Question

A body of mass 4kg is being rotated with 120 rev per minute in a horizontal circular path of radius 2m its kinetic energy is

Answer 1

Frequency  = 200 rev/minute.

⇒ f = 200/60

∴ f = 10/3 s⁻¹.

∴ ω = 2πf

= 2 × 22/7 × 10/3

= 440/21 rads⁻¹.

Now,

v = ω/r

⇒ v = (440/21) ÷ 2

∴ v = 10.48 m/s.

∴ K.E. = 1/2 × mv²

⇒ K.E. = 1/2 × 4 × (10.48)²

∴ K.E. = 2 × 109.75

∴ K.E. = 219.5 J.

Hope it helps.

Answer 2

Answer:

You answer is............. 219.5