Centripetal force for non-uniform circular motion

Question

For uniform circular motion, centripetal force is given by $$\dfrac{mv^2}{r}.$$ But what will be the centripetal force if the circular motion is non-uniform in the sense that linear velocity is changing its magnitude? Will the above relation still be valid for this case?

Answer 1

If the tangential velocity is changing in magnitude, that implies a tangential acceleration, and thus a tangential force in addition to the centripetal force.

If the motion of the object is in a circle of constant radius, then the instantaneous centripetal force is given by the expression you wrote.

The argument is not restricted to motion in a circle, but the analysis is easier for circular motion. For any point on any curved path one can find the circle that is tangent to the curve at that point. The instantaneous "centripetal" (perpendicular to the velocity) force is given by the same formula, with $r$ being the radius of the tangent circle (the "osculatory" or "kissing" circle).

Answer 2

Let's go: This is a circular motion: $$ x(t) = R\cos\theta $$ $$ y(t) = R\sin\theta $$

Hence we can derive it: $$ \dot x(t) = -R\dot\theta\sin\theta, \quad\quad \ddot x(t) = -R\ddot\theta\sin\theta - R\dot\theta^2\cos\theta $$ $$ \dot y(t) = R\dot\theta\cos\theta, \quad\quad \ddot y(t) = R\ddot\theta\cos\theta - R\theta^2\sin\theta $$

Notice that the position of the body in circular motion is $\mathbf r = (x(t), y(t))$. If we compute: $$ \mathbf r \cdot\mathbf{\dot{r}} = R^2\dot\theta\cos\theta\sin\theta - R^2\dot\theta\cos\theta\sin\theta = 0 $$

Thus, the position vector is perpendicular to the velocity vector. Then, an acceleration parallel to $\mathbf r$ will be a centripetal one, and an acceleration parallel to $\mathbf{\dot r}$ will be a tangential one. The acceleration can be re-writen: $$ \mathbf{\ddot r} = \ddot\theta\mathbf{\dot r} - \dot\theta^2\mathbf r $$

Therefore, we decomposed the acceleration in a centripetal component and a tangential component. We can compute the magnitude of each one: $$ a_c = |\dot\theta^2\mathbf r| = |\dot\theta|^2\cdot|\mathbf r| = \dot\theta^2\sqrt{R^2\cos^2\theta + R^2\sin^2\theta} = \dot\theta^2 R $$ $$ a_t = |\ddot\theta\mathbf{\dot r}| = \ddot\theta\sqrt{R^2\dot\theta^2\cos^2\theta + R^2\dot\theta^2\sin^2\theta} = \ddot\theta\dot\theta R $$

If we want uniform motion, we just need to set $\theta = \omega t$. Notice that if $\theta = \omega t$ we will have $a_c = \omega R$ and $a_t = 0$. Now we just need to plug speed somehow. But $|\mathbf{\dot r}| = v = \dot\theta R$. Therefore: $$ a_c = \dot\theta^2 R = \frac{v^2}{R}$$ $$ a_t = \ddot\theta v $$

We now see, centripetal acceleration remains the same formula for all possible circular motions, uniforms or not, since $\theta(t)$ is a generic function. Hence, since $v(t) = \dot\theta(t) R$ for a generic $\theta(t)$, $v(t)$ also is not fixed with time. But centripetal acceleration remains the same formula.

Answer 3

Just my two cents to add to the other answers a more concrete example. Imagine an initially circular motion due to a mass being pulled towards the center by the tension of a rope. If you keep the tension constant, you have circular motion. Now, if you start increasing the tension, the mass will move in a spiral fashion towards the center, then, if you stop increasing the tension, the motion will be circular again, but at a smaller radius. In both circular motions, the initial and the final one, you will have $T=\dfrac{mv^2}{r}$ (with both $T$ and $v$ larger for the smaller circle). And for the intermediate times when the tension was variable, it still will be valid that $T=\dfrac{mv_p^2}{r}$, where $v_p$ is the component of the speed perpendicular to the tension (the rope).

In the above example there was no tangential force to change the speed (the radius was not constant). But you can also have a pure circular motion with changing speed. Here the change in speed is due to a tangential force. One example is a pendulum that has enough energy to undergo a circular motion such that the tension at the top (vertical position) is zero. At the top, the only force is centripetal: $F_c=mg$. At the bottom it is also only centripetal: $F_c=T-mg$. At other positions you do have a tangential force, that is maximum at the horizontal position: $F_t=mg$ and $F_c=T$. At any point in the trajectory you still have $F_c=\dfrac{mv^2}{r}$