Question

A 792 kg rollercoaster car is on theinside of a loop of radius 18.7 m.At the top of the loop, the car movesat 17.3 m/s. What is the normalforce on the car at that point?(Unit = N)Enter

Answer 1

mass = 792 kg

r = 18.7 m

v = 17.3 m/s

Centripetal force is a force in a circle with a center. Generally, the normal force plus the weight of the car(mg) equals the net force(centripetal force). Therefore,

`\begin{gathered} f=\frac{mv^2}{r^{}} \\ f=(792*17.3)/(18.7) \\ f=(237037.68)/(18.7) \\ f_c=12675.8117647N \end{gathered}`

To find the normal force

`\begin{gathered} mg+\text{normal force=732.71} \\ 792*9.8+normal\text{ force=732}.71 \\ 7761.6+normal\text{ force=}12675.8117647 \\ \text{normal force=}4914.21176471N \end{gathered}`