Asked by missy
A child ties a rock to a string and whirls it around in a horizontal circle.
Assuming a 1.50 m -long string making 20.0 ∘ angle below the horizontal, find the speed of the rock.
Find the period of its uniform circular motion.
Assuming a 1.50 m -long string making 20.0 ∘ angle below the horizontal, find the speed of the rock.
Find the period of its uniform circular motion.
Answers
Answered by
bobpursley
ok, use the 20 degrees.
tan20=vertical force/horizontal force
horizontal force=vertical force/tan20
mv^2/r=mg*ctn20
solve for v.
tan20=vertical force/horizontal force
horizontal force=vertical force/tan20
mv^2/r=mg*ctn20
solve for v.
Answered by
Damon
Tension = T
T sin 20 = weight = m g
so
T = m g/sin 20 = 28.7 * m if g = 9.81
m v^2/R = T cos 20 = 28.7 m (.94)
so
v^2/R = 27
v^2 = 27 R
R = 1.5 cos 20 = 1.41 meter
so
v^2 = 27(1.41)
v = 6.17 m/s
circumference = 2 pi R = 8.86 meters
time around = 8.86 meters/6.17 m/s
= 1.44 seconds
T sin 20 = weight = m g
so
T = m g/sin 20 = 28.7 * m if g = 9.81
m v^2/R = T cos 20 = 28.7 m (.94)
so
v^2/R = 27
v^2 = 27 R
R = 1.5 cos 20 = 1.41 meter
so
v^2 = 27(1.41)
v = 6.17 m/s
circumference = 2 pi R = 8.86 meters
time around = 8.86 meters/6.17 m/s
= 1.44 seconds
Answered by
missy
The answers given was v=6.36m/s and t=1.39s.
Any insight where I calculated wrong. I got basically the same as Damon but not rounded until the final steps.
Any insight where I calculated wrong. I got basically the same as Damon but not rounded until the final steps.
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