You’ve taken your neighbor’s young child to the carnival to ride the rides. She wants to ride The Rocket. Eight rocket-shaped cars hang by chains from the outside edge of a large steel disk. A vertical axle through the center of the ride turns the disk, causing the cars to revolve in a circle. You’ve just finished taking physics, so you decide to figure out the speed of the cars while you wait. You estimate that the disk is $5m$ in diameter and the chains are $6m$ long. The ride takes $10s$ to reach full speed, then the cars swing out until the chains are $20°$ from vertical. What is the cars’ speed?

The ride takes $10s$to reach full speed, then the cars swing out until the chains are $20°$ from vertical.

The disk rotates and makes the chair move a distance $d\sin \theta$from the disk. So, the chair rotates in an orbital radius

Where $L$is the length of wire. and $r_{\text{disk}}$is the radius of the disk. Plug the values for $r_{\text{disk}},\theta$, and $L$to get $r$

Now, we analyze the situation by draw a free body diagram for the problem. In the horizontal axis the centripetal force which is given by equation (8.6) acts on the chair in opposite direction to the horizontal component of tension force $T\sin \theta$, so the net force in the$x$-direction is

$F_x=T\sin \theta -F_c$

$F_x=T\sin \theta -\frac{m{v}^2}{r}$

$0=T\sin \theta -\frac{m{v}^2}{r}$

$T\sin \theta =\frac{m{v}^2}{r}$

In the vertical axis, the chair is under two forces $T\cos \theta$and $mg$. so the net force in the $y$-direction on $m$is

$F_y=T\cos \theta -mg$

Our target is to find the velocity $v$, so we can divide equation (1) by (2) to find an expression for $v$

$\frac{T\sin \theta }{T\cos \theta }=\frac{m{v}^2/r}{mg}$

$\tan \theta =\frac{v^2}{gr}$

$v=\sqrt{rg\tan \theta }$

Now, plug the values for $r,g$and $\theta$into equation (3) to get $v$

$v=\sqrt{rg\tan \theta }$

$=\sqrt{(4.54\mathrm{m})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\tan 20°}$

$=4m/s$

The car's speed is$4m/s$.