A 3.00 kg ball is on the end of a 3.72 m rope. What is the tension (kN) in the rope if the ball is moving in a horizontal circular motion of 2.00 rotations every second.
To find the tension in the rope, we need to consider the forces acting on the ball.
In circular motion, there is a centripetal force that keeps the object moving in a curved path. This force is directed towards the center of the circular motion and is equal to the mass of the object multiplied by its acceleration towards the center.
The acceleration can be obtained using the formula:
a = (v^2) / r
Where:
- a is the acceleration
- v is the velocity (in this case, rotations per second)
- r is the radius of the circular motion (in this case, the length of the rope)
First, let's determine the velocity of the ball.
Since the ball completes 2.00 rotations every second, we can find the angular velocity (ω) using the formula:
ω = 2πf
Where:
- ω is the angular velocity
- f is the frequency (number of rotations per second)
ω = 2π * 2.00 = 4π rad/s
Next, we can find the velocity of the ball using the formula:
v = ωr
Where:
- v is the linear velocity
- ω is the angular velocity
- r is the radius
v = (4π rad/s) * (3.72 m) = 14.78 m/s
Now that we have the velocity, we can calculate the acceleration:
a = (v^2) / r
= (14.78 m/s)^2 / 3.72 m
= 58.70 m/s^2
Since the mass of the ball is given as 3.00 kg, we can find the centripetal force using the formula:
F = ma
Where:
- F is the force
- m is the mass
- a is the acceleration
F = (3.00 kg) * (58.70 m/s^2)
= 176.1 N
Finally, to convert the force from Newtons to kilonewtons, we divide by 1000:
176.1 N / 1000 = 0.1761 kN
Therefore, the tension in the rope is approximately 0.1761 kN.