Question

a special electronic sensor is embedded in the seat of a car that takes riders around a circular loop - the - loop ride at an amusement park. the sensor measures the magnitude of the normal force that the seat exerts on a rider. the loop - the - loop ride is in the vertical plane and its radius is 23 m. sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 840 n. at the top of the loop, the rider is upside down and moving, and the sensor reads 390 n. what is the speed of the rider at the top of the loop?

Explanation:

Step1: Find the mass of the rider

Before the ride starts, the normal force $N_0$ equals the weight of the rider $mg$. Given $N_0 = 840$ N, and $N_0=mg$, so $m=\frac{N_0}{g}$. Taking $g = 9.8$ m/s², $m=\frac{840}{9.8}\text{ kg}$.

Step2: Analyze forces at the top of the loop

At the top of the loop, the net - force towards the center of the circle provides the centripetal force. The forces acting on the rider are the normal force $N$ and the gravitational force $mg$. The centripetal - force formula is $F_c=\frac{mv^{2}}{r}$, and $F_c = N + mg$. We know $N = 390$ N, $r = 23$ m, and $m=\frac{840}{9.8}$ kg. Substituting into $F_c=\frac{mv^{2}}{r}=N + mg$, we get $v^{2}=\frac{r(N + mg)}{m}$.

Step3: Calculate the speed

First, substitute $m=\frac{840}{9.8}$ kg, $N = 390$ N, $r = 23$ m, and $g = 9.8$ m/s² into $v^{2}=\frac{r(N + mg)}{m}$.

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Then $v=\sqrt{\frac{23\times1230\times9.8}{840}}\approx18.1$ m/s.

Answer:

$v\approx18.1$ m/s