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.

Answer Details

To find the tension in the rope, we can first use the formula for centripetal force, which is given by:
F_centripetal = (m * v^2) / r
where: - F_centripetal is the centripetal force - m is the mass of the object - v is the velocity of the object - r is the radius of the circular path
In this case, the mass of the object (m) is given as 2 kg and the radius (r) is given as 1.2 m.
Now, to find the velocity (v), we need to convert the given value of 120 rev/min to m/s.
Here's how we can do that:
1. First, convert the revolutions per minute (rev/min) to revolutions per second (rev/s) by dividing by 60 (since there are 60 seconds in a minute):
120 rev/min = 120/60 rev/s = 2 rev/s
2. Next, we need to convert the revolutions per second to the linear velocity in meters per second (m/s). To do this, we need to find the circumference of the circular path.
The circumference of a circle is given by the formula:
C = 2πr where r is the radius of the circular path.
Substituting the value of the radius (r = 1.2 m) into the formula, we have:
C = 2π * 1.2 = 2.4π Now, to find the linear velocity (v), we can multiply the circumference (C) by the number of revolutions per second (2 rev/s):
v = C * rev/s = 2.4π * 2 = 4.8π m/s
Now that we have the values of m (2 kg) and v (4.8π m/s), we can substitute them into the centripetal force formula to find the tension in the rope:
F_centripetal = (m * v^2) / r = (2 * (4.8π)^2) / 1.2
Simplifying further:
F_centripetal = (2 * 23.04π^2) / 1.2
F_centripetal = 38.4π^2
Finally, to get a numerical value for the tension in the rope, we can approximate the value of π to 3.14 and calculate the centripetal force:
F_centripetal ≈ 38.4 * 3.14^2 ≈ 379 N
Therefore, the tension in the rope is approximately 379 N.
Therefore, the correct answer is 379.