PHYSICS| SPEED 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. A. 400 B. 288 C. 240 D. 379

Given angular speed = 120 rev/min
We know that 1 revolution = 2π radians and 1 minute = 60 seconds.

Angular speed (ω) = 120 (rev)/(min) × (2π rad)/(1 rev) × (1 min)/(60 s)
ω = (120 × 2π)/(60) rad/s
ω = 2 × 2π rad/s
ω = 4π rad/s

The tension in the rope provides the centripetal force (F_c).
The formula for centripetal force using angular speed is:
F_c = mω²r

Where:
m = mass of the object = 2 kg
ω = angular speed = 4π rad/s
r = radius of the circular path (length of the rope) = 1.2 m

Tension (T) = (2 kg) × (4π rad/s)² × (1.2 m)
T = 2 × (16π²) × 1.2 N
T = 32π² × 1.2 N
T = 38.4π² N

Using the approximate value of π ≈ 3.14159:
π² ≈ (3.14159)² ≈ 9.8696
T = 38.4 × 9.8696 N
T ≈ 378.99 N

Rounding to the nearest whole number, the tension is approximately 379 N.

Comparing this to the given options:
A. 400
B. 288
C. 240
D. 379

The calculated tension matches option D.

The tension in the rope is 379 N.

D. 379