Problem 200 — You need a good set of brakes to go the distance, as this month’s problem by Jerry Reckelhoff of Cincinnati demonstrates.

It was a sunny Sunday afternoon, and Finagel J. Wurme was enjoying an exhilarating ride in the country on his motorcycle. The picturesque lane he traveled at 60 mph had a downhill gradient of 5 deg. and crossed a babbling brook. Suddenly a sign, “Bridge Out 500 ft” jarred him out of his reverie. Wurme immediately hit the brakes, generating a constant braking force of 250 lb. If Wurme and motorcycle weigh a total of 750 lb, does Wurme end up in the brook? How many feet does it take him to come to a complete stop?

Send your answer to:

Fun With Fundamentals
POWER TRANSMISSION DESIGN
1100 Superior Ave.
Cleveland, OH 44114-2543

Deadline is January 10. Good luck!

Technical consultant, Jack Couillard, Menasha, Wis.

Solution to last month’s problem 199 — You are no Doubting Thomas if you answered 10 ft. Here’s how McGroundbound saw the light:

For the main-rotor blade to appear to be still, its tangential velocity must equal the helicopter’s linear velocity, but be in the opposite direction. Let:

v = Helicopter’s linear velocity, given as 150 knots or 253 fps
ω = Tangential velocity of main-rotor blade, radians/sec
r
= Radius of main-rotor blade that appears to be standing still, ft

From physics,

McGroundbound heard eight “slaps” per second from a two-bladed main rotor.

That means the rotor was turning at 4 rps. Plug these values into (1):

Thus the part of the rotor “standing still” is on the retreating blade at a point 10 ft from the rotor’s hub. Skylark helped himself to the dessert while McGroundbound pondered these mysteries of perception.