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The Correct Answer and Explanation is:
It looks like your question contains fragmented and jumbled text, making it difficult to understand the problem clearly. However, from what I can interpret, it seems to be related to a physics problem involving circular motion, particularly a ball attached to a rope and swinging in a circle, possibly in a conical pendulum setup.
To help you accurately, I’ll reconstruct a likely version of the problem based on the words and phrases you provided (e.g., “tension,” “vertical direction,” “angle,” “component of force,” “radius,” “period,” etc.).
🧮 Reconstructed Problem Statement:
A ball is tied to a rope and swings in a horizontal circle, forming a conical pendulum. The rope makes an angle θ with the vertical. The tension in the rope is 30 N.
(a) What is the gravitational force on the ball?
(b) What is the vertical component of the tension?
(c) What is the horizontal component of the tension?
(d) What angle does the rope make with the vertical?
(e) What is the radius of the circular path?
(f) What is the speed of the ball?
(g) What is the period of the motion?
Let’s assume the mass of the ball is m = 2 kg and the angle θ = 30° (since you mentioned “30”).
✅ Correct Answers:
(a) Gravitational force =
Fg=mg=2×9.8=19.6 NF_g = mg = 2 \times 9.8 = 19.6 \, \text{N}
(b) Vertical component of tension =
Ty=Tcosθ=30cos(30∘)≈25.98 NT_y = T \cos\theta = 30 \cos(30^\circ) \approx 25.98 \, \text{N}
(c) Horizontal component of tension =
Tx=Tsinθ=30sin(30∘)=15 NT_x = T \sin\theta = 30 \sin(30^\circ) = 15 \, \text{N}
(d) Angle with vertical = 30° (given)
(e) Radius of the circle =
Let rope length be L = 1.5 m (not given, assuming).
r=Lsinθ=1.5sin(30∘)=0.75 mr = L \sin\theta = 1.5 \sin(30^\circ) = 0.75 \, \text{m}
(f) Speed of the ball:
Horizontal tension provides centripetal force:
Tsinθ=mv2rT \sin\theta = \frac{mv^2}{r}
15=2v20.75⇒v2=15×0.752=5.625⇒v≈2.37 m/s15 = \frac{2v^2}{0.75} \Rightarrow v^2 = \frac{15 \times 0.75}{2} = 5.625 \Rightarrow v \approx 2.37 \, \text{m/s}
(g) Time period:
T=2πrv=2π×0.752.37≈1.99 secondsT = \frac{2\pi r}{v} = \frac{2\pi \times 0.75}{2.37} \approx 1.99 \, \text{seconds}
📘 300-Word Explanation:
This problem describes a conical pendulum, where a ball tied to a rope swings in a horizontal circle while the rope traces out a cone shape. In this setup, the forces acting on the ball are:
- Gravitational force (downward),
- Tension in the rope (angled upward),
- Centripetal force, provided by the horizontal component of tension, keeping the ball moving in a circle.
To find the gravitational force, we multiply mass by gravity:
Fg=mgF_g = mg. This helps us balance the vertical forces, since the vertical component of the tension must exactly cancel the ball’s weight (assuming uniform circular motion and no vertical motion).
Next, using trigonometry, we split the tension force into vertical and horizontal components:
- Ty=TcosθT_y = T \cos\theta balances gravity.
- Tx=TsinθT_x = T \sin\theta provides the centripetal force.
This horizontal component allows us to use the formula for centripetal force: Tx=mv2rT_x = \frac{mv^2}{r}
Solving for speed involves rearranging this and substituting known values. The radius of the circular path depends on the rope’s length and the angle: r=Lsinθr = L \sin\theta.
Finally, the period of the motion (the time for one complete circle) is calculated using the circumference of the circle divided by speed: T=2πrvT = \frac{2\pi r}{v}
This type of problem illustrates key concepts in uniform circular motion, vector resolution, and force balancing in 2D motion. It’s commonly found in introductory physics and helps build understanding of how forces interact to produce circular trajectorie