Loop-the-Loop with rotational motion

In summary, the conversation discusses the calculation of the height above the top of a loop-the-loop that a small spherical ball must be released from in order to just make it around the loop. The conversation includes equations and formulas, and the final answer is determined to be 7/10 times the radius of the loop, due to a mistake in the initial calculation of the moment of inertia for the sphere.
  • #1
nahya
27
0
A small spherical ball of radius r = 1.5 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.6 m. The ball has mass M = 358 g.

How high above the top of the loop must it be released in order that the ball just makes it around the loop?
---
i just calculated the height above zero potential energy (ground) and subtract 2R from it after my calculations.
the initial potential energy, then, is p = mgH.
at the top of the loop, the potential energy is all converted to translational and rotational KE, so...
mgH = 1/2(mv^2 + Iw^2).
the minimum speed needed to barely complete the loop is Sqrt(gR), which, in this case, is ~5.05 m/s.
I = 2/5mR^2 and w = v/R, so Iw^2 = 2/5mv^2
gH = 1/2(v^2 + 2/5v^2) = 1/2v^2(1 + 2/5) = 0.7v^2
so i get H = 1.8216, which is less than the radius of the loop!
this can't be right...
what am i doing wrong?
the equation for the minimum vel is correct, right?
then i should be doing this right...
 
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  • #2
You equation is right, before you put numbers into your equations, use algebra. This gives a nice equation that cancels down to:
h = (19/10) R
But the delta 'h' is the change in height from where the sphere is released to the top of the loop. You don't need to subtract 2R.

Regards,
Sam
 
  • #3
hmm... h = 1.9 * 2.6 = 4.94 m above 2R, but it says that it's incorrect.
bleh TT
 
  • #4
At the top of the loop there will still be some potential energy.
[tex]E_p = 2mgR[/tex]
 
  • #5
nahya said:
hmm... h = 1.9 * 2.6 = 4.94 m above 2R, but it says that it's incorrect.
bleh TT

Sorry, I got the moment of inertia for a sphere wrong. It should be:
[tex]\Delta h=\frac{7}{10} R[/tex].

Is that what your answer should be. Sorry about the mistake.

Regards,
Sam
 
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  • #6
[tex]mg\Delta h= \frac{1}{2} (mv^2 + I \omega^2)[/tex]
[tex]v= \sqrt{gR} \ \ I = \frac{2}{5} mr^2[/tex]

These are the formulae I used
 
Last edited:

1. What is "Loop-the-Loop with rotational motion"?

Loop-the-Loop with rotational motion is a physical phenomenon where an object, such as a roller coaster or a car, moves in a circular loop while also rotating around an axis. This motion combines both circular motion and rotational motion.

2. What factors affect the success of a "Loop-the-Loop with rotational motion"?

The success of a Loop-the-Loop with rotational motion depends on several factors, including the speed of the object, the radius of the loop, and the mass of the object. The object must have enough speed to overcome the force of gravity and maintain the circular motion without falling off the track. The radius of the loop must also be large enough to prevent the object from experiencing too much centripetal force, which can cause it to veer off track. Additionally, the mass of the object affects its inertia and how easily it can maintain the rotational motion.

3. How is "Loop-the-Loop with rotational motion" different from regular circular motion?

In regular circular motion, an object moves in a circular path at a constant speed. However, in Loop-the-Loop with rotational motion, the object is also rotating around an axis, which adds a new dimension to its motion. This rotational motion requires additional forces, such as centripetal force and torque, to maintain the object's circular path and rotation simultaneously.

4. What is the importance of the centripetal force in "Loop-the-Loop with rotational motion"?

The centripetal force is crucial in Loop-the-Loop with rotational motion as it is responsible for keeping the object moving in a circular path. Without enough centripetal force, the object will not be able to maintain the loop and will fall off the track. The centripetal force is created by the track's curvature and is proportional to the object's mass, speed, and radius of the loop.

5. How does "Loop-the-Loop with rotational motion" relate to real-world applications?

"Loop-the-Loop with rotational motion" can be observed in various real-world applications, such as roller coasters, car stunts, and amusement park rides. These applications utilize the principles of physics, including centripetal force, to create a thrilling and safe experience for riders. Understanding the mechanics of Loop-the-Loop with rotational motion can also help engineers design and build better and more efficient structures, such as bridges and highways.

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