Question

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest on a horizontal tabletop (Fig. 14.29). A frictionless ring at the center of the rod is attached to a spring with force constant k; the other end of the spring is fixed. The cylinders are pulled to the left a distance x, stretching the spring, and then released from rest. Due to friction between the tabletop and the cylinders, the cylinders roll without slipping as they oscillate. Show that the motion of the center of mass of the cylinders is simple harmonic, and find its period.

Answer 1

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

PROOF:

rotational inertia for cylinders  is given as:

                                  $I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       $RF=\frac{1}{2}MR^{2}\alpha$

                                       $RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   $f_{s}=\frac{1}{2}MR\alpha$------ (5)

As angular acceleration is related to linear by:

                                          $a= R\alpha$

Eq (5) becomes

                                    $f_{s}=\frac{1}{2}Ma_{x}$---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  $\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

                                  $f_{s} - kx = Ma_{x}$[/tex] -----(8)

Using (6) in (8)

                               $\frac{1}{2}Ma_{x} - kx =Ma_{x}$

                                     $a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

                                  $a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

                                  $\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

                                $a_{x} = - \omega^{2}x$

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

B) Time Period

Time period is related to angular frequency as:

                                   $T=\frac{2\pi }{\omega}$

                                  $T = 2\pi \sqrt{\frac{3M}{2k}$