Question

In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from the axis of rotation. The cables swing out at a constant angle of 45.0° when the ride is operating. What is the angular speed of rotation?

Answer 1

Answer:

The angular speed of rotation is 1.34 rad/s.

Explanation:

Given that,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We need to calculate the angular speed of rotation

Using balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Put the value of T

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Put the value into the formula

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Hence, The angular speed of rotation is 1.34 rad/s.