An aerialist on a high platform holds on to a trapeze attached to a support by an 8.0-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle that the trapeze cord makes with the vertical at this instant.
The correct answer and explanation is :
To solve this, we need to calculate the angle $\theta$ that the trapeze cord makes with the vertical when the aerialist releases the trapeze, which occurs when the trapeze is 0.75 m lower than its initial height. We’ll approach this using energy conservation and basic trigonometry.
Solution Overview:
Let:
- $L = 8.0 \, \text{m}$ be the length of the trapeze cord.
- $h = 0.75 \, \text{m}$ be the vertical height difference between the starting point and the release point.
We need to find the angle $\theta$ with respect to the vertical.
Step-by-step Solution:
Since the aerialist swings along a circular arc (radius = cord length), we can use the geometry of a circle.
Let’s define:
- Initial height $h_0 = 0$
- Final height $h = -0.75 \, \text{m}$
Let $y$ be the vertical distance from the lowest point of the swing. At the top (initial position), $y_1 = L$. At the release point, $y_2 = L – 0.75$.
So:
$$
y_2 = 8.0 – 0.75 = 7.25 \, \text{m}
$$
Now apply trigonometry using the definition of cosine in a right triangle:
$$
\cos(\theta) = \frac{\text{adjacent (vertical height)}}{\text{hypotenuse (cord length)}} = \frac{7.25}{8.0}
$$
$$
\cos(\theta) = 0.90625
$$
$$
\theta = \cos^{-1}(0.90625)
$$
$$
\theta \approx 24.6^\circ
$$
Final Answer:
$$
\boxed{\theta \approx 24.6^\circ}
$$
Explanation (300+ words):
This problem involves the conservation of mechanical energy and the geometric relationship between a swinging pendulum and its arc of motion. The aerialist begins on a platform holding a trapeze, which is attached to a support by an 8.0-meter cord. As he swings down and then up to the other side, he eventually releases the trapeze when it is 0.75 meters lower than the starting height. Since there is no friction or air resistance, the only forces doing work are conservative forces (gravity), meaning mechanical energy is conserved.
The motion of the trapeze can be modeled as a pendulum, where the aerialist swings in a circular arc with a radius equal to the length of the cord. The key to solving this problem lies in finding the vertical position of the trapeze at the point of release, then using trigonometric relationships in a right triangle to find the angle.
We know the vertical height from the pivot to the trapeze at release is $8.0 – 0.75 = 7.25 \, \text{m}$. In a right triangle formed by the cord, the vertical line, and the swing’s deviation from vertical, the cosine of the angle is the ratio of the adjacent side (vertical leg = 7.25 m) to the hypotenuse (cord length = 8.0 m). Taking the arccosine gives us the angle between the cord and the vertical, which calculates to approximately $24.6^\circ$.
This approach combines understanding of energy conservation (to validate height differences) with basic geometry and trigonometric identities, which are crucial in analyzing pendulum-like systems in physics.