#$&*
Phy 121
Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_26.1_labelMessages.txt **
A simple pendulum has length 2 meters. It is pulled back 10 cm from its equilibrium position and released. The tension in the string is 5 Newtons.
• Sketch the system with the pendulum mass at the origin and the x axis horizontal.
answer/question/discussion: ->->->->->->->->->->->-> :
We have pendulum making an angle with the y-axis of arctan (2 / .1) = 2.9 degrees as it is held .1 m from equilibrium position with a 2 m length. The tension along this pendulum's string is shown to be 5 N.
#$&*
• 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 2.9 pointing towards the equilibrium position.
#$&*
• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension is exerted upwards on the mass in the direction of 180 - 90 - 2.9 = 87.1 degrees, which is how we find the remaining angle of the right triangle formed by moving the pendulum away from equilibrium position.
#$&*
• 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 2.9 = 1.196 N as the x-component, and 5 N * cos 2.9 = -4.855 as the y-component.
#$&*
@&
sin(2.9 degrees) is about .05, so 5 N * sin(2.9 deg) is about 0.25 N.
*@
@&
Considering your later result, in which you use 0.25 N as the horizontal force, I suspect this was a typo.
*@
• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
With 4.855 N as its weight and 4.855 N / (g * cos 2.9) = .51 kg as the mass.
#$&*
• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
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
#$&*
*#&!
&#Good work. See my notes and let me know if you have questions. &#