Question

The Round Up carnival ride below has a radius of 3.62 meters and rotates 0.537 times per second. As shown, riders can be held up by only friction. What coefficient of friction is needed to keep the riders from sliding down? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

The free body diagram for the problem is shown below:

If the people don't slide down this means that the friction has to be equal to the Weight, then we have:

`\begin{gathered} F_f-W=0 \\ F_f=W \\ \mu F_n=W \\ \mu=(W)/(F_n) \end{gathered}`

Now, from newton's second law we have that:

`F_n=ma_c`

but

`a_c=(4\pi^2r)/(T^2)`

then:

`F_n=(4\pi^2mr)/(T^2)`

And then we have:

`\begin{gathered} \mu=(mg)/((4\pi^2mr)/(T^2)) \\ \mu=(gT^2)/(4\pi^2r) \end{gathered}`

Plugging the values given we have:

`\begin{gathered} \mu=((9.8)((1)/(0.537))^2)/(4\pi^2(3.62)) \\ \mu=0.238 \end{gathered}`

Therefore the coefficient of friction is 0.238