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Chapter 2: Problem 104

A ship which moves at a steady 20 -knot speed \((1 \mathrm{knot}=1.852 \mathrm{km} / \mathrm{h})\) executes a turn to port by changing its compass heading at a constant counterclockwise rate. If it requires 60 seconds to alter course \(90^{\circ}\), calculate the magnitude of the acceleration a of the ship during the turn.

Short Answer

Expert verified
The magnitude of the acceleration is approximately 0.2698 m/s虏.

Step by step solution

01

Convert Ship Speed to m/s

The ship's speed is given as 20 knots. First, we convert this speed to meters per second. Since 1 knot = 1.852 km/h, we have:\[ 20 \text{ knots} = 20 \times 1.852 \text{ km/h} = 37.04 \text{ km/h}. \] Converting this to m/s, we use the conversion 1 km/h = 1000/3600 m/s:\[ 37.04 \text{ km/h} \times \frac{1000}{3600} = 10.29 \text{ m/s}. \]
02

Calculate the Angular Velocity

The ship changes course by 90 degrees (or \(\frac{\pi}{2}\) radians) in 60 seconds, so we calculate the angular velocity \(\omega\):\[ \omega = \frac{\Delta \theta}{\Delta t} = \frac{\frac{\pi}{2}}{60} = \frac{\pi}{120} \text{ rad/s}. \]
03

Determine the Centripetal Acceleration

To find the magnitude of the acceleration of the ship during the turn, we use the formula for centripetal acceleration:\[ a = v \times \omega, \]where \(v\) is the linear speed of the ship, and \(\omega\) is the angular velocity we found. Therefore:\[ a = 10.29 \text{ m/s} \times \frac{\pi}{120} \text{ rad/s} = \frac{10.29\pi}{120} \text{ m/s}^2. \]
04

Simplify the Acceleration Expression

Simplifying the expression for \(a\), we find:\[ a = \frac{10.29\pi}{120} \approx 0.2698 \text{ m/s}^2. \]

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Key Concepts

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

Angular Velocity
Angular velocity is a concept used to describe how quick an object rotates or turns. It tells you the rate of change of its angular position. Think of a ship changing its direction while moving on a circular path. Angular velocity comes with both magnitude and direction. The units are usually radians per second (rad/s).
To calculate angular velocity (\(\omega\)), we use the formula \(\omega = \frac{\Delta \theta}{\Delta t}\), where \(\Delta \theta\) is the angular displacement in radians, and \(\Delta t\) is the time taken for this change. In our case, the ship changes its course by \(90^\circ\), which is equal to \(\frac{\pi}{2}\) radians, in a time period of 60 seconds. So, the angular velocity would be \(\omega = \frac{\pi}{120}\) rad/s.
This means every second, the ship turns by \(\frac{\pi}{120}\) radians, helping us understand the rapidity of its maneuver. Knowing the angular velocity is essential for determining how sharply and swiftly a vehicle, like a ship, can turn safely without veering off its path.
Centripetal Acceleration
Centripetal acceleration is what keeps an object moving in a circle rather than flying off straight. Imagine spinning a ball tied to a string; this type of acceleration keeps the ball on its circular path. The word 'centripetal' means 'center-seeking,' which describes the direction of the acceleration: always pointing toward the center of the circle.
To calculate centripetal acceleration (\(a\)), we use the formula \(a = v \times \omega\), where \(v\) is the linear velocity and \(\omega\) is the angular velocity. Linear velocity is how fast the ship is moving along its path, not counting the speed due to turning or spinning. In this problem, the ship's speed is 10.29 m/s, and its angular velocity is \(\frac{\pi}{120}\) rad/s. Multiplying these gives \(a = \frac{10.29\pi}{120}\) m/s\(^2\), or approximately 0.2698 m/s\(^2\).
The centripetal acceleration helps us understand how much force is required to keep the ship on its curve. It's crucial for the stability and control of turning vehicles, as too much can lead to loss of control.
Velocity Conversion
Before solving many dynamics problems, converting velocity units to a consistent format is crucial. Here, the ship's speed provided as knots needed to be converted to meters per second (m/s) for standard SI units use.
The conversion process involves:
  • Recognizing 1 knot equals 1.852 kilometers per hour (km/h).
  • Then using it to convert 20 knots into km/h: \(20 \times 1.852 = 37.04\) km/h.
  • Knowing 1 km/h can be written as \(\frac{1000}{3600}\) m/s for further conversion.
  • Thus, converting 37.04 km/h into m/s, we have \(37.04 \times \frac{1000}{3600} = 10.29\) m/s.
Learning how to convert units effectively is handy for ensuring all elements in a calculation align correctly. It brings clarity to computations and avoids errors, especially when combining various physical quantities.

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Most popular questions from this chapter

A game requires that two children each throw a ball upward as high as possible from point \(O\) and then run horizontally in opposite directions away from O. The child who travels the greater distance before their thrown ball impacts the ground wins. If child \(A\) throws a ball upward with a speed of \(v_{1}=70 \mathrm{ft} / \mathrm{sec}\) and immediately runs leftward at a constant speed of \(v_{A}=16\) ft/sec while child \(B\) throws the ball upward with a speed of \(v_{2}=64 \mathrm{ft} / \mathrm{sec}\) and immediately runs rightward with a constant speed of \(v_{B}=18 \mathrm{ft} / \mathrm{sec},\) which child will win the game?

A block of mass \(m\) rests on a rough horizontal surface and is attached to a spring of stiffness \(k .\) The coefficients of both static and kinetic friction are \(\mu\) The block is displaced a distance \(x_{0}\) to the right of the unstretched position of the spring and released from rest. If the value of \(x_{0}\) is large enough, the spring force will overcome the maximum available static friction force and the block will slide toward the unstretched position of the spring with an acceleration \(a=\mu g-\frac{k}{m} x,\) where \(x\) represents the amount of stretch (or compression) in the spring at any given location in the motion. Use the values \(m=5 \mathrm{kg}, k=150 \mathrm{N} / \mathrm{m}, \mu=0.40,\) and \(x_{0}=200 \mathrm{mm}\) and determine the final spring stretch (or compression \(x_{f}\) when the block comes to a complete stop.

When the effect of aerodynamic drag is included, the \(y\) -acceleration of a baseball moving vertically upward is \(a_{u}=-g-k v^{2},\) while the acceleration when the ball is moving downward is \(a_{d}=-g+k v^{2},\) where \(k\) is a positive constant and \(v\) is the speed in meters per second. If the ball is thrown upward at \(30 \mathrm{m} / \mathrm{s}\) from essentially ground level, compute its maximum height \(h\) and its speed \(v_{f}\) upon impact with the ground. Take \(k\) to be \(0.006 \mathrm{m}^{-1}\) and assume that \(g\) is constant.

A train which is traveling at \(80 \mathrm{mi} / \mathrm{hr}\) applies its brakes as it reaches point \(A\) and slows down with a constant deceleration. Its decreased velocity is observed to be \(60 \mathrm{mi} / \mathrm{hr}\) as it passes a point \(1 / 2 \mathrm{mi}\) beyond \(A\). A car moving at \(50 \mathrm{mi} / \mathrm{hr}\) passes point \(B\) at the same instant that the train reaches point \(A\) In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration \(a\) that the car must have in order to beat the train to the crossing by 4 seconds and find the velocity \(v\) of the car as it reaches the crossing.

A batter hits the baseball \(A\) with an initial velocity of \(v_{0}=100 \mathrm{ft} / \mathrm{sec}\) directly toward fielder \(B\) at an angle of \(30^{\circ}\) to the horizontal; the initial position of the ball is \(3 \mathrm{ft}\) above ground level. Fielder \(B\) requires \(\frac{1}{4}\) sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder \(B\) chooses his running speed so that he arrives at the "catch position" simultaneously with the baseball. The catch position is the field location at which the ball altitude is \(7 \mathrm{ft}\). Determine the velocity of the ball relative to the fielder at the instant the catch is made.
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