Question

Bridging Problem: Oscillating and Rolling 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 (Figure 1). A frictionless ring at the rod's center is attached to a spring of force constant k, the spring's other end is fixed. The cylinders are pulled to the left a distance I, stretching the spring, 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

The spring is stretched by x.

spring force = -kx , force acts to the right.

radius of the cylinder R

The force acts at the center of the cylinders and produces a torque about the point of contact, friction at contact prevents it from slipping.

Torque on the cylinder $\tau$ = -kx*R

MI of the cylinder about its axis I = MR²/2

I$\alpha$ = $\tau$ = -kxR

angular acceleration of the cylinder  $\alpha$ = -kxR/(MR²/2) = -k/2MR *x

$\alpha$ = a/R , a - linear acceleration of the center of the cylinders

a = -(k/2M) x

acceleration of the center is proportional to the displacement

=> the center of the cylinder is in SHM with

$\omega$² = k/2M

period of oscillation T = 2$\pi$/$\omega$ = (8M$\pi$²/k)^1/2  

E) translational motion of the cylinder

Ma = kx -f

frictional force f = kx- Ma

F) torque by the frictional force = fR = I$\alpha$

f = 1/R *(MR²/2) * a_cm/R = 1/2 Ma_cm