A mass m = 78 kg slides on a frictionless track that has a drop,followed by a loop-the-loop with radius R = 15.4 m and finally aflat straight section at the same height as the center of the loop(15.4 m off the ground). Since the mass would not make it aroundthe loop if released from the height of the top of the loop (do youknow why?) it must be released above the top of the loop-the-loopheight. (Assume the mass never leaves the smooth track at any pointon its path.)
What is the minimum speed the block must have at the top of theloop to make it around the loop-the-loop without leaving thetrack?
What height above the ground must the mass begin to make itaround the loop-the-loop?
If the mass has just enough speed to make it around the loopwithout leaving the track, what will its speed be at the bottom ofthe loop?
If the mass has just enough speed to make it around the loopwithout leaving the track, what is its speed at the final flatlevel (15.4 m off the ground)?
Now a spring with spring constant k = 1.8 × 104 N/mis used on the final flat surface to stop the mass. How far doesthe spring compress?
It turns out the engineers designing the loop-the-loop didn’treally know physics – when they made the ride, the first drop wasonly as high as the top of the loop-the-loop. To account for themistake, they decided to give the mass an initial velocity right atthe beginning.
How fast do they need to push the mass at the beginning (now ata height equal to the top of the loop-the-loop) to get the massaround the loop-the-loop without falling off the track?
The work done by the normal force on the mass (during theinitial fall) is:
