# Centripetal force for non-uniform circular motion

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?

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).

• +1. Sir, that's great...that means this formula is valid for both uniform & non- uniform motion. But one thing that's bothering me is that if during non-uniform circular motion, centripetal force varies due to changing of linear velocity, doesn't it produce spiral motion?? Confused about varying centripetal force:-C
– user36790
Commented Nov 22, 2014 at 4:58
• Well ... you specified circular motion. That automatically excludes spiral motion. If the motion is circular but the speed is not uniform, then both the tangential and centripetal forces will be changing in order to maintain the circular motion. Commented Nov 22, 2014 at 12:32
• That's the point. If centripetal force had not altered, the motion wouldn't be circular. Thus varying centripetal force is necessary for the circular motion, which is non-uniform. ... Spiral motion only occurs when the body loses kinetic energy like the Rutherford's flawed atom's model.... This is mine & yours opinion. But the others here speak otherwise. And why have you been downvoted without reason?? Confused.
– user36790
Commented Nov 22, 2014 at 15:39

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.

• I don't know why centripetal accln. should be constant always. $$a_c = \dfrac{v^2}{r}$$ ... If $v$ varies, obviously $a_c$ varies. Think your reasoning again.
– user36790
Commented Nov 22, 2014 at 16:16
• @user36790 You are right. Completely right. By constant I meant, the formula is still the same. I have chosen bad words to say it. I'll edit. Commented Nov 22, 2014 at 16:48
• Then ok, sir. Thanks for the rigorous proof. Just one mistake in one of the relations and absence of necessary words...nevertheless thanks:-)
– user36790
Commented Nov 22, 2014 at 16:56

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 $F_c=\dfrac{mv^2}{r}$

• My book told that when there is non-uniform circular motion, a tangential force acts along with the inward centripetal force. Since, centripetal force depends on linear velocity, it will vary but still forms circular motion. Is my book wrong as you are telling it will be spiral motion??
– user36790
Commented Nov 22, 2014 at 6:48