A doubt on “velocities” of a body on an incline

I'm stuck in understanding a few stupid concepts (stupid because they are easy but I need to expertise) about mechanics of incline.

Suppose I have a sphere, rolling on an incline, and there is a friction $\mu$.

Applying the conservation of energy $\mathcal{L} = -\Delta E$ (where $\mathcal{L}$ is the friction work), I can get, via simple calculations, the final velocity as

$$v_f = \sqrt{\frac{10}{7} g\ell (\sin(\alpha) - \mu\cos(\alpha))}$$

Where $\ell$ is the length of the incline.

But when I adopt Newton's law, I get the following:

$$\begin{split} \mathbf{F}_{net} & = m\mathbf{a} \\ ma &= W - f \\ ma &= mg\sin(\alpha) - m\mu g \cos(\alpha) \\ a &= g\sin(\alpha) - \mu g\cos(\alpha) \end{split}$$

Treating $a$ as $\dot v$ I get

$$v(t) = gt(\sin(\alpha) - \mu\cos(\alpha))$$

Which is a velocity as a function of time.

How can I manage to obtain the final velocity starting from $v(t)$ ?

By the obvious law of the accelerated motion

$$\ell = \frac{1}{2}at^2$$

whence

$$t = \sqrt{\frac{2\ell}{ g (\sin(\alpha) - \mu\cos(\alpha))}}$$

Which should be ok.

Question: if the angle is zero degrees, this root do no more exist. How then can we get the horizontal plane condition back?

Also if that $t$ is the total time, then using it into $v(t)$ shall give me the final velocity at that time, but I would get

$$v_f' = \sqrt{2\ell g(sin\alpha - \mu \cos \alpha)}$$

Which is not really the first final velocity I got...

• Look at the $t$ dimensionally. You have seconds on the left and square root of meters on the right – Dominik Car Apr 12 '18 at 14:23
• Can you elaborate on how you got the first result using "the work done by friction"? If the sphere is rolling without slipping, then the contact point is always at rest with respect to the surface, which means that the frictional force does no work. – Michael Seifert Apr 12 '18 at 14:30
• Yeah as Michael says, static friction does no work. In both cases (Energy conservation and Newton), the answer should be $v = \sqrt(10/7\; g h)$ where $h = \ell \sin\alpha$. Friction is not maximum. It's whatever it has to be to achieve smooth rolling. With Newton, you can figure out that $F_f = -2/7 mg\sin\alpha$ (positive x is down the ramp so the negative indicates up the ramp). Also with Newton, you figure out that $a = 5/7 g\sin\alpha$. $t = \sqrt(14\ell / (5g\sin\alpha))$. It all works out – DWade64 Apr 12 '18 at 15:38
• @MichaelSeifert I used this: $$\mathcal{L} = -\Delta E$$ $$f\cdot \ell = mgh_i - (U_f + T_f)$$ $$m\mu g \ell \cos(\alpha) = mg \ell\sin(\alpha) - \frac{1}{2}mv^2 - \frac{1}{2}I\omega^2$$ And from here I got $v$. – Les Adieux Apr 12 '18 at 15:48

If you are assuming that the sphere is rolling without slipping (and I think you are), then both results are incorrect.

The key realization to correct the first method (using energy) is to note that the sphere is rolling without slipping; you used this statement implicitly when you used the equation $v = \omega R$, where $R$ is the radius of the sphere. This means that the surface of the sphere never "rubs" against the table; in other words, the contact point of the sphere is momentarily at rest with respect to the table. This implies that the frictional force does no work. Thus, $\mathcal{L} = 0$.

The key realization to correct your second method is to note that if the sphere is rolling without slipping, then the distance the sphere rolls must be equal to $R$ times the angle it turns: $x = R \theta$, where $x$ is the distance the sphere has moved, and $\theta$ is the distance it has rotated. Taking the second derivative of this equation, we must have $a = R \alpha$, where $a$ and $\alpha$ are the linear and angular acceleration respectively. But since the torque on the sphere (measured about the CM) is $R f$, and the torque is also equal to $I \alpha$, we conclude that $$\tau = Rf = I \alpha = \frac{2}{5} m R^2 \left( \frac{a}{R} \right) \quad \Rightarrow \quad f = \frac{2}{5} m a.$$ In other words, the frictional force cannot be any old value (as DWade64 pointed out in the comments); for the sphere to roll without slipping, the frictional force must have a particular value.

If you make both of the above corrections to your methods, you will find that you get the same result in both cases. On the other hand, if you assume that the frictional force is equal to some value other than $f = \frac{2}{5} m a$ (which is what you're implicitly doing throughout), then you're no longer rolling without slipping. In this case, your second method will yield a correct. It may also be possible to correct your first method to account for the fact that $\omega \neq v/R$ in this case, but it's not immediately evident to me how you'd go about this.

• One thing I don't understand: if there is friction, then the ball rolls. But without friction the ball would just slip, but this slip would cause the ball to rub, hence a "friction". It's the friction that makes the ball to roll, and yet because no rubbing is present, there is no work done by friction. What am I missing? How can I image a slipping motion with friction, in a ball? I cannot. Rub exists only when I am dealing with boxes or books or blocks... Probably it's stupid, it's just a concept question! P.s. thank you for the answer. – Les Adieux Apr 12 '18 at 18:03
• I have to add that then my professor made a mess, because she treated the problem of a solid sphere rolling, including the friction work. And the sphere is just pure rolling... Meh! – Les Adieux Apr 12 '18 at 18:04
• @VonNeumann: In the case of rolling without slipping, the frictional force is static friction, not dynamic friction. By definition, static friction doesn't involve "rubbing", and can't do any work. Kinetic friction, on the other hand, can do work. – Michael Seifert Apr 12 '18 at 19:33

Using energy is the easiest way to solve this problem. Your kinetic energy of the sphere is $E_{KE} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ and the potential energy is $E_{POT} = mgh$ where $\omega = v/r$
Substituting and solving you get $\sqrt{\frac{10}{7}gh}$
That's it. If there were no friction you wouldn't get the rotational member, that is, the ball would just be slipping (the contact point at rest at all times).

Using the Newton's method the equation is: $$ma = mgsin(\alpha) - F$$
Where $F=I\dot{\omega}/r$ and $\dot{\omega}=\frac{a}{r}$
Substitute that into the equation and solve it.

$\textit{This is an ideal case!}$ In real world applications there is energy lost due to friction.