Question

Problem 6: The angle θ is slowly increased from zero. Find the value of θ at which t he slotted cyllinder of mass m will begin to slip.

Answer 1

Let us examine the equations of motion of a cylinder, of mass and radius, rolling down a rough slope

there are three forces acting on the cylinder. Firstly, we have the cylinder's weight,
M g
which acts vertically downwards. Secondly, we have the reaction, of the slope, which acts normally outwards from the surface of the slope. Finally, we have the frictional force, which acts up the slope, parallel to its surface.

Thus, applying the three forces,
M g
and, to the cylinder's center of mass, and resolving in the direction normal to the surface of the slope, we obtain

$\begin{displaymath} R = M g \cos\theta. \end{displaymath}$

Furthermore, Newton's second law applied to the motion of the center of mass parallel to the slope, yields

$\begin{displaymath} M \dot{v} = M g \sin\theta - f, \end{displaymath}$

where is$$\dot{v}$$ the cylinder's translational acceleration down the slope?

examine the cylinder's rotational equation of motion. First, we must evaluate the torques associated with the three forces acting on the cylinder. Recall, that the torque associated with a given force is the product of the magnitude of that force and the length of the level arm--i.e., the perpendicular distance between the line of action of the force and the axis of rotation. Now, by definition, the weight of an extended object acts at its center of mass. However, in this case, the axis of rotation passes through the center of mass. Hence, the length of the lever arm associated with the weight is
M g
zero. It follows that the associated torque is also zero. It is clear, from Fig. 84, that the line of action of the reaction force, passes through the center of mass of the cylinder, which coincides with the axis of rotation. Thus, the length of the lever arm associated with is zero, and so is the associated torque. Finally, according to Fig. 84, the perpendicular distance between the line of action of the friction force, and the axis of rotation is just the radius of the cylinder, --so the associated torque is.
f b
We conclude that the net torque acting on the cylinder is simply

$\begin{displaymath} \tau = f b. \end{displaymath}$

It follows that the rotational equation of motion of the cylinder takes the form,

$\begin{displaymath} I \dot{\omega} = \tau=f b, \end{displaymath}$

where is its moment of inertia, and is$$\dot{\omega}$$ its rotational acceleration.

It follows from Eqs. (403) and (405) that

$$\displaystyle \dot{v}$$	$$\textstyle =$$	$$\displaystyle \frac{g \sin\theta}{1 + I/M b^2},$$
$$\displaystyle f$$	$$\textstyle =$$	Mg sin θ

Since the moment of inertia of the cylinder is actually , the above expressions simplify to give

$\begin{displaymath} \dot{v} = \frac{2}{3} g \sin\theta, \end{displaymath}$

and

3 M g sin θ

Note that the acceleration of a uniform cylinder as it rolls down a slope, without slipping, is only two-thirds of the value obtainedwhen the cylinder slides down the same slope without friction. that, in the former case, the acceleration of the cylinder down the slope is retarded by friction. Note, however, that the frictional force merely acts to convert translational kinetic energy into rotational kinetic energy, and does not dissipate energy

Now, in order for the slope to exert the frictional force specified in Eq. (410), without any slippage between the slope and cylinder, this force must be less than the maximum allowable static frictional force, $$\mu R (=\mu M g \cos\theta)$$where $$\mu$$ is the coefficient of static friction. In other words, the condition for the cylinder to roll down the slope without slipping is,$$f< \mu R$$ or

tan θ 〈 3μ.

This condition is easily satisfied for gentle slopes, but may well be violated for extremely steep slopes (depending on the size of $$\mu$$). Of course, the above condition is always violated for frictionless slopes, for which.$$\mu=0$$

Suppose, finally, that we place two cylinders, side by side and at rest, at the top of a frictional slope of inclination $$\theta$$. Let the two cylinders possess the same mass, , and the same radius, . However, suppose that the first cylinder is uniform, whereas the second is a hollow shell. Which cylinder reaches the bottom of the slope first, assuming that they are both released simultaneously, and both roll without slipping? The acceleration of each cylinder down the slope is given by Eq. (407). For the case of the solid cylinder, the moment of inertia is , and so

Usolid g sin θ

For the case of the hollow cylinder, the moment of inertia is (i.e., the same as that of a ring with a similar mass, radius, and axis of rotation), and so

2 Uhollow g sin θ