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A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

Short Answer

Expert verified
(a) Zero for both; (b) Zero by tension, 25.088 J by gravity.

Step by step solution

01

Understand the Problem

We have a 0.800-kg ball tied to a 1.60 m long string, swung in a vertical circle. We need to calculate the work done by the tension in the string and the gravitational force in two scenarios: a complete circle (part a) and a semicircle from the lowest to the highest point (part b).
02

Calculate Work Done by Tension in Complete Circle

Work done by tension is calculated with the formula \( W = F \, \cos(\theta) \, d \) where \( F \) is the force, \( \theta \) is the angle between force and displacement, and \( d \) is displacement. Since tension is perpendicular to displacement through the complete circle, \( \theta = 90^\circ \) and \( \cos(90^\circ) = 0 \). Therefore, work done, \( W = 0 \).
03

Calculate Work Done by Gravity in Complete Circle

The gravitational force does work, \( W = mgh \), where \( h \) is the change in height. In a complete circle, the ball returns to the starting point, so the change in height is zero. Consequently, the work done by gravity is zero, \( W = 0 \).
04

Calculate Work Done by Tension for Semicircle

In a vertical motion from the lowest to the highest point of the semicircle, the tension remains perpendicular to the instantaneous displacement at every point. Thus, similar to the full circle, the work done by the tension remains \( W = 0 \).
05

Calculate Work Done by Gravity for Semicircle

From the lowest to the highest point, the vertical height change is twice the radius, \( 2r = 2 \times 1.60 \times \sin(\theta) \). At the top, \( \theta = 90^\circ \), so the height change is 3.20 m. Work done by gravity is \( W = mgh = 0.800 \times 9.8 \times 3.20 \). Calculate \( W = 25.088 \text{ J} \).

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Key Concepts

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

Circular Motion
Circular motion occurs when an object moves in a path in the shape of a circle. This type of motion is characterized by a constant change in direction, even if the speed of the object remains constant.
In a circular motion, the velocity of the object is always tangential to the circle's path. This means the direction of the velocity vector changes as the object moves, even if the speed (magnitude of the velocity) remains constant.
To maintain this motion and change direction continuously, a centripetal force is needed, which always points toward the center of the circle.
  • The centripetal force is not a separate force but the resultant of other forces acting on the object.
  • In the case of a ball tied to a string swung in a circle, the tension in the string contributes to this centripetal force.
Understanding circular motion is key to studying how forces interact in a rotating system and how different scenarios, like vertical circles, affect these dynamics.
Vertical Circle
Vertical circular motion involves an object moving in a circle located in a vertical plane. This adds complexity because, as the object moves, the gravitational force affects it differently at varying points.
One key factor in vertical circular motion is how gravitational force interacts with other forces like tension. Unlike horizontal motion, gravity contributes to the dynamics by accelerating or decelerating the object as it rises and falls.
When the object is at the highest point in the circle, gravity acts downward, potentially decreasing tension in the string. Conversely, when the object is at the lowest point, gravity increases tension since it acts in the same direction.
  • Gravitational force is crucial because it can lead to changing tension forces throughout the motion.
  • Understanding these dynamics enables the calculation of forces and predictability in motion outcomes, especially when calculating work done by gravity and tension.
Vertical circles provide an excellent real-world application, whether in amusement park rides or understanding planetary motion.
Work Done by Gravity
Work done by gravity plays a crucial role in many physics problems involving motion. It is the energy transferred by the gravitational force over a distance.
In the context of a ball in circular motion, the work done by gravity depends on the vertical displacement of the ball. When the ball returns to its starting height, the work done by gravity over the complete circle is zero, as there is no net change in height.
  • Formula: The work done by gravity is given by \[ W = mgh \] where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the change in height.
  • During vertical motion through a semicircle, however, there is a net vertical displacement, hence work done is not zero.
  • This calculation helps determine how potential energy converts to kinetic energy and vice-versa through motion.
Understanding work done by gravity aids in analyzing the energy changes and forces at play in cyclic motion.
Tension in Strings
Tension in strings is a vital concept when studying objects in circular motion. It is the force exerted along the string that transmits a pulling force from one end to the other.
When an object moves in a circular path, the tension helps keep the object in motion by providing the necessary centripetal force toward the center of the circle.
In our scenario with the ball on a string, tension remains perpendicular to the displacement because the ball moves along the arc of the circle. As a result, the work done by tension is zero when the object completes a full cycle or semicircle.
  • The tension force acts radially, maintaining the grip needed for circular motion.
  • Its value changes depending on the ball's position in the circle, especially impacted by the gravitational pull.
  • The perpendicular nature of tension regarding motion direction means that it does no work (i.e., transfers no energy over a distance).
Comprehending how tension works in strings forms the basis for solving complex motion and force problems in physics.

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Most popular questions from this chapter

A physics professor is pushed up a ramp inclined upward at 30.0\(^\circ\) above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and chair is 85.0 kg. She is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.00 m/s. Use the work\(-\)energy theorem to find her speed at the top of the ramp.

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

A net force along the \(x\)-axis that has \(x\)-component \(F{_x}= -12.0\mathrm{N} +(0.300\mathrm{N/m{^2}})x{^2}\) is applied to a 5.00-kg object that is initially at the origin and moving in the \(-x\)-direction with a speed of 6.00 m/s. What is the speed of the object when it reaches the point \(x = 5.00\) m?

An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (\(\textbf{Fig. P6.94}\)). The upward force is the lift force that keeps the airplane aloft, and the backward force is called \(induced \, drag\). At flying speeds, induced drag is inversely proportional to \(v^2\), so the total air resistance force can be expressed by \(F_air = \alpha v^{2} + \beta /v{^2}\), where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single-engine airplane, \(\alpha = 0.30 \, \mathrm{N} \cdot \mathrm{s^{2}/m^{2}}\) and \(\beta = 3.5 \times 10^5 \, \mathrm{N} \cdot \mathrm{m^2/s^2}\). In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum \(range\) (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum \(endurance\)(that is, remain in the air the longest time).

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40.0\(^\circ\) above the horizontal. The glider has mass 0.0900 kg. The spring has \(k\) = 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

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