Phy 121
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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: the pendulum at equilibrium will be at the origin, when pulled back .1m to the left will put the mass in quadrant 2 with the x component negative
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: tan-1(2/.1) = -87deg + 180 = 93deg, x component = -.1m and the y component = 2m
What is the direction of the tension force exerted on the mass?
answer/question/discussion: the tension force is directed toward the direction of motion
the tension force in the direction of the string; the horizontal component (which you calculate correctly below as -.26 N) is toward the direction of motion
What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: x = 5 cos(93) = -.26, y = 5sin(93) = 4.99
What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: m = F/a = 5N / 9.8m/s^2 = .51kg, then weight = mg = .51kg * 9.8m/s^2 = 5N
What is its acceleration at this instant?
answer/question/discussion: a = v^2 / r = 1.4^2 / 2 = .98m/s^2
The pendulum is released from rest so its velocity is zero, as is its centripetal acceleration.
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25min
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&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#