Question

In this problem you will consider the motion of a cylinder ofradius $r_cyl$ that is rolled from a certain height so that it "loops the loop," that is, rolls around thetrack with a loop of radius $r_loop$ shown in the figure without losing contact with thetrack.

Unless otherwise stated, assume that friction is sufficient thatthe cylinder rolls without slipping. The radius $r_cyl$ of the cylinder is much smaller than the radius $r_loop$ of the loop.

Q.

Compared to an object that does not roll, butinstead slides without friction, should a rolling object bereleased from the same,a greater, or a lesser height in order justbarely to complete the loop the loop?

The rollingobject should be released from a greater height. The rollingobject should be released from a lesser height. The rollingobject should be released from exactly the sameheight. The answerdepends on the moment of inertia of the rollingobject.

Q. Find the minimum height that will allow a solid cylinder of mass and radius $r_cyl$ to loop the loop of radius $r_loop$ .

Express in terms of the radius $r_loop$ of the loop.

Answer 1

Concepts and reason

The concepts used to solve this problem are energy conservation, centripetal force, rotational kinetic energy, potential energy and kinetic energy of the system.

First apply the law of energy conservation which simply states that energy can neither be created nor be destroyed, it can only be transformed from one form to another. Another important concept used in this problem is the fact that an object that translates and rotates simultaneously has two energies associated with its motion, one being due to the linear motion and the other due to rotation.

Fundamentals

Write the expression for the centripetal force.

${F_c} = \frac{{m{v^2}}}{r}$

Here,${F_c}$is the centripetal force and $r$is the radius of the circular field.

Write the expression for the kinetic energy when an object is rolling in a loop.

$K{E_r} = mgh$

Here,$K{E_r}$is the rolling kinetic energy of the object,$m$is the mass of the object,$g$is the gravitational acceleration and $h$is the height.

Write the expression for the rotational kinetic energy.

$K{E_R} = \frac{1}{2}I{\omega ^2}$

Here,$K{E_R}$is the rotational kinetic energy, $I$is the momentum of inertia and $\omega$is the angular speed.

Write the linear kinetic energy of the system.

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

Here,$KE$is the liner kinetic energy and $v$is the speed of the object.

Write the expression for the weight of the system.

$W = mg$

Here,$W$is the weight of the system.

(1)

From the expression for the rolling kinetic energy of the system.

$K{E_r} = mgh$

Rolling energy of the system is equal to,

$K{E_r} = KE + K{E_R}$

Substitute$mgh$for$K{E_r}$and$\frac{1}{2}I{\omega ^2}$for$K{E_R}$.

$\begin{array}{l}\\mgh = KE + \frac{1}{2}I{\omega ^2}\\\\KE = mgh - \frac{1}{2}I{\omega ^2}\\\end{array}$

(2)

Form the calculation in step 1, the body which have certain linear velocity and this body attain the top of the loop. So, to maintain contact with the surface the necessary centripetal force will be provided by gravity.

So,

$\begin{array}{c}\\mg = \frac{{m{v^2}}}{{{r_{{\rm{loop}}}}}}\\\\{v^2} = g{r_{{\rm{loop}}}}\\\end{array}$

$v = \sqrt {g{r_{loop}}}$

Now apply energy conservation, initially the kinetic energy is zero.

${E_{{\rm{init}}}} = mgh$

Now as the cylinder will complete the loop it will possess both kinetic and potential energy,

So,

${E_{{\rm{final}}}} = mg\left( {2{r_{{\rm{loop}}}}} \right) + \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$

The kinetic energy here, has two components one simply due to translation velocity and the other due to the rotation of the cylinder about its own axis.

Substitute$g{r_{{\rm{loop}}}}$ for$v{\,^2}$ in the final expression of the energy.

$\begin{array}{c}\\{E_{{\rm{final}}}} = mg\left( {2{r_{{\rm{loop}}}}} \right) + \frac{1}{2}mg{r_{{\rm{loop}}}} + \frac{1}{4}mg{r_{{\rm{loop}}}}\\\\ = \frac{{11}}{4}mg{r_{{\rm{loop}}}}\\\end{array}$

From the energy conservation.

${E_{{\rm{init}}}} = {E_{{\rm{final}}}}$

Substitute $\frac{{11}}{4}mg{r_{{\rm{loop}}}}$ for ${E_{{\rm{final}}}}$,and $mgh$ for ${E_{{\rm{init}}}}$.

$\begin{array}{c}\\mgh = \frac{{11}}{4}mg{r_{{\rm{loop}}}}\\\\h = \frac{{11}}{4}{r_{{\rm{loop}}}}\\\end{array}$

Ans: Part 1

Therefore, the rolling object should be released from a greater height.