The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of $0.80 \, \text{m}$ above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is $0.35 \, \text{m}$. The mass of the marble is $0.050 \, \text{kg}$. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.

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The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of $0.80 \, \text{m}$ above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is $0.35 \, \text{m}$. The mass of the marble is $0.050 \, \text{kg}$. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.(a) Draw arrow on the diagram or explain the direction of the resultant force acting on the marble.
, (b) State the names of all the forces acting on the marble.
, (c) Deduce that the speed of the marble is $3.0 \, \text{m/s}$.
, (d) What would happen to the the speed in part (c) if the release height of the marble were to double. Justify your answer.
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Jason · Accepted Answer

The resultant force would point down (towards the center of the circle).
Weight (a.k.a. force of gravity) and normal force.
Using the law of conservation of energy, the potential energy at point A transforms into potential and kinetic energy at point C. So it can be deduced that:
$ mgh_A = mgh_C + \frac{1}{2}mv^2 $
This would give a value of $ 3 \, \text{m/s} $.
Using the same equation as above, we see that the velocity squared and the change in height are directly proportional. Therefore, if we double the initial height of $ 0.8 \, \text{m} $ to $ 1.6 \, \text{m} $, the velocity would have to increase proportionally by $ 1.67 $ times. So the original $ 3 \, \text{m/s} $ velocity would increase by $ 1.67 $ times to approximately $ 5 \, \text{m/s} $.

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