JAMB - Physics (2023 - No. 17)
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
Explanation
m=2kg, r=1.2m, f=120rev/min ; T=?
f= \(\frac{120rev}{1min}\) × \(\frac{1min}{60s}\) = 2rev\s
w=2πf= 2×π×2= 4πrad\s
T= \(\frac{mv^2}{r}\), but v = wr
⇒ T = \(\frac{(m)(wr)^2}{r}\)
⇒ T = \(mw^2r\)
= T = 2× (4π)^2 × 1.2
T = 379N ( to 3 s.f)
f= \(\frac{120rev}{1min}\) × \(\frac{1min}{60s}\) = 2rev\s
w=2πf= 2×π×2= 4πrad\s
T= \(\frac{mv^2}{r}\), but v = wr
⇒ T = \(\frac{(m)(wr)^2}{r}\)
⇒ T = \(mw^2r\)
= T = 2× (4π)^2 × 1.2
T = 379N ( to 3 s.f)
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