Question

A majorette in the Rose Bowl Parade tosses a baton into the air with an initial angular velocity of 2.5 rev/s. If the baton undergoes a constant acceleration while airborne of 0.2 rev/s2 and its angular velocity is 0.8 rev/s when the majorette catches it, how many revolutions does it make in the air?

Answer 1

Answer:

14 rev

Explanation:

$w_{o}$ = initial angular velocity = 2.5 revs⁻¹

$w$ = final angular velocity = 0.8 revs⁻¹

$\alpha$ = Angular acceleration = - 0.2 revs⁻²

$\theta$ = Angular displacement

Using the equation

$w^{2} = w_{o}^{2} + 2 \alpha \theta\\0.8^{2} = 2.5^{2} + 2 (- 0.2) \theta\\ \theta = 14 rev$

So the number of revolutions are 14