We shoot an object of mass $m$ with velocity in the horizontal direction, $u_x$ and with velocity in the vertical direction, $u_y$.

The trajectory of the object is the combined result of a free fall with initial velocity opposite to gravity($u_y$) and translational motion in the horizontal direction due to $u_x$.

The question of my exercise says to find the tangential and the centripetal acceleration. But the object is not moving in a circle so I don't understand what that means.


The intent of the question is for you to resolve the acceleration into components parallel and perpendicular to the velocity vector $\mathbf v$, which you can do by definining $\hat v = \frac{\mathbf v}{\Vert \mathbf v\Vert}$ and then writing

$$\mathbf a_\parallel = (\mathbf a \cdot \hat v)\hat v$$ $$\mathbf a_\perp = \mathbf a - \mathbf a_\parallel$$

The conceptual point is that $\mathbf a_\parallel$ is responsible for changing the speed of the object while $\mathbf a_\perp$ is responsible for changing the direction of the velocity.

You're right that centripetal is not a good adjective here because the object is not moving in a circle. A better way to phrase the question would be "resolve the acceleration into tangential and normal components," or something similar.

  • $\begingroup$ You can still have a centripetal component even if the motion is not circular. It's just not going to be $\mathbf a- (\mathbf a\cdot\hat v)\hat v$. $\endgroup$ – BioPhysicist Feb 22 at 17:30
  • $\begingroup$ @BioPhysicist If the motion is not circular, I don't think it's reasonable to expect new students to know what centripetal is supposed to mean. The fact that it's possible to construct a tangent circle to every point of the trajectory is amusing and a nice way to look at things, but unless that's been explicitly explained I prefer "normal" to "centripetal" in the general case. $\endgroup$ – J. Murray Feb 22 at 18:39
  • $\begingroup$ I wasn't saying anything about what one should or shouldn't expect of new students. The OP says that they were asked specifically about centripetal and tangential. To me the better guess would be to look at the $-\hat r$ and $\hat\theta$ components, or at least to mention them as a possible interpretation. $\endgroup$ – BioPhysicist Feb 23 at 3:11
  • $\begingroup$ @BioPhysicist I’m not sure how you’re defining $\hat r$ and $\hat \theta$ here. Where is the origin of that coordinate system? $\endgroup$ – J. Murray Feb 23 at 3:21

Though the path of the projectile is not a perfect circle of constant radius but at any instant of time, the path does have a radius of curvature. This radius of curvature is changing continuously for the projectile.

Path of a projectile

The resultant force acting on the particle during it's flight is it's weight (mg) due to the gravitational acceleration $(\vec{g})$. You can resolve this resultant force into two components. One, along the local tangent direction which is responsible to change the magnitude of the velocity. The other one is perpendicular to the local velocity vector which is responsible to change the direction of the velocity. This second component can be considered as centripetal force. \begin{align} & v_x(t) = v_{x0} \\ & v_y(t) = v_{y0} - gt \\ & \vec{v}(t) = v_x(t)\hat{i} + v_y(t)\hat{j} \\ & \text{direction of velocity is:} ~~ \hat{v}(t)= \frac{\vec{v}(t)}{|\vec{v}(t)|} \\ & \text{Tangential acceleration:} ~~ \vec{a}_{||}(t) = (\vec{g}\cdot\hat{v}(t))\,\hat{v}(t) \\ & \text{centripetal acceleration:} ~~ \vec{a}_{\perp}(t) = \vec{g} - \vec{a}_{||}(t) \\ \end{align}

  • $\begingroup$ The direction of the blue arrows confused me for a moment. I generally expect arrows connecting a label with what it's labeling to point from the label to the object, not the reverse. My initial assumption was that the blue arrows were part of the diagram, associated with labels by proximity - that the blue arrows were themselves the local tangent(s) and object trajectory, which made the diagram nonsensical. $\endgroup$ – Douglas Jan 27 at 19:29
  • $\begingroup$ Good suggestion! I will keep that in mind. $\endgroup$ – Subhendu Chakraborty Jan 28 at 1:52

If an object is not moving on a straight line (with respect to some reference frame, of course) and/or is changing speed, the object must accelerate. In your context, the path will be parabolic and changing speed. You know the acceleration.

If you choose the coordinate system to be attached to the object and you define tangential to be in the instantaneous direction of the velocity and centripetal to be perpendicular to the velocity toward the inside of the curved path, you can resolve the acceleration vector into those components.

Non-zero centripetal acceleration doesn't mean you are traveling in a circle. It tells you that you aren't traveling on a straight path. The centripetal acceleration might be a component of some other acceleration.

I would guess that there is some explanation in your textbook about non-linear motion. You should go re-read that information carefully.


Part 1: A New Basis

Let $\vec{x}(t)$ be the position vector of your object. Set $\vec{v}(t)= \vec{x}'(t)$ as usual.

We define $\hat{v}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|}$ (note that we require that $\vec{v}(t) \neq \vec{0}$ for this definition to work). This is the unit vector in the tangential direction. By construction $$ \hat{v}(t) \cdot \hat{v}(t) = 1 $$ Differentiating both sides of this equation, we find $$ \hat{v}(t) \cdot \hat{v}'(t) = 0 $$ This tells us that $\hat{v}'(t)$ is orthogonal to $\hat{v}(t)$, the tangential direction. Define $$ \hat{n}(t) = \frac{\hat{v}'(t)}{\|\hat{v}'(t)\|} $$ Then $$ \hat{v}(t) \cdot \hat{n}(t) = 0 $$ so $\hat{n}$ is a unit vector in a direction normal to the tangential direction.

Let's keep the story simple and say we're considering motion in the plane. Then $\hat{v}$ and $\hat{n}$ must form a basis at every point of the object's trajectory $\vec{x}(t)$ where $\vec{v}(t) \neq 0$ and $\vec{v}'(t) \neq 0$. Again, $\hat{v}$ is the tangential direction, and $\hat{n}$ is the "normal" direction perpendicular to the tangential direction.

Part 2: Acceleration in the New Basis

Let $v(t) = \|\vec{v}(t)\|$ denote the object's speed. Then $$ \vec{v}(t) = v(t) \hat{v}(t) $$ Differentiating both sides, we obtain $$ \vec{v}'(t) = v'(t) \hat{v}(t) + v(t) \hat{v}'(t) $$ Of course, $\vec{a}(t) = \vec{v}'(t)$ is the object's acceleration. We also recognize that $\hat{v}'(t) = \|\hat{v}'(t)\| \hat{n}(t)$. Substituting, we obtain $$ \vec{a}(t) = v'(t) \hat{v}(t) + v(t) \|\hat{v}'(t)\| \hat{n}(t) \label{a}\tag{1} $$ So the object accelerates with magnitude $|v'(t)|$ in the tangential direction and magnitude $v(t) \|\hat{v}'(t)\|$ in the normal (what you call "centripetal") direction. In other words: $$ a_{\parallel} = v'(t) \quad \text{ and }\quad a_{\perp} = v(t) \|\hat{v}'(t)\| $$ The interpretation of $a_{\parallel}$ is obvious. The tangential acceleration is just how fast you're changing speed. You might expect this.

The interpretation of $a_{\perp}$ is bit more subtle. In particular, it's not at all obvious what $\|\hat{v}'(t)\|$ is physically.

Part 3: Curvature

By definition, the distance we've traveled after time $t$ (assuming, for definiteness-sake, that we start at $t = 0$) is $$ \ell(t) = \int_0^t v(t) dt $$ Abusing notation slightly, let $T = \ell^{-1}$, so $T(\ell(t)) = t$ and $\ell(T(l)) = l$. In words, $T(l)$ is how long it takes the object to travel a distance $l$. Let $\vec{y}(l)$ be the position vector of the object after it has traveled a distance $l$. Translating the definition for $\vec{y}(l)$ into math, we have: $$ \vec{y}(l) = \vec{x}(T(l)) $$ Differentiating both sides: $$ \vec{y}'(l) = \vec{x}'(T(l)) T'(l) = \frac{\vec{x}'(T(l))}{\ell'(T(l))} = \frac{\vec{x}'(T(l))}{v(T(l))} = \frac{\vec{v}(T(l))}{\|\vec{v}(T(l))\|} = \hat{v}(T(l)) $$ Summarizing: $$ \vec{y}'(l) = \hat{v}(T(l)) $$ Differentiating again: $$ \vec{y}''(l) = \hat{v}'(T(l)) T'(l) = \frac{\|\hat{v}'(T(l))\|\hat{n}(T(l))}{v(T(l))} \equiv k(l) \hat{n}(T(l)) \label{k}\tag{2} $$ where we have introduced the scalar quantity $k(l)$. This $k(l)$ is known to mathematicians as the "signed curvature" of the curve $\vec{y}(l)$. Notice $$ \vec{y}''(l) = k(l) \hat{n}(T(l)) $$ That is, the "acceleration" of $\vec{y}$ is entirely in the normal direction. By parametrizing the path of the object by distance ($\vec{y}$) instead of time ($\vec{x}$), we had $\|\vec{y}'(l)\| = \|\hat{v}(T(l))\| = 1$. The "speed" of $\vec{y}$ doesn't change, so all of its "acceleration" is in the normal direction. Somewhat informally, we can say that curvature is the magnitude of the acceleration of a path parametrized by the distance traveled (what mathematicians call "arc-length").

You should check that $k(l) = \frac{1}{R}$ for a circle of radius $R$ (you have all the tools you need if you've been following along). This is evidence of the fact that curvature $k$ is the reciprocal (up to a sign) of the radius of curvature $R$. If you like, this is actually a definition. In any case, we will use the equation $$ k(l) = \frac{1}{R(l)} \label{r}\tag{3} $$ to finish our analysis

Part 4: Putting It All Together

We see from \ref{k} that $$ k(l) = \frac{\|\hat{v}'(T(l))\|}{v(T(l))} $$ So $$ k(\ell(t)) = \frac{\|\hat{v}'(t)\|}{v(t)} $$ Substituting this into \ref{a}, we obtain $$ \vec{a}(t) = v'(t) \hat{v}(t) + v(t)^2 k(\ell(t)) \hat{n}(t) $$ Using \ref{r}, we finally obtain $$ \vec{a}(t) = v'(t) \hat{v}(t) + \frac{v(t)^2}{R(\ell(t))} \hat{n}(t) $$ We see that $$ a_{\parallel} = v'(t) \quad \text{ and }\quad a_{\perp} = \frac{v(t)^2}{R(\ell(t))} $$ which is the familiar formula from physics, except now $R$ is the radius of curvature of the path. If you like, you can define $r(t) = R(\ell(t))$ to be the radius of curvature at time $t$, which lets us write $$ a_{\parallel} = v'\quad \text{ and }\quad a_{\perp} = \frac{v^2}{r} $$


This (somewhat lengthy) analysis has led us to the conclusion that for general motion in the plane, the tangential acceleration is how fast your speed changes, and the centripetal acceleration is $\frac{v^2}{R}$, where $R$ is the radius of curvature.


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