#$&*
Phy 231
Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A simple pendulum has length 2 meters. It is pulled back 10 cm
from its equilibrium position and released. The tension in the
string is 5 Newtons.
Sketch the system with the pendulum mass at the origin and the x
axis horizontal.
answer/question/discussion: ->->->->->->->->->->->-> :
#$&*
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: ->->->->->->->->->->->-> :
From the origin up and to the left to the point (-10 cm, 200 cm).
angle = tan^-1(200 cm / (-10 cm) ) = 2.9 deg + 90 deg = 92.9 deg
#$&*
What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
Along the string at 92.2 deg.
#$&*
What therefore are the horizontal and vertical components of the
tension?
answer/question/discussion: ->->->->->->->->->->->-> :
T_x = 5 N * cos(92.9 deg) = -0.25 N = 0.25 N
T_y = 5 N * sin(92.9 deg) = 4.99 N = 50 N
#$&*
What therefore is the weight of the pendulum, and what it its
mass?
answer/question/discussion: ->->->->->->->->->->->-> :
F = m * a
m = weight / g = 5 N / (9.8 m/s^2) = 0.51 kg
#$&*
What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
a = F_net / m = 0.25 N / (0.51 kg) = 0.49 m/s^2
#$&*
*#&!
@& Good work.*@