Let's go through this problem step by step, using the concepts of circular motion and forces to find the answers. We are given a conical pendulum, which consists of a pebble attached to a string that swings in a horizontal circle. We'll break down the problem into parts:
Given Data
- Mass of the pebble: 53 g = 0.053 kg,
- Length of the string: 1.4 m,
- Radius of the horizontal circle: 25 cm = 0.25 m,
- Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \).
Part (a): Speed of the Pebble
In a conical pendulum, the pebble moves in a horizontal circular path, and the forces acting on it include the tension in the string and gravity. The tension provides the centripetal force required for circular motion. Let's start by analyzing the forces acting on the pebble:
- The vertical component of the tension \( T \) balances the weight of the pebble: \( T \cos \theta = mg \),
- The horizontal component of the tension provides the centripetal force: \( T \sin \theta = \frac{mv^2}{r} \), where \( v \) is the speed of the pebble and \( r \) is the radius of the circle.
We can find the angle \( \theta \) using trigonometry. The length of the string and the radius form a right triangle, so we can use:
tan \( \theta \) = \( \frac{r}{h} \)
where $h$ is the vertical distance from the pivot to the pebble. We can find $h$ by recognizing that the string forms a right triangle with the vertical component of the string. Thus:
h = \sqrt{L^2 - r^2} = \sqrt{(1.4)^2 - (0.25)^2} = \sqrt{1.96 - 0.0625} = \sqrt{1.8975} ≈ 1.378 \, \text{m}
Now, we can calculate $\theta$:
tan \theta = \( \frac{0.25}{1.378} \) ≈ 0.181
Thus, the angle $\theta$ is:
\( \theta = \tan^{-1}(0.181) ≈ 10.3^\circ \)
Now, we will use the horizontal component of the tension to solve for the speed $v$. The force equations are:
T \cos \theta = mg \quad \text{(vertical equilibrium)}
T \sin \theta = \frac{mv^2}{r} \quad \text{(horizontal force)}
From the vertical equilibrium equation, we can solve for $T$:
T = \frac{mg}{\cos \theta}
Substitute this expression for $T$ into the horizontal force equation:
\( \frac{mg}{\cos \theta} \sin \theta = \frac{mv^2}{r} \)
Simplifying:
\( \frac{g \sin \theta}{\cos \theta} = \frac{v^2}{r} \)
This simplifies further to:
\( v^2 = r \cdot g \tan \theta \)
Substitute the known values:
\( v^2 = 0.25 \times 9.8 \times \tan(10.3^\circ) \)
\( v^2 = 0.25 \times 9.8 \times 0.181 ≈ 0.444 \)
Thus, the speed of the pebble is:
\( v = \sqrt{0.444} ≈ 0.67 \, \text{m/s} \)
Part (b): Acceleration of the Pebble
The acceleration of the pebble is the centripetal acceleration, which is given by:
a = \frac{v^2}{r}
Substitute the known values:
a = \frac{(0.67)^2}{0.25} ≈ \frac{0.4489}{0.25} ≈ 1.80 \, \text{m/s}^2
So, the acceleration of the pebble is approximately 1.80 m/s².
Part (c): Tension in the String
We can now calculate the tension in the string using the vertical force balance equation:
T = \frac{mg}{\cos \theta}
Substitute the known values:
T = \frac{0.053 \times 9.8}{\cos(10.3^\circ)} ≈ \frac{0.5194}{0.9848} ≈ 0.527 \, \text{N}
So, the tension in the string is approximately 0.527 N.
Summary of Results
- (a) Speed of the pebble: \( v ≈ 0.67 \, \text{m/s} \)
- (b) Acceleration of the pebble: \( a ≈ 1.80 \, \text{m/s}^2 \)
- (c) Tension in the string: \( T ≈ 0.527 \, \text{N} \)
