Phy 201
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.
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• 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.)
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• What is the direction of the tension force exerted on the mass?
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The direction of the tension on the mass would be upward and to the left at about 1.7 degrees off of vertical.
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• What therefore are the horizontal and vertical components of the tension?
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(horizontal) x component = T * cos (90 deg + 1.7 deg)
(vertical) y component = T * sin (90 deg + 1.7 deg)
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• What therefore is the weight of the pendulum, and what it its mass?
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The weight of the pendulum is that of the acceleration of gravity.
The mass of the pendulum is (Ft + Fg) / a.
Right ideas, but not quite put together correctly.
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• What is its acceleration at this instant?
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When the pendulum is being pulled back and tension is held, the system is not moving, so acceleration is 0.
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The pendulum has been released.
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30 minutes
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&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#