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Imagine that you swing about your head a ball attached to the end of a string. The ball moves at a constant speed in a horizontal circle. (a) Can the string be exactly horizontal? Why or why not? (b) If the mass of the ball is \(0.250 \mathrm{~kg}\), the radius of the circle is \(1.50 \mathrm{~m}\), and it takes 1.20 s for the ball to make one revolution, what is the ball's tangential speed? (c) What centripetal force are you imparting to the ball via the string?

Short Answer

Expert verified
(a) The string cannot be horizontal. (b) Tangential speed is 7.85 m/s. (c) Centripetal force is 10.3 N.

Step by step solution

01

Analyze Horizontal Tension

For part (a), consider the forces acting on the ball. The ball has a gravitational force pulling it downward due to its weight and a tension force from the string. In a perfectly horizontal circle, the string would have to provide both the centripetal force and counteract gravity. However, tension cannot have a vertical component to balance the weight if the string is horizontal, so a completely horizontal circle is not possible.
02

Calculate Tangential Speed

For part (b), use the formula for tangential speed: \[ v = \frac{2\pi r}{T} \]where \( r = 1.50 \mathrm{~m} \) and \( T = 1.20 \mathrm{~s} \). Plug in these values:\[ v = \frac{2\pi \times 1.50 \mathrm{~m}}{1.20 \mathrm{~s}} \approx 7.85 \mathrm{~m/s} \]Thus, the tangential speed of the ball is approximately \( 7.85 \mathrm{~m/s} \).
03

Determine Centripetal Force

For part (c), use the formula for centripetal force:\[ F_c = \frac{mv^2}{r} \]with \( m = 0.250 \mathrm{~kg} \), \( v = 7.85 \mathrm{~m/s} \), and \( r = 1.50 \mathrm{~m} \). Calculate the force:\[ F_c = \frac{0.250 \mathrm{~kg} \times (7.85 \mathrm{~m/s})^2}{1.50 \mathrm{~m}} \approx 10.3 \mathrm{~N} \]Therefore, the centripetal force imparted by the string is approximately \( 10.3 \mathrm{~N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangential Speed
In circular motion, tangential speed is the velocity of an object along the edge of the circle. Even if an object is moving in a circle, its speed can remain constant, and this particular speed is called tangential speed.
It is called 'tangential' because the direction of the speed is always tangent to the circle at the object's position. For an object moving at constant speed in a circle of radius \( r \), the tangential speed \( v \) can be determined using the formula:
  • \(v = \frac{2\pi r}{T}\)
Here, \( 2\pi r \) represents the circumference of the circle, and \( T \) is the time taken for one complete revolution.
For instance, if a ball takes 1.20 seconds to make a complete circle of radius 1.50 meters, its tangential speed would be approximately 7.85 meters per second. This value means the ball is moving with a speed of 7.85 m/s along the circular path. Understanding tangential speed is essential because it combines both speed and direction, crucial in circular motion applications.
Centripetal Force
Centripetal force is the necessary force that keeps an object moving in a circular path. Without it, an object would continue in a straight line due to inertia. The word 'centripetal' means "center-seeking," which refers to the fact that this force always points toward the center of the circle.
When calculating centripetal force, we use the formula:
  • \(F_c = \frac{mv^2}{r}\)
where \( m \) is the mass, \( v \) is the tangential speed, and \( r \) is the radius of the circle.
In our scenario, the ball has a mass of 0.250 kg, a tangential speed of 7.85 m/s, and a circle radius of 1.50 meters. By plugging in these values, we can calculate the centripetal force as approximately 10.3 N. This force is crucial because it is the string's pull that acts as the centripetal force, keeping the ball moving in the circle. Without this force, the ball would fly off in a straight line, as per Newton's first law of motion.
Horizontal Circle Tension Analysis
When a ball is swung in a horizontal circle, multiple forces come into play to maintain the circular motion.
### Components of Forces
  • Gravitational force: Acts downward due to the ball's weight.
  • Tension force: Acts along the string.
To have a perfectly horizontal circle, the string tension would need to provide both the centripetal force and oppose gravity simultaneously.
The gravitational force pulls downward, thus a horizontal string cannot have a vertical component to balance this.
This makes a perfectly horizontal tension impossible, as both forces have to be in equilibrium to sustain circular motion. ### Tension Angle The reality is that the string will be slightly tilted. This tilt ensures the vertical component of tension balances gravity. The horizontal component of the string's tension furnishes the necessary centripetal force. In summary, understanding this analysis is imperative for recognizing the dynamics of forces in circular motion and why a purely horizontal tension isn't feasible.

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Most popular questions from this chapter

A car with a constant speed of \(83.0 \mathrm{~km} / \mathrm{h}\) enters a circular flat curve with a radius of curvature of \(0.400 \mathrm{~km} .\) If the friction between the road and the car's tires can supply a centripetal acceleration of \(1.25 \mathrm{~m} / \mathrm{s}^{2},\) does the car negotiate the curve safely? Justify your answer.

Consider a simple pendulum of length \(L\) that has a small mass (the bob) of mass \(m\) attached to the end of its string. If the pendulum starts out horizontally and is released from rest, show that (a) the speed at the bottom of the swing is \(v_{\max }=\sqrt{2 g L}\) and \((b)\) the tension in the string at that point is three times the weight of the bob, or \(T_{\max }=3 m g .\) [Hint: Use conservation of energy to determine the speed at the bottom and centripetal force ideas and a free-body diagram to determine the tension at the bottom.

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