/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A stone is tied to a string (len... [FREE SOLUTION] | 91Ó°ÊÓ

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A stone is tied to a string (length 1.10 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

Understanding the Problem

We have a stone tied on a string with length 1.10 m. It's whirled in two different circular paths: horizontally and vertically. The tension is 15.0% larger in the vertical circle than in the horizontal circle. We need to find the speed of the stone.
02

Write the Equations for Horizontal Circle

In the horizontal circle, the tension in the string provides the centripetal force. Let the tension in the horizontal circle be \( T_h \). The centripetal force formula is: \[ T_h = \frac{mv^2}{r} \] where \( m \) is the mass of the stone, \( v \) is the speed, and \( r = 1.10 \) m is the radius of the circle.
03

Write the Equations for Vertical Circle

In the vertical circle, the tension is maximum at the lowest point. Let the tension in the vertical circle be \( T_v \). At the lowest point, tension is the sum of the centripetal force and weight of the stone: \[ T_v = \,\frac{mv^2}{r} \, + \, mg \] We are given \( T_v = 1.15 \, T_h \).
04

Relate the Tensions

Since \( T_v = 1.15 T_h \), substitute the expressions for \( T_h \) and \( T_v \): \[ \frac{mv^2}{r} + mg = 1.15 \times \frac{mv^2}{r} \] Simplifying gives: \[ mg = 0.15 \times \frac{mv^2}{r} \]
05

Solve for the Speed

Cancel out the mass \( m \) from both sides of the equation and solve for \( v \): \[ v^2 = \frac{rg}{0.15} \] \[ v = \sqrt{\frac{rg}{0.15}} \] Substitute \( r = 1.10 \) m and \( g = 9.81 \) m/s² to find \( v \): \[ v = \sqrt{\frac{1.10 \times 9.81}{0.15}} \approx 8.48 \, \text{m/s} \]

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Key Concepts

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

Tension in Circular Motion
When an object is moving in a circular path, tension plays a key role in keeping it on that path. Tension is a force that pulls the object inward, towards the center of the circle. It acts through the string or whatever medium is facilitating the circular motion. The tension must be strong enough to counteract opposing forces trying to pull the object away from the circular path, such as gravity or friction.

In circular motion, tension varies depending on the object's speed and the radius of the circle. It is also affected by the position of the object along its path. For example, in vertical circles, the tension changes from point to point because of the varying influence of gravity. Understanding tension is essential to solve many physics problems involving circular motion.
Horizontal and Vertical Circular Paths
Circular paths can be horizontal or vertical, each affecting the forces involved differently. In a horizontal circular path, the forces acting are typically balanced horizontally, and the centripetal force is provided entirely by tension.

For vertical circular paths, however, the situation is a bit more complex. Here, the gravity also plays a significant role and must be considered along with tension. As the object moves through the bottom of the circle, both the weight of the object and the tension are working together to provide the necessary centripetal force. This is why the tension in vertical motion is usually more at the lowest point, as it has to counteract gravity and still maintain the required centripetal force.

Clearly distinguishing between these path types helps in proper analysis and solution of physics problems involving circular motion.
Centripetal Force
Centripetal force is the inward-directed force that causes an object to follow a circular path. It is not a separate force itself but is supplied by other forces such as tension, gravity, or friction, depending on the situation.

Mathematically, the centripetal force required for circular motion is given by the formula: \[ F_c = \frac{mv^2}{r} \]where \( F_c \) is the centripetal force, \( m \) is the mass of the object, \( v \) is its speed, and \( r \) is the radius of the circle.

In scenarios like the one in the exercise, where the stone is whirled around, the tension in the string provides the centripetal force. Understanding centripetal force is vital because it's at the core of determining how different forces interact to create and maintain circular motion.
Physics Problem Solving
Solving physics problems often involves a clear understanding of the concepts and clever application of mathematical formulas. Begin by identifying the forces at play and how they interact to result in the given phenomena, like circular motion for this exercise.

Next, clearly outline the problem as if telling a story: what is known, what needs to be found, and any relations or formulas that link them. Organizing information like this can simplify complex problems.

Substitute known values into your equations as much as possible and simplify where you can before solving for the unknowns. This strategy of breaking down the problem into smaller, manageable steps can make finding solutions much more straightforward.
  • Identify forces and relationships.
  • Organize the given data.
  • Simplify and substitute in equations.
  • Solve step-by-step for unknowns.
Mastering these fundamental techniques will enhance your problem-solving skills across all areas of physics.

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Most popular questions from this chapter

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