Answer :
Answer:
To solve this problem, we can use the following equation:
T = 2π√(r/g)
where T is the period (time for one complete revolution), r is the length of the rope, and g is the acceleration due to gravity.
First, we need to find the length of the rope. We can use trigonometry to do this:
sin(12°) = r/1.55m
r = 1.55m * sin(12°) = 0.319m
Next, we need to find g. We can use the value of 9.81 m/s^2 for the acceleration due to gravity.
Now we can plug in our values and solve for T:
T = 2π√(0.319m/9.81 m/s^2)
T ≈ 0.807 s
Therefore, the period of the tetherball's motion is approximately 0.807 seconds.
Final answer:
The question deals with the physics of a 2.0 kg tetherball in uniform circular motion, attached to a 1.5 m rope at a 30-degree angle from vertical and moving at 3.5 m/s. A free-body diagram is crucial for visualizing the forces at play, and a mathematical analysis determines the specifics of the motion.
Explanation:
To solve the problem, we need to determine different properties of the tetherball's motion. This may include finding the height of the pivot point above the ground, the horizontal circular path radius of the ball, the angular or linear velocity of the ball, or the tension in the rope. Since the specific question isn't provided, I'll explain how to find some of these properties given the information provided.
1. **Height of the pivot point above the ground:**
Let's call this height 'h'. To find 'h', we'll use the trigonometric relationship involving the sine of the angle, because we're dealing with the height (opposite side) and the hypotenuse (the length of the rope).
The sine of the angle (12.0°) is equal to the opposite side (height 'h') over the hypotenuse (the length of the rope, which is 1.55 m).
So we write:
sin(12.0°) = h / 1.55
To solve for 'h', we multiply both sides by 1.55:
h = 1.55 * sin(12.0°)
Now we calculate this using a calculator (making sure it is set to degrees):
h ≈ 1.55 * 0.207912 = 0.3223654 meters (rounded to 0.32 meters)
2. **Radius of the horizontal circular path:**
Let's call the radius 'r'. The radius corresponds to the adjacent side in the right triangle formed by the rope, the height of the pivot point, and the horizontal path of the ball. To find 'r', we'll use the cosine of the angle (12.0°):
cos(12.0°) = r / 1.55
Solving for 'r', we get:
r = 1.55 * cos(12.0°)
Using a calculator to find this value:
r ≈ 1.55 * 0.9781476 = 1.51602918 meters (rounded to 1.52 meters)
3. **Tension in the rope:**
To calculate the tension in the rope, we would need additional information about the mass of the ball and its velocity. The tension can be found using the centripetal force formula which is required to keep the ball moving in a circle:
Tension (T) = (Mass (m) x Velocity (v)^2) / Radius (r) + m * g * cos(θ)
Where 'm' is the mass of the ball, 'v' is the linear velocity of the ball, 'g' is the acceleration due to gravity (approx. 9.81 m/s^2), and θ is the angle of the rope with the horizontal.
4. **Linear or angular velocity of the ball:**
Again, to find the velocity of the ball, we would need additional information such as the frequency of rotation or the time for one revolution.
Once you have the specific detail of what needs to be calculated, you can expand on the appropriate formula and solve for it.
