Minimum Height for Object to Travel Around Loop-the-Loop

[NickCherryJiggz]

Question: A small object of mass m slides without friction around a loop-the-loop aparatus. It starts from rest at point A, at height h above the bottom of the loop...What is the minimum value of h (in terms of R [the radius of the loop] such that the object moves around the loop without falling off at the loop's highest point, B.

The solution to this question is probably a very simple one...I'm familar with similar problems concerning potential/kinetic energy, but one point confuses me...I'm not sure what the requirement is for the block to not fall off.

 

[xanthym]

Question: A small object of mass m slides without friction around a loop-the-loop aparatus. It starts from rest at point A, at height h above the bottom of the loop...What is the minimum value of h (in terms of R [the radius of the loop] such that the object moves around the loop without falling off at the loop's highest point, B.

The solution to this question is probably a very simple one...I'm familar with similar problems concerning potential/kinetic energy, but one point confuses me...I'm not sure what the requirement is for the block to not fall off.

The object would not fall off if, at the highest point z=(2*R), the centripetal force required to remain in circular motion is greater than or equal to the object's weight. That is:
mv²/R = mg ::: (minimum requirement at highest point)
::: ⇒ v² = {g*R} ::: Eq #1
By conservation of energy, if the object begins from rest at height z=h, then at height z=(2*R):
m*g*h = m*g*(2*R) + (1/2)*m*v²
Using Eq #1 above for (v²), we get:
m*g*h = m*g*(2*R) + (1/2)*m*{g*R}
::: ⇒ h = (5/2)*R


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[thepatientmental]

The minimum value of h for the object to travel around the loop-the-loop without falling off at the highest point, B, can be determined by considering the conservation of energy. At point A, the object has only potential energy, given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the bottom of the loop. At point B, the object has both kinetic and potential energy, given by 1/2mv^2 and mgh, respectively.

For the object to remain on the loop at point B, the centripetal force (provided by the normal force of the loop) must be equal to or greater than the force of gravity pulling the object downwards. This can be expressed as:

mv^2/R ≥ mg

Where v is the velocity of the object at point B and R is the radius of the loop.

To determine the minimum value of h, we can equate the potential energy at point A with the sum of the kinetic and potential energy at point B:

mgh = 1/2mv^2 + mgh

Canceling out the mass and solving for h, we get:

h = R/2

Therefore, the minimum value of h in terms of R is R/2. This means that the object must start at a height of at least half the radius of the loop in order to complete the loop without falling off at point B.