Phy 231
Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A simple pendulum has length 2 meters. It is pulled back 10 cm from its equilibrium position and released. The tension in the string is 5 Newtons.
Sketch the system with the pendulum mass at the origin and the x axis horizontal.
answer/question/discussion: We have pendulum making an angle with the y-axis of arctan (2 / .1) = 0.29 degrees as it is held .1 m from equilibrium position with a 2 m length. The tension along this pendulum's string is shown to be 5 N.
arctan(2/.1) is about 2.9 degrees, not .29 degrees
Sketch a vector representing the direction of the pendulum string at this instant. As measured from a horizontal x axis, what is the direction of this vector? (Hint: The y component of this vector is practically the same as the length; you are given distance of the pullback in the x direction. So you know the x and y components of the vector.)
answer/question/discussion: With a direction tangent to the string at this instant we find the direction to be downwards at angle equal that of the angle found away from the y-axis; therefore, this direction would be along the degree of .29 pointing towards the equilibrium position.
What is the direction of the tension force exerted on the mass?
answer/question/discussion: The tension is exerted upwards on the mass in the direction of 180 - 90 - .29 = 89.71 degrees, which is how we find the remaining angle of the right triangle formed by moving the pendulum away from equilibrium position.
What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: with 5 N tension along the string a tension on the mass of 5 N * sin .29 = 0.127 N as the x-component, and 5 N * cos .29 = 4.9999 as the y-component.
At 2.9 deg from the vertical I believe the x component of the tension would be about .25 N.
What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: With 4.9999 N as its weight and 4.9999 N / (g * cos .29) = 0.51 kg as the mass.
What is its acceleration at this instant?
answer/question/discussion:The acceleration is as I calculated above is very nearly that of normal Earth gravity, and only a small percentage is exerted in the x-direction.
The net force on the pendulum is the x component of the tension, which is about .25 N, so its acceleration is
a = F_net / m = .25 N / (.51 kg) = .49 m/s^2, approx..
Note that this is about 1/20 the acceleration of gravity, and that the 10 cm x displacement is about 1/20 of the 2 meter length.
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25 minutes
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Right procedures; I think you had an order-of-magnitude error in calculating the angle.