Question

A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exerted by the seats on the passengers is equal to ¼ of their weight.

Answer 1

Answer:

$v=10m/sec$

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

$F_N+Fg$

where

$F_N =forceof the normal$

$Fg= force due to gravity$

Generally the net force is given to be $FC(force towards center)$

$F_C =F_N + Fg$

$F_N =F_C -Fg$

$F_N=F_C-F_g$

$F_N=\frac{mv^2}{R} -mg$

Mathematical we can now derive V

$m_g + \frac{8m}{4}= \frac{mv^2}{8}$

$\frac{5mg}{4} =\frac{mv^2}{8}$

$v^2 =\frac{40*10}{4}$

$v=10m/sec$

Therefore the speed of the roller coaster is given ton be $v=10m/sec$

Answer 2

$v=10m/s$

Given:

Radius of vertical r= 8m

To find:

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

$F_N+F_g$

where

$F_N=\text{Force of Normal}$ and $F_g=\text{Force due to gravity}$  

Generally the net force is given to be $FC- \text{Force towards centre}$

$F_C=F_N+F_g\\\\F_N=F_C+F_g\\\\F_N=\frac{mv^2}{r} -mg$

Mathematically we can now derive V

$mg-\frac{8m}{4} =\frac{mv^2}{8}\\\\\frac{5mg}{4}=\frac{mv^2}{8}\\\\v^2=\frac{40*10}{4} \\\\v=10m/s$

Therefore, the speed of the roller coaster is given to be 10m/s.

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