A mass m = 73 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 17.9 m and finally a flat straight section at the same height as the center of the loop (17.9 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 (17.9 m off the ground)?
Now a spring with spring constant k = 15500 N/m is used on the final flat surface to stop the mass. How far does the spring compress?
To find the minimum speed the block must have at the top of the loop to make it around without leaving the track, we can use the principle of conservation of energy. At the top of the loop, the block has potential energy and kinetic energy. At the bottom of the loop, all of the potential energy is converted into kinetic energy.
1. Find the height from which the mass needs to be released to make it around the loop-the-loop:
At the top of the loop, the kinetic energy is at its minimum (zero) since the speed is zero. The total mechanical energy (sum of potential energy and kinetic energy) of the block must be equal at both the top and bottom of the loop. At the top, the only form of energy is potential energy (mgh), where m is the mass, g is the acceleration due to gravity, and h is the height.
mgh = 0.5mv²
Solve for h to find the height above the ground from which the mass should be released.
2. Determine the minimum speed at the top of the loop:
To find the minimum speed, we need to use the concept of circular motion and centripetal force. At the top of the loop, the gravitational force acts downward, while the normal force (which provides the centripetal force) acts upward. The net force is the difference between these two forces.
At this point, the normal force is equal to zero because it is at the top of the loop. The force equation is:
mg - N = mv²/R
Solve for v to find the minimum speed at the top of the loop.
3. Determine the speed at the bottom of the loop:
Using conservation of energy, we can set the potential energy at the top equal to the kinetic energy at the bottom.
mgh = 0.5mv²
Solve for v to find the speed at the bottom of the loop.
4. Determine the speed at the final flat level:
Since the height at the final flat level is the same as the center of the loop, the speed will remain the same as at the bottom of the loop due to conservation of energy.
5. Determine how far the spring compresses:
To find the distance the spring compresses, we can use the concept of work done by the spring. The work done by the spring is equal to the change in kinetic energy of the mass. We can set this work equal to the potential energy of the spring when compressed.
0.5kx² = 0.5mv²
Solve for x to find the distance the spring compresses, where k is the spring constant and x is the distance the spring compresses.
You can plug in the given values to calculate the answers to the specific questions.