Question

A train has an acceleration of magnitude 0.90 m/s2 while stopping. A pendulum with a 0.55-kg bob is attached to a ceiling of one of the cars. Determine the angle that the pendulum makes with the vertical during the decceleration of the train

Answer 1

The angle of the pendulum with the vertical is $5.2^{\circ}$

Explanation:

As the train decelerates, the bob of the pendulum will feel a force given by

$F=ma$

where

m = 0.55 kg is the mass of the bob

$a=0.9 m/s^2$ is the magnitude of the acceleration

In the horizontal direction.

The pendulum will be inclined at an angle $\theta$ from the vertical, so it will be in equilibrium, and therefore the horizontal component of the tension in the string must be equal to the net force F of the previous equation:

$T sin \theta = ma$ (1)

where T is the tension in the string.

We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,

$T cos \theta = mg$ (2)

where $g=9.8 m/s^2$ is the acceleration of gravity.

Dividing eq.(1) by eq(2), we get:

$tan \theta = \frac{a}{g}$

And therefore, we find the angle:

$\theta=tan^{-1}(\frac{a}{g})=tan^{-1}(\frac{0.90}{9.8})=5.2^{\circ}$

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