However, I find that the net torque of the cylinder is 0 if I set the axis of rotation as the bottom of the cylinder (the contact point between the ground and the cylinder); my first question is, **what accounts for this discrepancy?** I suspect there should be a force about the center of mass, but I don't see what could be the source of such force.

You're making an elemental error: the axis of rotation is through the CoG of the round object, not the contact point with the floor.

So your torque is the *friction force times the radius* of the round object, as explained in better detail below. This torque causes angular acceleration, assuming the object wasn't already rolling at $\omega={v}{r}$. If the object was rolling without slippage on a horizontal plane at *constant* speed and thus also constant angular speed then no friction is needed because there's no angular acceleration (or deceleration) and thus no torque needed.

I understand how the velocity at the contact point relative to the ground is zero. My second question is, **would this mean that there is no friction force other than rolling friction, as I marked on the diagram?** I suspect that there should be static friction force at the contact point, but that wouldn't make any sense because then the static friction would provide net torque, resulting in continual angular acceleration of the object.

Yes, it means there is only one type of friction in your idealised set up.

For the inclines, I've adapted one of your diagrams as follows:

Let's look at the case without slippage first.

The weight of the rolling object $mg$ can be decomposed into two components, one parallel to the slope:

$$F_x=mg\sin\theta,$$

and one vertical to it:

$$F_y=F_n=mg\cos\theta.$$

The vertical (to the slope) component, also referred to as *the normal force* $F_n$ is the one that gives rise to a friction force $F_f$, usually somewhat simplistically modelled as:

$$F_f=\mu F_n,$$

with $\mu$ some friction coefficient. If $\mu$ is large enough there will be no slippage. When there is no slippage then the linear speed $v$ is given by:

$$v=\frac{2\pi R}{T},$$

and with $\omega=\frac{2\pi}{T}$ ($T$ is the time needed to complete one rotation),

$$v=\omega R.$$

The net force on the object in the $x$-direction (parallel to the slope) can now be used to set up a Newtonian equation of motion:

$$F_x-F_f=ma,$$

with:

$$a=\frac{dv}{dt}=\frac{d^2 x}{dt^2}.$$

So with substitutions, the equation of motion is the following differential equation:

$$(\sin\theta - \mu \cos\theta)g = \frac{dv}{dt}.$$

Integration between $0,0$ (for example but you can set your own initial conditions) and $t,v$ then gives:

$$v=(\sin\theta - \mu \cos\theta)gt.$$

In the case of no slippage, the velocity increase can also be determined from an energy balance. As the object rolls down the slope potential energy $U$ is converted to kinetic energy $K$, so that $\Delta U = \Delta K$. $K$ is made of two components, *translational* and *rotational* kinetic energy and we can write:

$$\Delta U=\Delta K_t+\Delta K_r.$$

$\Delta U$, for a **vertical** distance $h$ travelled is given by $\Delta U=mgh$ and with the expressions for $\Delta K_t$ and $\Delta K_r$ we get:

$$mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2},$$

where $I$ is the *moment of inertia* and $\omega={v}/{R}$. Substitution then allows to isolate $v^2$.

Now let's briefly look at the case where $\mu=0$, then $F_f=0$.

In the case where there was no slippage, $F_f$ provided a moment (torque) around the centre of the object and the equation of rotation is:

$$I\dot{\omega}=F_fR,$$

where $\dot{\omega}=\frac{d \omega}{dt}$ is the *angular acceleration*. With $F_f=0$, $\dot{\omega}=\frac{d \omega}{dt}=0$. In plain English this means that there is no angular acceleration and if the object wasn't rotating to begin with, it will not start to do so. In that case the motion is purely translational and there is no rotation.

It also means that when there's no friction the energy balance is reduced to:

$$mgh=\frac{mv^2}{2},$$

and this means that a *purely sliding* object will gain more translational speed than a rotating one.

My main, and third, question is what would happen after the cylinder starts rolling without slipping in this scenario; **would it continue rolling without slipping, the angular velocity and velocity of the center of mass increasing by the same factor, or would one overtake the other?**

No, if it wasn't slipping to begin with it will never slip.

*Edit:*

**Derivation of the critical friction coefficient $\mu_c$:**

The equation of rotation for the round object on a slope is:

$$I\dot{\omega}=F_fR,$$

or:

$$I\dot{\omega}=\mu mgR\cos\theta,$$

$$I\frac{d \omega}{dt}=\mu mgR\cos\theta,$$

assuming $\omega=0, t=0$ then integrated we get:

$$\omega=(\mu \frac{mgR}{I}\cos\theta)t.$$

Now we can spot a snake in the grass: it appears that for large values of $\mu$, $\omega$ would also become large and this is not correct. This *apparent error* arises from the fact that $F_f$ is a *reactive force* that cannot become arbitrarily large.

Now assume again no slippage at all. We know from above that in that time $t$ the object has also acquired translational speed $v$:

$$v=(\sin\theta - \mu \cos\theta)gt.$$

and also with $v=\omega R$, we get:

$$(\mu \frac{mgR^2}{I}\cos\theta)t=(\sin\theta - \mu \cos\theta)gt,$$

$$\mu \frac{mR^2}{I}\cos\theta=(\sin\theta - \mu \cos\theta),$$

Reworked we get:

$$\large{\mu_c=\frac{I}{I+mR^2}\tan\theta}.$$

What does this mean? In order to have no slippage at all we need:

$$\mu \geq \mu_c$$

Then for the general case (where at $t=0, \omega=\omega_0$) of no slippage:

$$\large{\omega=\omega_0+(\mu_c \frac{mgR}{I}\cos\theta)t}.$$

**Secondly**, for the case where there is no friction at all:

$$\mu=0$$

$$\large{\omega=\omega_0}.$$

In other words, it it was already rotating at $t=0$ it will keep rotating at that angular speed $\omega_0$. If it wasn't rotating it won't start doing so at any time.

**Thirdly**, for the intermediate case (*some* slippage}:

$$\mu_c > \mu > 0$$

$$\large{\omega=\omega_0+(\mu \frac{mgR}{I}\cos\theta)t}.$$

Well, that was a lot of algebra and some simple calculus, enjoy and I hope it helps!

## Best Answer

Neither of the 2 answers is true. You ask about the velocity

between the points 2 and 3.Answer 1. is false because the direction of the velocity is horizontal and to the left at all points between 2 and 3.

Answer 2 is false because the magnitude of velocity is constant between 2 and 3.

I think what you are really concerned about is

how the velocity changes at point 2. This cannot be decided without further information about the apparatus.If there is a

smooth changein the direction of the ramp at point 2 then the magnitude of velocity will be constant, only the direction changes (Answer 1). The velocity between 2 and 3 is $v_1$, the same as it was at the bottom of the incline.If there is a

sudden changein direction, then there is a collision between the object and the horizontal surface. If the object does not rebound and jump up after colliding with the horizontal surface at point 2, then we must assume that its vertical component of momentum is absorbed by the horizontal surface. Only the horizontal component of velocity remains, so the velocity between 2 and 3 is $v_1 \cos\theta$, where $\theta$ is the angle of inclination of the slope between 1 and 2. So in this case both the magnitude and direction of velocity change at 2 (Answer 2).Judging by the gap I would be inclined to think there is a

sudden changein direction, in which case Answer 2 is correct.