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A stone is tied to a string (length \(=1.10 \mathrm{m}\) ) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is \(15.0 \%\) larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

Short Answer

Expert verified
The speed of the stone is approximately 8.48 m/s.

Step by step solution

01

Understand the Problem

We have a stone tied to a string, which is being whirled in a circle. First, the stone moves horizontally, and the tension in the string is due to centripetal force. Next, the stone moves vertically in a circle, where tension varies and reaches a maximum at the bottom of the circle. We need to find the speed in both cases, knowing the maximum tension in the vertical case is 15% more than the tension in the horizontal case.
02

Identify Forces in Horizontal Circle

In the horizontal circle, we can assume the tension provides the centripetal force to keep the stone moving in a circle. Thus, the tension in the horizontal case, denoted as \( T_h \), is given by the centripetal force formula: \[ T_h = \frac{m v^2}{r} \] where \( m \) is the mass of the stone, \( v \) is the speed, and \( r = 1.10 \text{ m} \) is the radius of the circle.
03

Identify Forces in Vertical Circle

When the stone is swung in a vertical circle, the tension varies, reaching a maximum at the bottom of the circle. The maximum tension \( T_v \) at the lowest point is given by: \[ T_v = T_h + m g \] where \( g = 9.81 \text{ m/s}^2 \) is the gravitational acceleration. The new tension \( T_v \) is 15% greater than \( T_h \): \[ T_v = 1.15 T_h \].
04

Relate Tensions and Solve for Speed

From the earlier relationship, \( T_v = T_h + m g \) and \( T_v = 1.15 T_h \), we equate these: \[ 1.15 T_h = T_h + m g \]. Solving for \( m g \), we get: \[ 0.15 T_h = m g \]. Using \( T_h = \frac{m v^2}{r} \), substitute into the equation and solve for \( v \):\[ 0.15 \frac{m v^2}{r} = m g \] \[ \frac{0.15 v^2}{r} = g \] \[ v^2 = \frac{g r}{0.15} \] \[ v = \sqrt{\frac{g r}{0.15}} \].
05

Calculate the Speed

Substitute \( g = 9.81 \text{ m/s}^2 \) and \( r = 1.10 \text{ m} \) into the velocity equation from Step 4: \[ v = \sqrt{\frac{9.81 \times 1.10}{0.15}} \]Calculate this to find \( v \).
06

Compute the Result

Evaluating the expression: \[ v = \sqrt{\frac{10.791}{0.15}} = \sqrt{71.94} \approx 8.48 \text{ m/s} \]. Therefore, the speed of the stone is approximately 8.48 m/s.

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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 shaped like a circle. This motion is common in various physical phenomena and can occur both in horizontal or vertical orientations. In circular motion, there needs to be a constant force that acts toward the center of the circle. This force is called "centripetal force."

A crucial aspect of circular motion is the concept of centripetal acceleration. This is the acceleration experienced by an object moving in a circular path at constant speed because the direction of its velocity is continually changing. The expression for centripetal acceleration is given by:
  • Formula: \( a_c = \frac{v^2}{r} \)
where \( v \) is the velocity and \( r \) is the radius of the circle.

It is important to note that if the path is not perfectly horizontal and instead is vertical, additional forces such as gravity come into play, which will affect the dynamics of the circular motion. This interplay between different forces leads to variations in the tension of the string or whatever is keeping the object in its circular path.
Tension in Physics
Tension is a force exerted by a string, rope, or in this context, a string tied to a stone, that transmits a pulling force from one end to the other. In the case of circular motion, tension is one component of the centripetal force that keeps the stone moving in a path along the circle. It is crucial to understand that tension can change based on the orientation of the motion (horizontal or vertical).

In horizontal circular motion, the tension is straightforwardly providing all the centripetal force, which is described by the equation:
  • \( T_h = \frac{m v^2}{r} \)
where \( T_h \) is the tension, \( m \) is the mass of the object, \( v \) is the velocity, and \( r \) is the radius of the circle.

In vertical circular motion, gravity also influences the motion of the object. The tension at the bottom of the circle becomes greatest because gravity adds to the tension required to maintain circular motion. The relationship becomes:
  • \( T_v = T_h + m g \)
where \( m g \) is the gravitational force on the stone, and \( T_v \) is the tension at the lowest point of the motion.
Vertical and Horizontal Motion
When we talk about vertical and horizontal motion, we're usually referring to how forces are aligned in different planes, and how they affect the motion of objects. In the given exercise, we see a practical application of these principles in the context of a stone being swung in a circle at different orientations.

**Horizontal Motion:**
  • In horizontal circular motion, the tension in the string is primarily responsible for providing the necessary centripetal force.
  • This force acts inwards horizontally, keeping the stone on its circular path.
**Vertical Motion:**
  • In vertical motion, however, gravity also plays a significant role. The gravitational force continuously influences the motion of the object, varying the tension as the stone moves along the circular path.
  • At the bottom of the circle, gravity and tension work together to provide the centripetal force needed.
Understanding the distinction between these two orientations is crucial in physics as it involves analyzing how forces such as gravity affect the resultant motion.

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Most popular questions from this chapter

Pilots of high-performance fighter planes can be subjected to large centripetal accelerations during high-speed turns. Because of these accelerations, the pilots are subjected to forces that can be much greater than their body weight, leading to an accumulation of blood in the abdomen and legs. As a result, the brain becomes starved for blood, and the pilot can lose consciousness ("black out"). The pilots wear "anti-G suits" to help keep the blood from draining out of the brain. To appreciate the forces that a fighter pilot must endure, consider the magnitude \(F_{\mathrm{N}}\) of the normal force that the pilot's seat exerts on him at the bottom of a dive. The magnitude of the pilot's weight is \(W\). The plane is traveling at \(230 \mathrm{m} / \mathrm{s}\) on a vertical circle of radius \(690 \mathrm{m} .\) Determine the ratio \(F_{\mathrm{N}} / W .\) For comparison, note that blackout can occur for values of \(F_{\mathrm{N}} / W\) as small as 2 if the pilot is not wearing an anti-G suit.

A car is safely negotiating an unbanked circular turn at a speed of \(21 \mathrm{m} / \mathrm{s} .\) The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

A child is twirling a \(0.0120-\mathrm{kg}\) plastic ball on a string in a horizontal circle whose radius is \(0.100 \mathrm{m}\). The ball travels once around the circle in 0.500 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

In a skating stunt known as crack-the-whip, a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of \(80.0 \mathrm{kg}\) and is \(6.10 \mathrm{m}\) from the pivot. He is skating at a speed of \(6.80 \mathrm{m} / \mathrm{s}\). Determine the magnitude of the centripetal force that acts on him.

A satellite is in a circular orbit about the earth \(\left(M_{\mathrm{E}}=5.98 \times 10^{24} \mathrm{kg}\right)\). The period of the satellite is \(1.20 \times 10^{4}\) s. What is the speed at which the satellite travels?

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