# Acceleration of body rolling down inclined plane

Acceleration of a body rolling down an inclined plane is given by:

$$\frac{g\sin\theta}{1+\frac{k^2}{r^2}}$$

$g$=acceleration due to gravity

$\theta$=angle of inclined plane

$k$=radius of gyration

$r$=radius of abject

How come acceleration is independent of coefficient of friction of plane when friction itself is causing the torque? I am not able to understand this although I am able to derive the formula on my own.

For this equation to be valid you also have to assume that the body rolls without slipping. In that case the friction is static and defined by the no-slipping constraint, but for this the coefficient of friction has to be bigger than a certain value.

The reason that the acceleration is independent of the friction for an object rolling down a plane (assuming no slipping) is because the friction in this system is static friction, and it does no work on a rolling object. Consider the following diagram:

The point of contact with the surface denoted by the red dot only comes into contact with a single point on the surface while rolling. Since friction only acts when it is in contact with the surface, and the distance the point travels while in contact with the surface is $\Delta x = 0$, the friction acting on the object can be considered to be static. The static frictional force is generally written as $F_s = \mu_s N$, where $\mu_s$ is the static frictional coefficient, and $N$ is the normal force. This isn't strictly true, and should be more generally written as $F_s \leq \mu_s N$. So, the magnitude of the force can change depending on the incline to a value necessary to supply the needed torque to get the object rolling.

Now, lets consider the work done by a static frictional force. The work done on a system can be considered to be how much you change the energy of the system. Since static friction does not change the energy, the conversion of gravitational potential energy to rotational and translational kinetic energy remains the same regardless of the coefficient of friction, $\mu_s$. Therefore, the acceleration is not affected by it. So, how do we know it does no work. Consider the following definition for work.

$W = F\Delta x$

where $F$ in this case is the force of friction, and $\Delta x$ is the distance over which the friction acts. Now, you might say that since the object is rolling, the frictional force acts over distance the object rolls, but this is not the case as we've established, because the point of contact with the surface for the rolling object does not actually move relative to the surface when it is in contact with the surface.

• Thank you. That was very clear. I had assumed the friction acting was kinetic because the object was moving. – Mahathi Vempati Oct 26 '15 at 2:15
• I'm glad that it helped. That is a common mistake that is easy to make given the dynamic nature of the system. – tmwilson26 Oct 26 '15 at 2:18

Imagine a really rough surface with infinite coefficient of friction. The body will still roll just fine. So the amount of friction does not affect the rolling or not. Only the presence of friction does.

In general, try to consider the extreme situations in order to understand how and why an equation works (or not).

The forces perpendicular to the incline surface need to be balanced $$mg\cos {\theta}=N\quad \quad (1)$$ Forces parallel to the incline generate the parallel acceleration$$mg\sin{\theta}-f_r=m\frac{\mathrm{d}v}{\mathrm{d}t}\quad \quad(2)$$ playing around with equation 2, we get $$\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{mg\sin{\theta}-f_r}{m}\\ \int g\sin{\theta}\mathrm{d}t-\frac{1}{m}\int f_r\mathrm{d}t=v(t)\quad\quad(3)$$ The torque is defined as $$\tau=I\frac{\mathrm{d}\omega}{\mathrm{d}t}\quad \quad (4)$$But also $\tau=\vec{r}\times\vec{F}$. Thus here $\tau=f_rR$ and the Moment of inertia $I$ is $I=mK^2$. Plugging all this into equation 4 and playing around, we get...$$\frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{f_rR}{mK^2}\\\omega(t)=\frac{R}{K^2}\frac{1}{m}\int f_r\mathrm{d}t\\ \frac{1}{m}\int f_r\mathrm{d}t=\frac{K^2}{R}\omega(t)\quad \quad (5)$$ Plugging eq5. into eq.3, we get $$\int g\sin{\theta}\mathrm{d}t-\frac{K^2}{R}\omega(t)=v(t)\quad \quad (6)$$

Now is the important part. For, PURE ROLLING, $v(t)=\omega(t)R$. Thus using this fact in eq.6, eq.6 can be rewritten as $$\int g\sin{\theta}\mathrm{d}t-\frac{K^2}{R^2}v(t)=v(t)\\v(t)=\frac{1}{1+\frac{K^2}{R^2}}\int g\sin{\theta}\mathrm{d}t$$ Hence the acceleration can be written as $$a(t)=\frac{\mathrm{d}v(t)}{\mathrm{d}t}=\frac{g\sin{\theta}}{1+\frac{K^2}{R^2}}$$