A mass m = 88 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.5 m and finally a flat straight section at the same height as the center of the loop (15.5 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.)
3) 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?
6) 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?
3) Well, if the mass has just enough speed to make it around the loop without leaving the track, its speed at the bottom of the loop will be... exhilaratingly fast! Hang on tight, because this ride is about to get wild!
6) Ah, those bumbling engineers! It seems they made quite the blunder. To make up for their mistake and ensure the mass doesn't go flying off the track, they'll need to push it with a speed that is fast enough to counteract gravity and centripetal force at the top of the loop. How fast? Well, let's just say they'll need a turbo boost button and maybe even a rocket engine. Strap in tight, folks! It's going to be one crazy ride!
To solve these problems, we can use the principles of conservation of energy and circular motion.
3) The speed of the mass at the bottom of the loop can be determined using the conservation of mechanical energy. The total mechanical energy of the mass at the top of the loop should be equal to the total mechanical energy at the bottom of the loop. At the top of the loop, the only energy the mass has is potential energy, given by mgh, where h is the height of the loop. At the bottom of the loop, the mass has both potential energy and kinetic energy. The potential energy is mgh, and the kinetic energy is (1/2)mv^2, where v is the speed of the mass at the bottom of the loop.
Since energy is conserved, we can set the potential energy at the top equal to the sum of potential and kinetic energy at the bottom:
mgh = mgh + (1/2)mv^2
Simplifying the equation:
0 = mgh - mgh - (1/2)mv^2
0 = (1/2)mv^2
As a result, the speed of the mass at the bottom of the loop will be zero. This means the mass will momentarily come to a stop at the bottom of the loop.
6) To find the initial velocity necessary for the mass to complete the loop, we need to consider two points: the height of the loop and the top of the loop.
First, we can calculate the speed required at the top of the loop using the conservation of mechanical energy. At the top, the mass has both potential and kinetic energy. The potential energy is mgh, and the kinetic energy is (1/2)mv^2. The total mechanical energy should be equal to the sum of potential and kinetic energy at the top.
mgh = mgh + (1/2)mv^2
Simplifying the equation:
0 = mgh - mgh - (1/2)mv^2
0 = (1/2)mv^2
As a result, the speed at the top of the loop should be zero.
Next, we need to consider the height at the start of the ride. We know that the mass will not make it around the loop if released from the height of the top of the loop. Therefore, we need to find the velocity at the top of the loop that will give the mass enough kinetic energy to complete the loop.
Using the equation for gravitational potential energy, mgh, we can solve for the velocity:
mgh = (1/2)mv^2
Simplifying the equation:
2gh = v^2
v = √(2gh)
The initial velocity required to get the mass around the loop is given by √(2gh), where h is the height from which the mass is released just at the beginning (equal to the top of the loop height).
To solve these questions, we will use the principle of conservation of energy, which states that the total energy of a system remains constant if no external forces are acting on it. We will assume that there is no friction or other external forces acting on the mass.
Let's start with question 3.
3) When the mass is at the bottom of the loop, it will have both kinetic energy and potential energy. The maximum potential energy is at the top of the loop, where the height is 15.5 m, and the potential energy at the bottom of the loop is zero. Therefore, the entire potential energy is converted into kinetic energy.
The potential energy at the top of the loop is given by the formula:
PE = mgh
Where m is the mass (88 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the top of the loop (15.5 m).
Therefore, the potential energy at the top of the loop is:
PE = (88 kg) x (9.8 m/s^2) x (15.5 m) = 129,424 J
Since all the potential energy is converted into kinetic energy at the bottom of the loop, the kinetic energy at the bottom of the loop is also 129,424 J.
The equation for kinetic energy is:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass, and v is the velocity.
To find the velocity at the bottom of the loop, we can rearrange the equation:
v = sqrt(2KE/m)
Substituting the values, we get:
v = sqrt(2 * 129424 J / 88 kg) ≈ 16.4 m/s
So, the speed of the mass at the bottom of the loop is approximately 16.4 m/s.
Now, let's move on to question 6.
6) In this case, we need to find the initial velocity the mass must have at the beginning of the ride in order to make it around the loop without falling off the track.
At the top of the loop, the mass will have maximum potential energy, which is given by:
PE = mgh
Where m is the mass (88 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the top of the loop (15.5 m).
The potential energy at the top of the loop is:
PE = (88 kg) x (9.8 m/s^2) x (15.5 m) = 129,424 J
To make it around the loop, the mass must have enough kinetic energy and enough speed to counteract gravity and maintain contact with the track.
At the bottom of the loop, all the potential energy is converted into kinetic energy. Therefore, the kinetic energy at the bottom of the loop must be equal to the potential energy at the top of the loop.
Using the formula for kinetic energy:
KE = (1/2)mv^2
We can rearrange the equation to find the velocity:
v = sqrt(2KE/m)
Substituting the potential energy at the top of the loop (129,424 J) as the kinetic energy at the bottom of the loop, we get:
v = sqrt(2 * 129424 J / 88 kg) ≈ 16.4 m/s
So, the mass needs an initial velocity of approximately 16.4 m/s at the beginning to make it around the loop without falling off the track.