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Chapter 3: Problem 27

A small ball is attached to the lower end of a 0.800 -m-long string, and the other end of the string is tied to a horizontal rod. The string makes a constant angle of \(37.0^{\circ}\) with the vertical as the ball moves at a constant speed in a horizontal circle. If it takes the ball \(0.600 \mathrm{~s}\) to complete one revolution, what is the magnitude of the radial acceleration of the ball?

Short Answer

Expert verified
The magnitude of the radial acceleration of the ball can be calculated by substituting the calculated values of \(v\) and \(r\) into the formula \(a_{r}= \frac{v^{2}}{r}\).

Step by step solution

01

Determine the radius of the circle

Since the string makes a constant angle of \(\theta = 37.0^{\circ}\) with the vertical, we can use basic trigonometry to find the radius (\(r\)). The radius is the horizontal component of the length of the string (\(L = 0.800 m\)), so we have\[r = L \sin(\theta) = 0.800 \sin(37.0^{\circ})\]
02

Calculate the magnitude of the speed

Given that the time to complete one revolution (\(t\)) is \(0.600 s\), we calculate speed using the formula \(v = \frac{Circumference}{t}\). Considering the circumference of a circle is given by \(2\pi r\), we calculate the speed as \(v = \frac{2\pi r}{t}\)
03

Calculate the radial acceleration.

Finally, to find the magnitude of the radial acceleration (\(a_{r}\)), we can use the formula \(a_{r}= \frac{v^{2}}{r}\). By substituting the calculated values of \(v\) and \(r\) into the formula, we can get the magnitude of the radial acceleration.

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Key Concepts

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

Understanding Circular Motion
Circular motion occurs when an object moves along a curved path or a circular trajectory. Imagine a ball on a string swinging around in a perfect circle.
The ball continually changes direction, even if its speed remains constant. This change in direction involves a special type of acceleration called radial acceleration or centripetal acceleration.

In this problem, the ball moves in a horizontal circle while being constrained by a string. The string creates a force that alters the ball's path, allowing it to keep moving in a circle. This force is essential to create radial acceleration, keeping the ball on its circular path.
The Role of Trigonometry
Trigonometry helps us deal with angles and relationships within triangles. In circular motion problems, it's key for breaking down vectors into components.
If the ball's string makes an angle of \(37.0^{\circ}\) with the vertical, we use the sine function to find the horizontal component, which is the radius of the circle.
  • Using the formula: \[r = L \sin(\theta)\]
  • Here, \(L\) is the length of the string, and \(\theta\) is the angle given.

This radius is critical because it defines the size of the circle the ball is moving in, and it's used in further calculations to find speed and acceleration. Trigonometry makes it easier to solve these complex-looking geometry problems.
The Concept of Constant Speed
Though speed remains consistent, the direction changes as the ball goes around in a circle. Constant speed in circular motion still involves acceleration, due to the continuous change in direction.
The formula \(v = \frac{2\pi r}{t}\) helps calculate the magnitude of the speed, where:
  • \(v\) is the speed
  • \(r\) is the radius
  • \(t\) is the time per revolution

This formula calculates how fast the ball is moving along the circular path. Knowing this speed is necessary to further determine the radial acceleration of the ball.
Physics Problem Solving Technique
Solving physics problems necessitates a step-by-step methodology. Start by understanding what is given and what is required. Break down the problem using known formulas.
  • Identify all known values and relevant physics principles.
  • Use proper formulas for calculations, like \(a_{r} = \frac{v^{2}}{r}\) for radial acceleration.
  • Substitute known values, keeping consistent units.

In our exercise, we found the radius using trigonometry, calculated speed, and finally used these to find radial acceleration. Each step relies on solid understanding and application of physics concepts.

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Most popular questions from this chapter

An object moves in a horizontal circle at constant speed \(v\) (in units of \(\mathrm{m} / \mathrm{s}) .\) It takes the object \(T\) seconds to complete one revolution. Derive an expression that gives the radial acceleration of the ball in terms of \(v\) and \(T,\) but not \(r .\) (a) If the speed doubles, by what factor must the period \(T\) change if \(a_{\text {rad }}\) is to remain unchanged? (b) If the radius doubles, by what factor must the period change to keep \(a_{\text {rad }}\) the same?

A snowball rolls off a barn roof that slopes downward at an angle of \(40^{\circ}\) (Fig. \(\mathbf{P 3 . 6 1}\) ). The edge of the roof is \(14.0 \mathrm{~m}\) above the ground, and the snowball has a speed of \(7.00 \mathrm{~m} / \mathrm{s}\) as it rolls off the roof. Ignore air resistance. (a) How far from the edge of the barn does the snowball strike the ground if it doesn't strike anything else while falling? (b) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion in part (a). (c) A man \(1.9 \mathrm{~m}\) tall is standing \(4.0 \mathrm{~m}\) from the edge of the barn. Will the snowball hit him?

Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 60 minutes, reaching a point \(2.0 \mathrm{~km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) \(5.0 \mathrm{~km}\) downstream from the turnaround point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

A cannon, located \(60.0 \mathrm{~m}\) from the base of a vertical \(25.0-\mathrm{m}-\)tall cliff, shoots a \(15 \mathrm{~kg}\) shell at \(43.0^{\circ}\) above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of \(25.0 \mathrm{~m}\) above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

On level ground a shell is fired with an initial velocity of \(40.0 \mathrm{~m} / \mathrm{s}\) at \(60.0^{\circ}\) above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell's initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.
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