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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Assignment 26 Seed Question 1
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: My diagram looks like a straight vertical line hanging from a horizontal line, that forms a perpendicular relationship. The pendulum then has an arrow emanating from the right side, pulling up the pendulum. The horizontal displacement of this line is 10 cm.
• 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: If the pullback is in the x direction, then the vector direction should be positive, before the pendulum is released. This is because we know it is moving in a positive x direction, and if the pendulum is being pulled up to the right, then it is moving in toward the positive y direction as well before it is let go. (However, we know that the y direction itself is still negative force, since we find such a calculation for the y component below).
• What is the direction of the tension force exerted on the mass?
answer/question/discussion: We know that the direction of the tension has to be opposite the pull, because this is one of the forces (along with gravity) that will pull down the pendulum. Thus, if the direction we specified above is positive, then the force of tension must be negative (it has to be opposite that direction).
• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: The horizontal component of tension equals Fx = - m g cos (theta) = - 0.51 kg * 9.8 m/s^2 * cos (92.86 degrees) = 0.249 N. We found the mass by setting up F = ma so that m = F/a = 5 N / 9.8 m/s^2 = 0.51 kg. In addition, the degree angle was found by tan ^-1 (10/200) = 2.86 degrees, which was added to 90 degrees to get 92.86 degrees. The y component was found by – m g * sin (theta) = - 0.51 kg * 9,8 m/s^2 * sin (92.86 degrees) = -4.992 N.
• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: The weight of the pendulum is dependent on the mass, which we found to be F = ma so that m = F/a = 5 N / 9.8 m/s^2 = 0.51 kg. The weight then is 0.51 kg * 9.8 m/s^2 = 5lbs.
• What is its acceleration at this instant?
answer/question/discussion: The acceleration at this instant is that of gravity, 9.8 m/s^2, since this is the force that the earth exerts to give the mass its weight on our planet.
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This took me about 30 minutes to do.
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You never did say how you got the .51 kg from the given information. That is the correct mass, to 2 significant figures, but it's not clear that you got it by the correct process.
However everything else holds together and you appear to understand the situation well.
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.
** **
Assignment 26 Seed Question 1
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: My diagram looks like a straight vertical line hanging from a horizontal line, that forms a perpendicular relationship. The pendulum then has an arrow emanating from the right side, pulling up the pendulum. The horizontal displacement of this line is 10 cm.
• 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: If the pullback is in the x direction, then the vector direction should be positive, before the pendulum is released. This is because we know it is moving in a positive x direction, and if the pendulum is being pulled up to the right, then it is moving in toward the positive y direction as well before it is let go. (However, we know that the y direction itself is still negative force, since we find such a calculation for the y component below).
• What is the direction of the tension force exerted on the mass?
answer/question/discussion: We know that the direction of the tension has to be opposite the pull, because this is one of the forces (along with gravity) that will pull down the pendulum. Thus, if the direction we specified above is positive, then the force of tension must be negative (it has to be opposite that direction).
• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: The horizontal component of tension equals Fx = - m g cos (theta) = - 0.51 kg * 9.8 m/s^2 * cos (92.86 degrees) = 0.249 N. We found the mass by setting up F = ma so that m = F/a = 5 N / 9.8 m/s^2 = 0.51 kg. In addition, the degree angle was found by tan ^-1 (10/200) = 2.86 degrees, which was added to 90 degrees to get 92.86 degrees. The y component was found by – m g * sin (theta) = - 0.51 kg * 9,8 m/s^2 * sin (92.86 degrees) = -4.992 N.
• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: The weight of the pendulum is dependent on the mass, which we found to be F = ma so that m = F/a = 5 N / 9.8 m/s^2 = 0.51 kg. The weight then is 0.51 kg * 9.8 m/s^2 = 5lbs.
• What is its acceleration at this instant?
answer/question/discussion: The acceleration at this instant is that of gravity, 9.8 m/s^2, since this is the force that the earth exerts to give the mass its weight on our planet.
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This took me about 30 minutes to do.
** **
You never did say how you got the .51 kg from the given information. That is the correct mass, to 2 significant figures, but it's not clear that you got it by the correct process.
However everything else holds together and you appear to understand the situation well.