Pendulum question

[arc5005]

The question <- click me


C) C

This question can be solved using physical intuition. As the pendulum is displaced by a greater angle it has more PE, so it will have greater KE when the PE is converted to KE at its lowest point. The mass is the same, so this means that the velocity at its lowest point is greater. This eliminates D. As the frequency increases, it cycles through its period faster, so it must move faster. This means that f cannot be in the denominator and choice B is eliminated. Choice A is eliminated, because it yields inccorect units.

Should you want to solve this problem exactly use conservation of energy. When the pendulum is displaced by an initial angle θ from the vertical, it also gains PE that we can write as mgh. Just before the pendulum strikes the nail, it is at its lowest vertical position; the PE has been converted to KE:

mgh = 1/2mv2, which leads to v = √(2gh)

However, we do not know what h is. In order to solve for h, we need to do a little trigonometry. From the following picture we can solve for h: Picture.

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I'm confused by the 1-cos theta.

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[AdaptPrep]

I'm confused by the 1-cos theta.

Hi, arc5005--
Do you understand where they are getting cos theta = (L-h)/L? If yes, do the following.

Multiply both sides by L:
L cos theta = L - h

add h to both sides:
h + L cos theta = L

subtract L cos theta from both sides:
h = L - L cos theta

now, factor out L from the right hand side:
h = L (1 - cos theta)

I hope that helps.

 

[BerkReviewTeach]

Excellent question (although that may seem self-serving as it is Question #21 on page 173 of BR Physics Book 1--for reference if anyone wishes to play along).

Before looking at the math needed to solve this question, let's consider the purpose of this question. Why did we put this in our book? Do we, or the MCAT test writers, expect you to do the math necessary to get an exact answer in about one minute? Definitely not. This is typical of their apparently hard questions that can be made easy with the right reasoning.

Let's start with v. It can be found from either kinematics or energetics. Because we are dealing pendulums, it is likely dealing with energetic and not kinematics. We know that the initial potential energy depends on height and mass and g. We know that the kinetic energy at the bottom depends on mass and velocity. Upon equating 1/2mv^2 to mgh, we see that m drops out and that there is still a g and h to consider. By this little bit, we can eliminate choices A, B, and D, resulting in just choice C still standing. That is a very good path to the best answer.

We could also consider units. Velocity is in m/s. Choice A has units of nothing/s, so it is out. Choice B has units of meters · nothing/meters per s^2 · per s, which is not m/s, so choice B is out. Choice C has units of square root of (m/s^2 · m) = m/s, giving choice C the correct units. Choice D has units of square root of (kg · m/s^2 · m) which cannot be m/s being that it has a kg term, so choice D is out. Only choice C passes the test.

Lastly, there is visualization (physical intuition as it is called here). From life experience, we know that lifting the bob to a greater height will result in a greater speed at the bottom. If the pendulum is longer, that it must be raised a greater amount to get the same displacement angle, so a greater h correlates with a greater L. The equation must contain L, eliminating choice A. We also know that if gravity went up, it would speed up, eliminating choice B. Lastly, as with free fall, a heavier object will fall the same as a lighter object, so there should be no m term. This eliminates choice D. Only choice C remains standing.

This question can be solved three different ways more quickly than traditional physics approaches. That is a major goal of our materials: to present questions that can be solved by different routes.