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Chapter 8: Problem 57

A 220 -kg speedboat is negotiating a circular turn (radius \(=32 \mathrm{~m}\) ) around a buoy. During the turn, the engine causes a net tangential force of magnitude \(550 \mathrm{~N}\) to be applied to the boat. The initial tangential speed of the boat going into the turn is \(5.0 \mathrm{~m} / \mathrm{s}\). (a) Find the tangential acceleration. (b) After the boat is \(2.0 \mathrm{~s}\) into the turn, find the centripetal acceleration.

Short Answer

Expert verified
(a) The tangential acceleration is 2.5 m/s虏. (b) The centripetal acceleration after 2 seconds is 3.125 m/s虏.

Step by step solution

01

Determine Tangential Acceleration

To find the tangential acceleration, use Newton's second law: \( F = ma \).Given that the net tangential force \( F = 550 \, \text{N} \) and the mass \( m = 220 \, \text{kg} \), the tangential acceleration \( a_t \) can be calculated as follows:\[ a_t = \frac{F}{m} = \frac{550}{220} \approx 2.5 \, \text{m/s}^2 \].
02

Calculate Velocity After 2 Seconds

Using the equation of motion for constant acceleration, \( v = u + at \), where:- \( v \) is the final velocity after time \( t \),- \( u = 5.0 \, \text{m/s} \) is the initial velocity,- \( a = 2.5 \, \text{m/s}^2 \) is the tangential acceleration, and- \( t = 2.0 \, \text{s} \) is the time in seconds.Calculate the velocity after 2 seconds:\[ v = 5.0 + (2.5 \times 2.0) = 5.0 + 5.0 = 10.0 \, \text{m/s} \].
03

Find Centripetal Acceleration

Centripetal acceleration \( a_c \) is given by the formula \( a_c = \frac{v^2}{r} \), where:- \( v = 10.0 \, \text{m/s} \) is the velocity after 2 seconds,- \( r = 32 \, \text{m} \) is the radius of the circular turn.Calculate the centripetal acceleration as follows:\[ a_c = \frac{10.0^2}{32} = \frac{100}{32} = 3.125 \, \text{m/s}^2 \].

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Key Concepts

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

Centripetal Acceleration
In circular motion, centripetal acceleration is crucial for keeping an object moving along a circular path. Even when a speedboat turns around a buoy, it needs a force directed toward the center of the circle to keep it from flying off straight. This force is responsible for changing the boat's direction, not its speed.
Centripetal acceleration is always pointed towards the center of the circle and is calculated using the formula: - \( a_c = \frac{v^2}{r} \) Here, \( v \) is the velocity at that point in the motion, and \( r \) is the radius of the circular path.
For example, after 2 seconds of turning, the speedboat's velocity is \( 10.0 \, \text{m/s} \) and radius is \( 32 \, \text{m} \). So, the centripetal acceleration is \( 3.125 \, \text{m/s}^2 \).
Always remember: without centripetal acceleration, the boat would not follow a curved trajectory, but rather continue in a straight line.
Newton's Second Law
Newton's Second Law lays the foundation for understanding forces and motion. This law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F = ma \)). This relationship helps us understand how forces impact motion.
In the case of our speedboat, we use this principle to find tangential acceleration. Given a net force of \( 550 \, \text{N} \) applied by the engine and a mass of \( 220 \, \text{kg} \), applying Newton's second law allows us to easily calculate the tangential acceleration.
- Simply divide the force by the mass: \( a_t = \frac{550 \, \text{N}}{220 \, \text{kg}} = 2.5 \, \text{m/s}^2 \).
This acceleration is what causes the boat to speed up as it goes around the turn, in addition to changing direction.
Circular Motion
When discussing circular motion, we refer to any object traveling along a circular path. A speedboat making a turn around a buoy exemplifies this motion perfectly. Key characteristics of circular motion include both centripetal acceleration and the object's linear or tangential acceleration.
Circular motion involves:
  • Centripetal acceleration, which changes the direction of the object moving in a circle
  • Tangential acceleration, which can change the speed of the object along the path
Here's an easy way to remember: - Tangential acceleration affects how fast the boat spins around. - Centripetal acceleration keeps it moving on the circular path.
The forces and accelerations in circular motion always work together to ensure the object stays along the curved path. This coordination is crucial, especially when cornering at high speed!

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Most popular questions from this chapter

An automobile tire has a radius of \(0.330 \mathrm{~m},\) and its center moves forward with a linear speed of \(v=15.0 \mathrm{~m} / \mathrm{s}\). (a) Determine the angular speed of the wheel. (b) Relative to the axle, what is the tangential speed of a point located \(0.175 \mathrm{~m}\) from the axle?

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 8 blades and rotates at an angular speed of \(1.25 \mathrm{rad} / \mathrm{s}\). The opening between successive blades is equal to the width of a blade. A golf ball of diameter \(4.50 \times 10^{-2} \mathrm{~m}\) is just passing by one of the rotating blades. What must be the minimum speed of the ball so that it will not be hit by the next blade?

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a constant tangential speed of \(1.20\mathrm{~m} / \mathrm{s}\) on its circular path. Find the linear speed with which the bucket moves down the well.

A bicycle is rolling down a circular hill that has a radius of \(9.00 \mathrm{~m}\). As the drawing illustrates, the angular displacement of the bicycle is 0.960 rad. The radius of each wheel is \(0.400 \mathrm{~m}\). What is the angle (in radians) through which each tire rotates?

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has a length of \(64 \mathrm{~cm}\) and is wound around the top at a spot where its radius is \(2.0 \mathrm{~cm}\). The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of \(+12 \mathrm{rad} / \mathrm{s}^{2}\). What is the final angular velocity of the top when the string is completely unwound?
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