So the radial and transverse acceleration are equal.

 

[Faiq]

If P goes around in a circle, the velocity vector is always at right angles to the radius through P. So the radial and transverse acceleration are equal.

In the next part, we are supposed to calculate radial force and transverse force and the answer is 3ma and 4ma respectively. If they are both equal then the force due to their acceleration should be same as well

 

[Chestermiller]

The radial acceleration is $v^2/a$, where v is the instantaneous tangential velocity. When they are talking about transverse acceleration, it think they mean tangential acceleration.

 

[Faiq]

The radial acceleration is $v^2/a$, where v is the instantaneous tangential velocity. When they are talking about transverse acceleration, it think they mean tangential acceleration.

I went through a post on a different website which acquired the following relation. Is it correct?
$$\frac{d}{dt}v=r\frac{d}{dt}\omega$$
$$a_t=ra_r$$

 

[Chestermiller]

I went through a post on a different website which acquired the following relation. Is it correct?
$$\frac{d}{dt}v=r\frac{d}{dt}\omega$$
$$a_t=ra_r$$

The first equation is correct. The second isn't.

 

[Faiq]

Why? I don't see anything wrong with the equation?

 

[Chestermiller]

Why? I don't see anything wrong with the equation?

What!? Since when is $a_r=\frac{d\omega}{dt}$?

 

[Faiq]

Isn't angular acceleration the time derivative of angular velocity?
https://en.wikipedia.org/wiki/Angular_acceleration

And further on the wiki page it mentions the formula tangential accel = r * angular accel

 

[Chestermiller]

$a_r$ is the conventional symbol for radial acceleration. The conventional symbol for angular acceleration is $\alpha$. If you are going to use unconventional symbols, you need to make that clear.

 

[Faiq]

Is there a difference between radial, angular and centripetal acceleration for a circular motion? In our class, we used them interchangeably

 

[Chestermiller]

Is there a difference between them? In our class, we used them interchangeably

Let me understand this correctly: In your class, they think that radial acceleration is the same thing as angular acceleration? Actually, they are entirely different quantities. The correct equation for radial acceleration is $a_r=\omega ^2r$ and angular acceleration is $\alpha=\frac{d\omega}{dt}$. They don't even have the same units.

 

[Faiq]

So
radial acceleration = centripetal acceleration
but
angular acceleration = tangential acceleration / radius.

 

[Chestermiller]

So
radial acceleration = centripetal acceleration
but
angular acceleration = tangential acceleration / radius.

For.a circular arc, yes.

 

[Faiq]

Okay thank you