Why do plane mirrors and convex mirrors form only virtual images?
2 answers:
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Answer:
Train accaleration = 0.70 m/s^2
Explanation:
We have a pendulum (presumably simple in nature) in an accelerating train. As the train accelerates, the pendulum is going move in the opposite direction due to inertia. The force which causes this movement has the same accaleration as that of the train. This is the basis for the problem.
Start by setting up a free body diagram of all the forces in play: The gravitational force on the pendulum (mg), the force caused by the pendulum's inertial resistance to the train(F_i), and the resulting force of tension caused by the other two forces (F_r).
Next, set up your sum of forces equations/relationships. Note that the sum of vertical forces (y-direction) balance out and equal 0. While the horizontal forces add up to the total mass of the pendulum times it's accaleration; which, again, equals the train's accaleration.
After doing this, I would isolate the resulting force in the sum of vertical forces, substitute it into the horizontal force equation, and solve for the acceleration. The problem should reduce to show that the acceleration is proportional to the gravity times the tangent of the angle it makes.
I've attached my work, comment with any questions.
Side note: If you take this end result and solve for the angle, you'll see that no matter how fast the train accelerates, the pendulum will never reach a full 90°!
Answer:
<u>The pendulum bob swing past the mean position because:</u>
When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is and the the acceleration decreases as
towards the mean position.
The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.