A 1 kg mass attached to a 1 m string is rolled around in a horizontal circular path with an angular velocity of 100 RPM what is the tension of the string?
Please show how you got your answer.
To find the tension in the string, we can apply Newton's second law of motion in a circular path. This law states that the net force acting on an object is equal to the mass of the object multiplied by its centripetal acceleration. In this case, the tension in the string is the centripetal force.
The centripetal acceleration can be calculated using the formula:
a = ω^2 * r
Where "a" is the centripetal acceleration, "ω" is the angular velocity in radians per second, and "r" is the radius of the circular path.
First, let's convert the angular velocity from RPM (revolutions per minute) to radians per second. We know that 1 revolution is equal to 2π radians, and 1 minute is equal to 60 seconds.
So, angular velocity (ω) = (100 RPM) * (2π radians/1 revolution) * (1 minute/60 seconds)
= (100 * 2π) / 60 radians per second
= (π/3) radians per second
Now, we can substitute the given values into the formula for centripetal acceleration:
a = (π/3)^2 * 1
= π^2 / 9 m/s^2
Next, applying Newton's second law:
F = m * a
Where "F" is the force/tension in the string, "m" is the mass of the object, and "a" is the centripetal acceleration.
Given that the mass of the object (m) is 1 kg, we can calculate the tension (F):
F = 1 kg * (π^2 / 9 m/s^2)
≈ 1.103 N
Therefore, the tension in the string is approximately 1.103 Newtons.