# Does car have more kinetic energy when turning?

I asked this Phys.SE question Does car lose kinetic energy when turning?

Assume a car turning without losing its speed by holding to a point by a rope.

IMO, while the car is turning, its kinetic energy is constant $E=(1/2)mv^2$, but it has additional rotation energy as it changes it direction $E=(1/2)Iw^2$ (while turning)

I feel this has something wrong, but I don't know. So whether or not the car has more energy when turning?

• In my answers I intrprete the moment of inertia $J$ as that one for rotations about the center of mass of the car. You should clarify what you mean by it. Commented Apr 4, 2014 at 14:04
• Thanks Tobias, that is an excellent answer which I have looked for Commented Apr 6, 2014 at 3:51

When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake.

For instance for a point like mass on a straight line we have:

$$E_l = \frac{1}{2}m v^2$$

Now we catch the mass with a lasso of length $R$ and we force it to a circular motion. Let's try the wrong thing: $$E_c = \frac{1}{2}m v^2 + \frac{1}{2}I \omega^2 = \frac{1}{2}m v^2 + \frac{1}{2} mR^2 \left(\frac{v}{R}\right)^2 = mv^2=2E_l\;???$$

But since the lasso generates a perpendicular force that does not do any work, is not possible for the energy to change. As Kvothe pointed out:

If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $\vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which means their product is always zero. The rope may change the direction of the car, but it cannot speed it up or slow it down.

So you can compute the energy with either the tangential quantities ($m,v$) or the angular quantities ($I,\omega$) as it is simpler, but do the job just one time!

• Basically: "You are counting double" Commented Apr 4, 2014 at 13:51
• I do not think that the original poster considers a car as a point mass. The motion of the car can be decomposed into the motion of the center of mass (CoM) and the rotation about the CoM. If the car moves in the (x,y) plane, $m$ and $J$ are the mass and the moment of inertia about the local z-axis through the CoM, $v$ the velocity of the CoM and $\omega$ the rotational speed about the z-axis you have the kinetic energy $\frac m2 v^2 + \frac J2\omega^2$. Commented Apr 4, 2014 at 14:18
• @Tobias Such a detailed explanation is not relevant to the big picture and (fair guess?) beyond the scope of the question/OP. @ DarioP feel free to copy my answer into yours. It's best to have it all in one place :) Commented Apr 4, 2014 at 17:47

No, it does not gain energy. The confusion arises because there's a force that does no work.

If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $\vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which means their product is always zero. The rope may change the direction of the car, but it cannot speed it up or slow it down.

$\def\om{\omega}\def\vr{{\vec r}}\def\l{\left}\def\r{\right}\def\ve{{\vec e}}\def\vom{{\vec\omega}}\def\ds{\,'}$ Let the car move in the (x,y)-plane, let $m$ be the car's mass and let $J$ be the moment of inertia for the rotation about the axis through the center of mass aligned parallel to the z-direction.

If you have a straight line and a circle with radius $R$ and a smooth transition in between the line and the circle and the car is lossless constrained to this curve then the balance of energy says for the velocity $v_1$ on the line and the velocity $v_2$ on the circle \begin{align} \frac m2 v_1^2 &= \frac m2 v_2^2 + \frac{J}{2R^2} v_2^2\\ \frac{v_1}{v_2} &= \sqrt{1+\frac{J}{mR^2}}. \end{align} So yes, without losses or energy transformations (like chemical to mechanical energy) the velocity in the circular section will be smaller than the velocity on the straight line.

Pityingly it is difficult to write down the equations of motion for such a path. If one tries it with a direct connection between the straight line and the circular path one ends up with what I have written below.

The resume is that the instantaneous transition from the linear path to the circular path corresponds to an impulsive start of the rotational motion. But there you have something like a collision.

Let us consider the curve \begin{align} (x(s),y(s),0)=\begin{cases} (1,s,0)&\text{ for } s<0\\ (\cos(s),\sin(s),0)&\text{ for } s\in[0,2\pi) \end{cases} \end{align} parameterized through arc length $\l(|d\vr| = \sqrt{\cos'(s)^2 + \sin'(s)^2} ds = ds\r)$.

The instantaneous translational velocity of the car is $|\dot\vr(t)| = \dot s(t)$. The instantaneous rotational velocity of the car is $\vom=\ve\times\dot\ve$ with the unit vector $$\ve = \vr\ds = \begin{cases} (0,1,0)&\text{ for }s<0\\ (-\sin(s),\cos(s),0)&\text{ for }s\in[0,2\pi) \end{cases}$$ (the derivative w.r.t. the arc length parameter). Thus, $$\vom(t) = \begin{cases} \vec{0}&\text{ for }s(t)<0\\ (0,0,\dot s(t))&\text{ for } s(t)\in[0,2\pi) \end{cases}$$ It is a pure rotation about the z-axis with rotational velocity $$\omega = \begin{cases} 0&\text{ for }s(t)<0,\\ \dot s(t)&\text{ for } s(t)\in[0,2\pi). \end{cases}$$

With the car mass $m$ and the moment of inertia $J$ the kinetic energy and also the Lagrangian is \begin{align} L(s,\dot s) &= \frac m2 {\dot s}^2 + \frac J2 \omega^2\\ &= \frac12(m+JH(s)){\dot s}^2 \end{align} where $$H=\begin{cases}0&\text{ for }s<0\\\frac12&\text{ for }s=0\\1&\text{ for }s> 0\end{cases}$$ is the Heaviside function. The equation of motion is \begin{align} \frac{d}{dt}\l(\frac{\partial}{\partial{\dot s}}L\r)-\frac{\partial}{\partial s}L &= 0 \end{align} which gives with \begin{align} \frac{\partial}{\partial{\dot s}}L&= (m+JH(s)){\dot s}\\ \frac{d}{dt}\l(\frac{\partial}{\partial{\dot s}}L\r) &= (m+JH(s)){\ddot s}+J\delta(s){\dot s}^2\\ \frac{\partial}{\partial{s}}L&= \frac12 J\delta(s){\dot s}^2 \end{align} the equation \begin{align} (m+JH(s)){\ddot s}+\frac{J}{2}\delta(s){\dot s}^2&=0\\ {\ddot s}&=-\frac{J}{2(m+JH(s))}\delta(s){\dot s}^2 \end{align} This says that the velocity will decrease at $s=0$ if $\dot s\neq 0$ there.

The coefficient of $\delta(s)$ is discontinuous. That means that the integral value of the product is not well defined. E.g., for the product $\frac{J}{2(m+JH(s))}\delta(s)$ the integral value can be between $\frac{J}{2 m}$ and $\frac{J}{2(m+J)}$. This expresses that at the transition from the straigt line to the circular curve you have something like an impact and this impact can be lossy or not.

Let us assume that the car passes $s=0$ at $t=0$ with $\dot s(t-)>0$. Then we have \begin{align} \int_{t=0-}^{0+}\ddot s dt &= \int_{t=0-}^{0+} -\frac{J}{2(m+JH(s))}\delta(s){\dot s}^2 dt\\ \dot s(0+)-\dot s(0-) &= \int_{s=0-}^{0+} -\frac{J}{2(m+JH(s))}\delta(s){\dot s} ds \end{align} If we use the symmetric dirac delta distribution with $\int H(s) \delta(s) ds = \frac12$ we get \begin{align} \dot s(0+)-\dot s(0-) &= -\frac{J}{2(m+\frac12J)}{\dot s}(0+)+\frac{J}{2m}{\dot s}(0-) \end{align} and we can calculate the speed after $t=0$ \begin{align} \dot s(0+)&= \frac{1+\frac{J}{2m}}{1+\frac{J}{2m+J}}{\dot s}(0-)\\ &=\frac{2m+J}{4m(m+J)}{\dot s}(0-) \end{align} But this corresponds not to the speeds that are necessary for energy conservation.

• @DanioP I interprete $J$ as moment of inertia of the car for rotations about the center of mass of the car. In this case my way is correct. If you model the car as point mass you get $J=0$. Consider a heavy Lorry... It will have a moment of inertia w.r.t. the center of mass. Commented Apr 4, 2014 at 14:02
• @DanioP Why did your comment disappear? Now, my previous comment is out of context. Commented Apr 4, 2014 at 14:09
• I deleted it because I realised what you were doing. It is correct but I think you misunderstood the question. However in the Lagrangian I would put $(RH)^2$ instead of $H$. I found it interesting to propagate the contribution of the bending radius. Commented Apr 4, 2014 at 14:28
• @DarioP In the example we have $R=1$ and if we omit pointwise evaluation of $H$ we have $H^2=H$. In fact, I wanted to use the definition $$H(s) = \begin{cases}0&@ s<0\\ 1 &@ s\geq 0\end{cases}$$ at first but adopted then en.wikipedia.org/wiki/Heaviside_step_function. In the distributional sense they are the same (modification on zero measure). What counts is the limit for the dirac impulse and that may vary for a product with an essentially discontinous function. One could interprete $\dot s$ there as multivalued. The right value must be selected by some other criterium (e.g. energy). Commented Apr 4, 2014 at 15:41
• What you did is correct for sure, I was just saying that if $R$ is kept through the computation the final result becomes more interesting. Commented Apr 4, 2014 at 15:47

The potential energy $P$ of the car doesn't change (the car stays on the ground the whole time), and because it moves uniformly in a circle, its speed $\left | \mathbf{V} \right |$ doesn't change. But the kinetic energy $K$ of the car is $\frac {1}{2} m\left | \mathbf{V} \right |^2$, so it doesn't change aswell. Hence, the total energy $E= K + P$ won't change also.