Phy201
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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:
After sketching the system my included angle was about 2.86 deg from the pendulum hanging straight down which would represent the 92.86 deg. Is this anywhere close?
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:
Vx = .1 cos 92.86
Vy = 2 sin 92.86
What is the direction of the tension force exerted on the mass?
answer/question/discussion:
The tension is directed away from the mass, opposite of the gravity
What therefore are the horizontal and vertical components of the tension?
The tension force must be directed along the string or thread. It is not possible for the end of a straight thread to exert a force perpendicular to its direction.
answer/question/discussion:
Vertical comp. is T sin 92.86
Horizontal comp. isT cos 92.86
What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion:
T sin 92.86 = 5 N, T = 5 N/sin 92.86 = 5.006 N
F = m * a, 5.006N = m * 9.8 m/s^2, m = 5.006N/9.8m/s^2 = .51 kg
w = m * g, w = .51kg * 9.8 m/s^2 = 5.006 kg
Question? If mass times gravity is the weight of an object I think, then how does the units evolve? If what I computed is correct which it probably isnt, the units should be Newtons for the weight. I know that the metric weight should be kg.
The kg is a measure of mass, not weight.
Pounds measure weight, and in common usage pounds and kilograms are treated as if they measure the same thing. It is not at all uncommon to hear people speak of kg of weight. However this is not correct, and the confusion often leads to errors when when it extends into physics.
In F = m * a the mass has MKS units of kg, a has units of m/s^2. These are the fundamental units of the quantities, so the fundamental unit of F must be the product of these units. That is, F has units of kg * m/s^2.
This unit is given its own name, the Newton.
Your calculation should be w = .51kg * 9.8 m/s^2 = 5.006 kg * m/s^2 = 5.006 Newtons, not 5.006 kg.
What is its acceleration at this instant?
answer/question/discussion:
m = .51 kg, F = 5.002, F = m * a, a = 5.002N / .51kg = 9.81 m/s^2
5.002N / .51kg = 5.002 kg m/s^2 / (.51 kg) = 9.81 m/s^2.
It's important to do the units calculation, which reinforces the conceptual meaning of these quantities and helps avoid errors and procedure.
However the net force on the mass is not the 5.002 N. This is the tension in the string. Gravity also pulls downward, and provides a force equal and opposite to the vertical component of the tension.
Once the pendulum is released, however, there is no force to resist the horizontal component of the tension, and the pendulum accelerates in response to the resulting horizontal net force.
What therefore is the net force, and what is the resulting acceleration?
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about 30 minutes
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This is similar to the problems I was having with the Asst 25 start q a
You've got most of it, and shouldn't have any problem putting it all together.
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