1. For the experiment done in lab, recall that applying F x = ma x to the ball g
ID: 1355166 • Letter: 1
Question
1. For the experiment done in lab, recall that applying Fx = max to the ball gives the equation Fstring = m42f2L. Combine this result with the result of applying Fy = may and derive an equation for sin in terms of f and L. is the angle between the string and the horizontal.
A)For L = 50.0 cm and f = 1.20 rev/s, what is the angle ? (Use g = 9.80 m/s2.)
2. A small object of mass m = 0.200 kg moves with constant speed in a horizontal circle on the end of a light string that has length l = 0.600 m, as shown in the sketch. The string makes a constant angle of = 30.0° with the vertical.
A) What is the tension ont the string?
B) What is the magnitude of the acceleration of the mass as it moves in its circular path?
(c) What is the period of the motion, the time its takes the ball to complete one revolution?
Explanation / Answer
2)
mass(m) = 0.200 kg
length(l) = 0.600m
angle(theta) = 30 degrees
A)
a*cos theta = g
a = g /cos theta
= 9.8 / cos30
= 11.316 m/s^2
Tension on the string = m*a
= 0.200*11.316
= 2.263 N
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B)
Acceleration = g * tan theta
= 9.8 * tan 30
= 5.658 m/s^2
================================================
C)
radius = L*sin theta
= 0.600*sin 30
= 0.3 m
Period of the motion = 2 * pi / sqrt[g*tan theta / r]
= 2*3.14 / sqrt(9.8*tan 30 / 0.3)
= 6.28 / 4.343
= 1.45 sec
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