# Fun With Fundamentals: Problem 214

April 1, 2000
There are a number of ways to be all wet, as this month's problem by Steve Wyatt of Beloit, Wis., demonstrates

Carnival of venue
PROBLEM 214
-- There are a number of ways to be all wet, as this month's problem by Steve Wyatt of Beloit, Wis., demonstrates.

It was a cloudy day, and deep within the recesses of the Sneeterville County Courthouse, a heated trial was underway.

"Your honor, I would like to point out that my client suffered severe emotional trauma and humiliation as a result of the negligence of the Fun Unlimited Carnival," intoned Rufus Filibuster, attorney at law, "not to mention damage to a \$1,000 business suit.

"Ladies and gentlemen of the jury, the facts of the case are these: My client, Barnaby McSnuff, stepped aboard the Whirl-Lee Bird ride expecting to have an enjoyable experience. Due to the carnival management's inane layout of the ride, the chair in which he sat passed through a nearby artificial waterfall, thus soaking my client to the skin, destroying his apparel, and humiliating him before his colleagues. We therefore are seeking no less than \$1,000,000 in damages."

"Why that's nonsense!" roared carnival owner Buford Buff. "He admitted he fell into a pond."

"Order in the court" thundered Judge Dockett as he banged his gavel.

The Whirl-Lee Bird ride consists of one-person seats attached with cables to the top of a revolving vertical pole. The waterfall is 30 ft away from the center of the pole. The ride turns at 10 rpm, and the cable is 40 ft long. Mc- Snuff and the chair together weigh 150 lb. For simplicity's sake, assume the rider and chair occupy a space equivalent to a 4-ft-diameter sphere and that the cable is weightless.

Compute the radius of the chair from the pole while the ride is in motion to see if McSnuff will get his settlement.

SOLUTION TO LAST MONTH'S PROBLEM 213 -- You put the proper spin on it if you answered 2.11 rps.

Here's the solution.

Since the initial energy imparted to the globes is constant, the kinetic energy of rotation is the same for both globes:

where

ϖ1 = Angular velocity of first 80- kg globe, given as 4π rad/sec

ϖ2 = Angular velocity of second 72-kg globe, rad/sec

I1 = Moment of inertia of first globe, kg-m2

I2 = Moment of inertia of second globe, kg-m2

The moment of inertia for a solid sphere is defined as:

where: m = Mass of sphere, kg r = Radius of sphere, m

Equation (1) reduces to:

ϖ12m1 = ϖ22m2 (2)

From the problem, we know that m1 is 80 kg, and m2 is 72 kg. Plug these values into (2) to get

Macrae’s artistic temperament frequently drove him into budget problems.

### Pumps Push the Boundaries of Low Temperature Technology

June 14, 2024
As an integral part of cryotechnology, KNF pumps facilitate scientific advances in cryostats, allowing them to push temperature boundaries and approach absolute zero.

### The entire spectrum of drive technology

June 5, 2024
Read exciting stories about all aspects of maxon drive technology in our magazine.

### MONITORING RELAYS — TYPES AND APPLICATIONS

May 15, 2024
Production equipment is expensive and needs to be protected against input abnormalities such as voltage, current, frequency, and phase to stay online and in operation for the ...

### Solenoid Valve Mechanics: Understanding Force Balance Equations

May 13, 2024
When evaluating a solenoid valve for a particular application, it is important to ensure that the valve can both remain in state and transition between its de-energized and fully...