Question

As the object rotates through a small angle Δθ in the short time Δt, the point P moves a distance Δs = r Δθ along the arc of its circular path. Thereforev =

Δs

Δt

=

rΔθ

Δt

.This becomes v = rω, whereω =

dθ

dt

.Therefore, the tangential speed of the point P is

v = rω = (80 m)(1 rad/s) =  m/s.

FINALIZE
Use the simulation to examine the relationship between angular rate of rotation and linear speed.

How does an increase or decrease in tangential speed translate into a change in angular speed?

What if the distance from the axis is multiplied by a factor of 1.5 with the angular rate of rotation ω kept the same? The small distance Δs would correspondingly be multiplied by a factor of  , and therefore the tangential speed v would be multiplied by a factor of  .

Answer 1

The equation relating the angular rate of rotation to the linear speed (tangential speed) is given by:

so, tangential speed at P will be: v = 80(1) = 80 m/s

If the tangential speed is increased, the angular speed at the same distance from the center (constant r) will increase and if the tangential speed decreases, the angular speed decreases for the same r.

Let the distance from the axis be increased by a factor of 1.5. Thus, r' = 1.5r

so, the small distance travelled along the circumference will increase by a factor of 1.5 since

so, in the given time , the tangential speed v will be:

therefore, the tangential speed v will be multiplied by a factor of 1.5.