A 1.5 kg mass is moving in a circle of radius 1.1 m on a flat frictionless table
ID: 1457849 • Letter: A
Question
A 1.5 kg mass is moving in a circle of radius 1.1 m on a flat frictionless table at the end of a string. The speed of the mass is 2.2 m/s. The string routes through a hole in the center of the table and is held by you underneath the table.
What is the angular momentum of the mass?
What is the tension in the string?
If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string?
How much work did you do to reduce the radius by a factor of one-half?
Explanation / Answer
here,
mass , m = 1.5 kg
radius , r = 1.1 m
speed of mass , v = 2.2 m/s
the angular momentum , L = m * v * r
L = 1.5 * 2.2 * 1.1
L = 3.63 kg .m^2 /s
the angular momentum is 3.63 kg .m^2 /s
the tension in the string , T = m*v^2/r
T = 1.5 * 2.2^2 /1.1
T = 6.6 N
the tension in the string is 6.6 N
when radius , r' = 0.55 m
the tension in the string , T' = m*v^2/r'
T = 1.5 * 2.2^2 /0.55
T = 13.2 N
the tension in the string is 13.2 N
let the new angular be v'
using conservation of energy
L = m * r'*v'
3.63 = 1.5 * 0.55 * v'
v' = 4.4 m/s
the work done is change in kinetic energy , w = KEf - KEi
w = 0.5 * m *( 4.4^2 - 2.2^2)
w = 10.89 J
the workdone to reduce the radius is 10.89 J
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