Asked by Eric G
a 10 kg mass is attached to one end of a string with a lenght of 5m and whirled in a horizontal circle overheadd at a rate of 1 revolution per second what is tension in the sring
Answers
Answered by
Damon
It is impossible of course to make that circle exactly horizontal since tan theta = g/Ac
so we must assume that g is small or Ac is much much bigger than g. Let's see:
Ac = v^2/R = w^2 r
where w is omega, radians/second
omega = 2 pi radians/second
so
Ac = (2 pi)^2 (5) = 197 m/s^2
yes, that is much bigger than g so circle is nearly horizontal
F = m Ac = 10*197
= 1970 Newtons in the string
so we must assume that g is small or Ac is much much bigger than g. Let's see:
Ac = v^2/R = w^2 r
where w is omega, radians/second
omega = 2 pi radians/second
so
Ac = (2 pi)^2 (5) = 197 m/s^2
yes, that is much bigger than g so circle is nearly horizontal
F = m Ac = 10*197
= 1970 Newtons in the string
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