Question

problem 10: (8% of assignment value) riders in an amusement park ride shaped like a viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. at each end of the swing the ship hangs motionless for a moment before the ship swings down under the influence of gravity. assume that this motionless point occurs when the bar connecting the pivot point and the ship is horizontal. randomized variables r = 16 m m = 55 kg part (a) assuming negligible friction, find the speed of the riders at the bottom of its arc, given the systems center of mass travels in an arc having a radius of 16 m and the riders are near the center of mass in m/s. vf = 17.70 correct! part (b) what is the centripetal acceleration at the bottom of the arc in m/s²? ac = 19.58 correct! part (c) how many times her weight is the normal force exerted by the ride on a 55 kg rider? f_n/w =

Explanation:

Step1: Identify the forces at the bottom of the arc

At the bottom of the arc, the net - force acting on the rider is given by $F_{net}=F_N - mg$, where $F_N$ is the normal force and $mg$ is the gravitational force. The net - force provides the centripetal force $F_c$, so $F_N - mg=ma_c$, and $F_N=m(a_c + g)$.

Step2: Recall the centripetal - acceleration formula

The centripetal acceleration formula is $a_c=\frac{v^{2}}{r}$. We are given $v = 17.70\ m/s$, $r = 16\ m$, and $m = 55\ kg$, $g=9.8\ m/s^{2}$. First, calculate the centripetal acceleration $a_c=\frac{v^{2}}{r}=\frac{(17.70)^{2}}{16}$.
\[a_c=\frac{313.29}{16}=19.58\ m/s^{2}\]

Step3: Calculate the normal force

Substitute $a_c = 19.58\ m/s^{2}$, $m = 55\ kg$, and $g = 9.8\ m/s^{2}$ into the formula $F_N=m(a_c + g)$.
\[F_N=55\times(19.58 + 9.8)\]
\[F_N=55\times29.38\]
\[F_N = 1615.9\ N\]
The weight of the rider is $W=mg=55\times9.8 = 539\ N$.
To find how many times the weight the normal force is, calculate $\frac{F_N}{W}=\frac{1615.9}{539}\approx3.0$.

Answer:

3.0