Question

homework 5 begin date: 8/17/2025 11:00:00 pm due date: 9/30/2025 11:59:00 pm end date: 9/30/2025 11:59:00 pm 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 = submission history all date times are displayed in central standard time red submission date times indicate late work. date time answer hints feedback part (a) 1 8/17/2025 11:00:00 pm degrees radians log10() atan() cotan() sin() exp() acotan() asin() cos() fact() ln() acos() tan() e π √() + / ^ ( ) backspace 0 1 4 7 2 5 8 3 6 9 del clear end - home hint 1 hint 2 hint 3 feedback 1 feedback 2 feedback 3 i give up! grade summary deductions potential 0% 100% submissions attempts remaining: 3 (4% deduction per attempt) detailed view

Explanation:

Step1: Apply conservation of mechanical energy

The initial - state is at the extreme position with only potential energy \(U = mgh\), and the final - state is at the bottom with only kinetic energy \(K=\frac{1}{2}mv^{2}\). At the extreme position, the height \(h\) of the riders (assuming the center - of - mass motion) with respect to the bottom of the arc is \(h = r\) (the radius of the arc of motion). According to the conservation of mechanical energy \(U = K\), \(mgh=\frac{1}{2}mv^{2}\).

Step2: Solve for the speed \(v\)

Canceling out the mass \(m\) from both sides of the equation \(mgh=\frac{1}{2}mv^{2}\), we get \(gh=\frac{1}{2}v^{2}\). Given \(g = 9.8\ m/s^{2}\) and \(h=r = 16\ m\), we can solve for \(v\). Rearranging the equation \(v=\sqrt{2gh}\). Substituting \(g = 9.8\ m/s^{2}\) and \(h = 16\ m\) into the formula, we have \(v=\sqrt{2\times9.8\times16}\).
\[v=\sqrt{313.6}\approx17.7\ m/s\]

Answer:

\(17.7\)