Kinematics for Non-Uniform Circular Motion

Question

I'm trying to understand how kinematics for non-uniform circular motion. I know that you can represent the net acceleration of an object in non-uniform circular motion with the following equation:

$$a_{net} = a_T + a_C$$

where $a_T$ is the tangential acceleration and $a_C$ is the centripetal acceleration. You can further simplify this equation to:

$$a_{net} = \left(\frac {dV}{dT}\right)\,\hat e_t + \frac {V^2}{R}\, \hat e_c$$

where $\hat e_t$ and $\hat e_c$ are unit vectors in the tangential and centripetal directions.

What I don't know how to do is a scenario where you know that an object has a certain constant net acceleration. The object is accelerating from some initial angular/tangential velocity to some final angular/tangential velocity. In this scenario, the higher the object's tangential velocity gets, the lower the object's tangential acceleration becomes. This is because the higher the velocity, the more force you need to change direction. So the centripetal component of net acceleration increases and the tangential component of net acceleration decreases.

How would you go about solving for the time it would take for the object to accelerate from $v_0$ to $v_f$ ?

My initial thoughts on this are that you would need to express tangential/angular acceleration as a function of time so that you can apply the following kinematics formula:

$$v_f = v_0 + \int_{t_0}^{t_f} a_t {\rm d}t $$

I'd appreciate some help on how to be able to find a function for tangential acceleration so I can do kinematics with these types of scenarios.

Answer 1

It is simply a matter of resolving the total acceleration vector $a$ into its components. Tangential acceleration $a_t=\ddot s$ is the component which lies along the circle or curve, while the centripetal acceleration $a_c=\dot s^2/r$ is the component normal to the curve. ($s$ is distance along the curve.) The centripetal acceleration has no effect on speed along the curve, so it can be ignored.

This procedure works for any curve and also for non-constant net acceleration $a$. So all you need to know is the angle between the total acceleration vector and the tangent to the curve at each point.

Your integration formula is not very useful because you do not know how $a_t=\ddot s$ varies with time, so you cannot easily integrate. What you can do is express $\ddot s$ in terms of distance along the arc, or angle around a circle :
$a_t = \ddot s =\frac{dv}{dt}=v\frac{dv}{ds}=\frac{v}{r}\frac{dv}{d\theta}$
where $v=\dot s$ is speed along the curve, $\theta$ is angle swept out by a radius vector, and $r$ is the radius.

Suppose the constant total acceleration $a$ and the tangent vectors are initially parallel, and suppose the particle's position on the circle is measured in terms of angle $\theta$ from this point. Then the tangential acceleration is $a_t=a\cos\theta$. The equation of motion is
$v\frac{dv}{d\theta}=ar\cos\theta$
which you can solve by integration to find the speed $v$ at any angle $\theta$ on the circle.

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However, I suspect that the scenario you wish to solve is not one in which the net acceleration is constant. For example, a particle sliding on a vertical circular wire in a gravitational field. The gravitational force on the particle is constant, but there is also a normal reaction force, and this is not constant, so the net acceleration of the particle is not constant in this scenario.

In this case you can define a potential energy for the particle-wire system. (This is possible for any conservative force, not only a constant force.) Then you can relate the speed of the particle along the wire to its height in the force field, using conservation of mechanical energy.