A man swung an object of mass 2 kg in a circular path with a rope 1.2 m long. If the object was swung at 120 rev/min, find the tension in the rope.
- 400
- 288
- 240
-
379
The correct answer is: D
Explanation
m=2kg, r=1.2m, f=120rev/min ; T=?
f= \(\frac{120rev}{1min}\) Γ \(\frac{1min}{60s}\) = 2rev\s
w=2Οf= 2ΓΟΓ2= 4Οrad\s
T= \(\frac{mv^2}{r}\), but v = wr
β T = \(\frac{(m)(wr)^2}{r}\)
β T = \(mw^2r\)
= T = 2Γ (4Ο)^2 Γ 1.2
T = 379N ( to 3 s.f)
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