Question

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 a flat 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 around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track?

What height above the ground must the mass begin to make it around the loop-the-loop?

If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (15.4 m off the ground)?

Now a spring with spring constant k = 1.8 × 10⁴ N/m is used on the final flat surface to stop the mass. How far does the spring compress?

It turns out the engineers designing the loop-the-loop didn’t really know physics – when they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning.

How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track?

The work done by the normal force on the mass (during the initial fall) is:

Answer 1

m = 78 kg

Radius of loop-the-lopp, R = 15.4 m

Let the minimum speed at top of loop be v

Then for it to stay in track,

=> v = 12.29 m/s

So minimum speed required at top = 12.29 m/s

Let the height of release be h. Then a drop of (h-2R) = (h-30.8) should produce a speed 12.29 m/s

By conservation of energy

0.5*m*12.29² = m*9.81*(h-30.8)

=> h = 38.5 m

So height above the ground the massmust begin to make it around the loop-the-loop = 38.5 m

Let the speed at bottom of loop be v

By conservation of energy between top and bottom

mgh = 0.5mv²

=> 9.81*38.5 = 0.5*v²

=> v = 27.48 m/s

So speed be at the bottom of the loop = 27.48 m/s

Let the speed at final track be v. Its height difference with point of release = 38.5 - 15 = 23.5m

By conservation of energy

mgh = 0.5mv²

=> 9.81*23.5 = 0.5*v²

=> v = 21.47 m/s

speed at the final flat level = 21.47 m/s

Spring constant,k = 18000 N/m

Let compression be x

By conservation of energy

0.5*m*v² = 0.5*k*x²

=>x =

So compression of spring = 1.413 m

Since the initial height is same as the highest point of loop the loop, the initial speed that needs to be imparted is equal to the speed at the highest point of loop the loop(conservation of energy) = 12.29 m/s

So initial velocity to be given to mass = 12.29 m/s

Work done by normal force = 0 J (since its direction is perpendicular to direction of motion)