# How can tangential acceleration from a radial force be explained?

A mass is attached to a rope, and put into a circular motion. If I pull the string from the center, the tangential speed of the mass will increase (by conservation of angular momentum).

I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force?

• It's not clear to me what you are asking - can you try to clarify your question? – ACuriousMind May 21 '15 at 18:34
• Why did you accept an answer and then post a bounty that a canonical answer is required? This seems inconsistent, since it indicates you did not find any answer good enough to be satisfied (which is what accepting an answer usually indicates). – ACuriousMind Jun 9 '15 at 15:21
• – BowlOfRed Jun 13 '15 at 5:01

The object will move in a curved path whose center is not where I am pulling from. This center, my hand and the mass form a triangle whose lead angle might be positive or negative depending if the speed of the mass is increasing or decreasing.

Consider the body above at B moving along the indicated curved path (like a closing spiral). While pulling from A with a force $F$, some of the force goes into rotating the mass about C (the $m v^2/r$ part) and some into accelerating the mass (the $m \dot{v}$) part.

• I really like the visual representation - this is a conceptually hard problem and this answer deserves a lot of upvotes. – Floris Jun 11 '15 at 13:43
• Personally I love it when geometry can explain concepts better than equations. – ja72 Jun 11 '15 at 15:52
• Completely agree. I often like your answers but this one in particular. Wait 24 hours and you will see just how much I like it... That's how long it takes before I can award the bonus. :-) – Floris Jun 11 '15 at 15:54

No. The easiest way to see this without invoking rotating reference frames is to write out Newton's Law's in polar coordinates, which work out to be: \begin{align*} F_r &= m \ddot{r} - m r \dot{\phi}^2 \\ F_\phi &= m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} \end{align*} From these, it's pretty easy to see that if we have $F_\phi = 0$ and $\dot{r} \neq 0$ (i.e., no tangential forces but a decreasing radius), we will still get $\ddot{\phi} = \dot{\omega} \neq 0$.

The question you might ask, then, is why do Newton's Laws look so strange in polar coordinates? Why are those extra terms present on the right-hand side of the equations? The answer is that the "real" equation is that $\vec{F} = m \ddot{\vec{r}}$. If we express $\vec{r}$ in, say, Cartesian components ($\vec{r} = x \hat{x} + y \hat{y}$), then the vectors $\hat{x}$ and $\hat{y}$ are constant with respect to time; thus, $\ddot{r} = \ddot{x} \hat{x} + \ddot{y} \hat{y}$. However, if express the particle's position in polar coordinates ($\vec{r} = r \hat{r}$), then $\hat{r}$ is not constant with respect to time; so when we take the derivative of $\vec{r}$ with respect to time, we get unexpected terms. See the above link for the full derivation.

• Nice mathematical derivation. The key here is that the radius unit vector changes with respect time. – Cicero Jun 9 '15 at 1:06

A mass is attached to a rope, and put into a circular motion. ... I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force?

For short, I'll call the mass attached to the rope a "rock". So how does the rock gain angular velocity? If you truly are applying a purely radial force, and if you aren't shortening the length of the rope between your hand and the rock, the answer is that it can't. Radial forces do zero work. If you keep the length of the rope constant and you apply the force radially toward a fixed location, the angular acceleration of the rock about that fixed location is identically zero.

The rock does gain angular velocity as you swing it, and presumably you aren't changing the length of the rope between your hand and the rock. That means you aren't applying a pure radial force when the rock's angular velocity is increasing.

Think of how you get the mass at the end of the rope (I'll call it a "rock" from now on) moving. At the start, your hand makes a big circle about your body / above your head. You gradually tighten up that circle your hand makes as the rock starts moving about you.

At any point in time, the acceleration on the rock is toward your hand, but that is not purely toward the center of the average motion of the rock, and it is not normal to the rock's velocity vector with respect to your body. You are imparting a tangential acceleration on the rock, and you are doing so without having to continuously shorten the length of the rope.

Only a radial force is applied so that angular momentum $L=rmv$ is conserved,

$$dL = m(dr\, v+dv\, r)=0.$$

The force imparts an impulse on the system. So the system is starting to move towards a new equilibrium with shorter radius.

The constraint $$r\, dv = -v\, dr$$ now means that for negative $dr$ the system picks up a positive angular acceleration.

• I think you would have to make your answer more quantitative for it to be useful i.e. show the maths behind your reasoning. – John Rennie Jun 9 '15 at 5:41

The centripetal force is directed radially inwards . Work done due to the centripetal force is 0 (as, S(displacement in one rotation)=0). Work done = F.S.cos(\theta)=m . a . cos(\theta).The accln. is thus directed tangentially outwards. Thus, tangential velocity increases.