Question

Ethan and Rebecca are riding on a merry-go-round. Ethan rides on a horse at the outer rim of the circlar platform, twice as far from the center of the circular platform as Rebecca, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, describe Ethans tangential speed?

Answer 1

Answer:

Ethan's tangential speed is twice as Rebecca's.

Explanation:

Let $r$ be the distance from the center of the platform to Rebecca's horse, $v_R$ Rebecca's tangential speed, $v_E$ the Ethan's tangential speed and $\omega$ the merry-go-round angular speed. The distance from the center of the platform to Ethan's horse is $2r$. The angular speed is related to the tangential speed by the equation:

$\omega =\frac{v}{R}$

Since the angular speed is the same for Ethan and Rebecca, we have that:

$\frac{v_E}{2r}=\frac{v_R}{r}$

Now, solving for $v_E$ we get:

$v_E=2v_R$

It means that Ethan's tangential speed is twice as Rebecca's.