Question

Figure 10-1 illustrates a simplified roller bearing. The inner cylinder has a radius of 1.0 cm and is stationary. The outer hollow cylinder has a radius of 1.2 cm and is rotating at 8.0 rpm. Between the two cylinders are several small cylinders with a radius of 0.10 cm, which roll without slipping on both the inner and outer cylinders. Only one of these cylinders is shown in the figure. What is the angular speed of the small cylinders?

Answer 1

In the figure of simplified roller bearing

Find-angular speed of the small cylinders

Explanation:

For calculating the angular speed of the small cylinders we will find velocity

omega=8×2pi=16 pi

so, we can find by the above formula

velocity of the point present on the outer circle=16 pi × 1.2=19.2 pi

19.2 pi=diameter of small cylinders × omega

so,

$omega=\dfrac{19.2 pi}{0.2}$

=96 pi rad/sec

=48 rpm