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Wednesday, 24 August 2011

Physics GRE - #23

A particle is initially at rest at the top of a curved frictionless track.

The x and y coordinates of the track are related in dimensionless units by $y=\frac{x^2}{4}$, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what is its tangential acceleration?
  1. $0$
  2. $g$
  3. $\frac{gx}{2}$
  4. $\frac{gx}{\sqrt{x^2+4}}$
  5. $\frac{gx^2}{\sqrt{x^2+16}}$

Solution :

Choice 4 is the solution.

First we note that choice 2 cannot be the answer as there is a normal force from the track.

Now as $x\rightarrow\infty$, we should expect that the acceleration goes to $g$ as the tangent becomes nearly vertical. This makes choice 4 is the only possible answer.

We can also solve this question elegantly without resorting to limits and boundary conditions (however it will take a good half page). Can you see how it can be done?


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