Rotational Kinetic energy in a loop

In summary, the conversation discussed the problem of finding the minimum height h for a marble to successfully make it around a loop-the-loop without falling off. The total mechanical energy at the top of the track and at the top of the loop were calculated, and the sum of forces at the top of the loop was also considered. A few mistakes were made in the calculations, but the problem was eventually solved. The question about the source of the problem was also raised.
  • #1
Metalsonic75
29
0
The marble rolls down a track and around a loop-the-loop of radius R. The marble has mass m and radius r (see picture). What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?


This is driving me nuts. I know that the total mechanical energy at the top of the track (starting point) is mgh, and at the top of the loop, it is mgR + 1/2mv^2 + 1/2I[tex]\omega[/tex], so mgh = mgR + 1/2mv^2 + 1/2I[tex]\omega[/tex]. I then substituted 2/5mr^2 for I (general I of a sphere), and v/r for [tex]\omega[/tex]. Leaving that for a moment, I found that the sum of forces at the top of the loop are F = ma_c = m*(v^2/r) = mg - n. Since I'm trying to find the point where n=0, I substituted 0 for n, and solved for g = v^2/r. Then I substituted v^2/r for g in my earlier equation that I obtained for total mechanical energy. Simplifying everything gave me h = R + 0.7r, which is incorrect. I would really like to know where my error lies. The answer must include the variables r and R. Any help would be appreciated. Thank you.
 

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  • #2
Metalsonic75 said:
This is driving me nuts. I know that the total mechanical energy at the top of the track (starting point) is mgh, and at the top of the loop, it is mgR + 1/2mv^2 + 1/2I[tex]\omega[/tex],

Should be omega^2.

Leaving that for a moment, I found that the sum of forces at the top of the loop are F = ma_c = m*(v^2/r) = mg - n. Since I'm trying to find the point where n=0, I substituted 0 for n, and solved for g = v^2/r.

Should be R here, not r.
 
  • #3
Everything is done properly except those mistakes which Shooting mentioned.

And by the way,is this a sum from HC Verma?
 
  • #4
FedEx said:
Everything is done properly except those mistakes which Shooting mentioned.

And by the way,is this a sum from HC Verma?

Thanks for the help, Shooting star. I was able to solve the problem. And, sorry FedEx, I have no idea what you're talking about.
 
  • #5
Metalsonic75 said:
Thanks for the help, Shooting star. I was able to solve the problem. And, sorry FedEx, I have no idea what you're talking about.

I was just asking whether the problem was from a book named Concepts Of Physics by HC Verma.
 

Related to Rotational Kinetic energy in a loop

1. What is rotational kinetic energy in a loop?

Rotational kinetic energy in a loop refers to the energy of an object that is rotating around an axis while following a curved path, such as in a loop-the-loop roller coaster. It is a type of kinetic energy that is associated with an object's rotational motion.

2. How is rotational kinetic energy in a loop calculated?

The formula for calculating rotational kinetic energy in a loop is K = 1/2 * I * ω^2, where K is the kinetic energy, I is the moment of inertia, and ω is the angular velocity. In a loop, the moment of inertia will depend on the mass distribution of the object and the axis of rotation.

3. What factors affect the rotational kinetic energy in a loop?

The rotational kinetic energy in a loop is affected by various factors, including the mass and shape of the object, the speed of rotation, and the radius of the loop. Other factors that can impact the energy include air resistance, friction, and the angle of the loop.

4. How does rotational kinetic energy in a loop relate to potential energy?

In a loop-the-loop roller coaster or other similar situations, the rotational kinetic energy is converted into potential energy as the object moves up the loop. This means that at the top of the loop, the object has the highest potential energy and the lowest kinetic energy. As it moves down the loop, this energy is converted back into kinetic energy.

5. How is rotational kinetic energy in a loop important in real-life applications?

Understanding and calculating rotational kinetic energy in a loop is crucial in designing and analyzing various objects and structures, such as roller coasters, spinning rides, and even satellite orbits. It is also important in industries such as manufacturing, where precise measurements of rotational energy are needed for machines and equipment to function properly.

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