Find the Minimum Height Needed to Complete a Loop-the-Loop Track

• Punchlinegirl
In summary, to roll a solid marble without slipping along the loop-the-loop track, it must be released from rest at a height above the bottom of the track. The radius of the loop-the-loop is R=1.30 m, so the marble must be released with a velocity of at least .5m/R to ensure it doesn't leave the track at the top of the loop.
Punchlinegirl
A solid marble of mass m=35 and radius r=7 cm will roll without slipping along the loop-the-loop track if it is released from rest somewhere on the straight section of the track. From what minimum height h above the bottom of the track must the marble be released to ensure that it doesn't leave the track at the top of the loop? The radius of the loop-the loop is R=1.30 m.

I don't know where to begin. Any advice?

there needs to be enough centripetal force on the top of the loop to even out the weight so it the mass doesn't fall

$$\frac{mv^2}{r}=mg$$

I don't know the velocity or acceleration

you find that by

$$mgh=.5mv^2$$

find v, plug that into the first equation and solve for h

Urban, your 2nd equation will not work because you have neglected the rotational kinetic energy of the marble.

Punchline, use Urban's first equation to determine the minimum velocity that the marble needs at the top of the loop. Then apply conservation of energy to find minimum h. You will have to use
$$\Delta PE = .5mv^2 + .5I\omega^2$$

I don't think they have learned rotational energy yet, so I did not include it, but yes you are correct

Skomatth said:
Urban, your 2nd equation will not work because you have neglected the rotational kinetic energy of the marble.

Punchline, use Urban's first equation to determine the minimum velocity that the marble needs at the top of the loop. Then apply conservation of energy to find minimum h. You will have to use
$$\Delta PE = .5mv^2 + .5I\omega^2$$
(Note: From Skomatth's comments, above equation applies at all points along marble's path referenced to the initial Potential Energy. See summary of solution hints below.)

SOLUTION HINTS (including summary of previous hints):
{TOTAL Energy Required by Marble at Loop's Top} =
= {Linear Kinetic Energy @ Top} + {Rotational Kinetic Energy @ Top} + {Potential Energy @ Top} =
= (1/2)*m*v2 + (1/2)*I*ω2 + m*g*(2*R) =
= (1/2)*m*v2 + (1/2)*{(2/5)*m*r2}*(v/r)2 + 2*m*g*R =
= (1/2)*m*v2 + (2/10)*m*v2 + 2*m*g*R =
= m*{(7/10)*v2 + 2*g*R}

The above TOTAL Energy must be obtained from the Potential Energy at an initial height "H" above Loop's bottom given by:
{Initial Required Potential Energy} = m*g*H* =
= m*{(7/10)*v2 + 2*g*R}
::: ⇒ H = (7/10)*v2/g + (2*R) ::: (Eq #1)

The velocity "v" is then determined from requiring centripetal force at loop's top to equal the marble's weight "m*g":
m*v2/R = m*g
::: ⇒ v = sqrt{g*R} ::: (Eq #2)

Place Eq #2 into Eq #1 and determine "H" to complete solution.

~~

Last edited:
All I wrote was that the change in potential energy was equal to the kinetic energy at some point in the marble's path. Indeed there is no term for the marble's potential energy at the top of the loop because the LHS side of the equation accounts for it. Were I to write the LHS in terms of m,g,h and R it would be identical to your equation, though I did not because I'm hesitant to work almost half the problem for punchline.

1. What is a loop-the-loop track?

A loop-the-loop track is a type of roller coaster track where the train travels in a complete loop, with the track forming a vertical circle.

2. What is the purpose of finding the minimum height needed for a loop-the-loop track?

The minimum height needed for a loop-the-loop track is important because it ensures the safety of the passengers on the roller coaster. If the track is not high enough, the train may not have enough speed to complete the loop and could potentially get stuck or derail.

3. How is the minimum height for a loop-the-loop track calculated?

The minimum height for a loop-the-loop track is calculated using the formula h = R(1 + cosθ), where h is the height of the loop, R is the radius of the loop, and θ is the angle of the track at the top of the loop. This formula takes into account the forces of gravity and centripetal force to determine the minimum height needed for the train to complete the loop without falling off the track.

4. Are there any other factors that affect the minimum height needed for a loop-the-loop track?

Yes, there are other factors that can affect the minimum height needed for a loop-the-loop track, such as the weight of the train and the speed at which it enters the loop. These factors can vary depending on the type of roller coaster and can impact the overall safety and experience of the ride.

5. What happens if the minimum height for a loop-the-loop track is not met?

If the minimum height for a loop-the-loop track is not met, the train may not have enough speed to complete the loop and could potentially get stuck or derail. In some cases, the ride may have to be modified or the loop may need to be removed in order to ensure the safety of the passengers.

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