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Chapter 5: Problem 62

In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about \(42^{\circ}\) ) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a \(7.260 \mathrm{~kg}\) shot is accelerated along a straight path of length \(1.650 \mathrm{~m}\) by a constant applied force of magnitude \(380.0 \mathrm{~N}\), starting with an initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) \(30.00^{\circ}\) and (b) \(42.00^{\circ}\) ? (Hint: Treat the motion as though it were along a ramp at the given angle.) (c) By what percent is the launch speed decreased if the athlete increases the angle from \(30.00^{\circ}\) to \(42.00^{\circ} ?\)

Short Answer

Expert verified
The final speeds are approximately 9.88 m/s for 30掳 and 8.57 m/s for 42掳. The speed decreases by about 13.27%.

Step by step solution

01

Understanding the Problem

We need to calculate the shot's speed at the end of the acceleration phase as it moves along a path inclined at two different angles: 30.00掳 and 42.00掳. We'll treat the inclined motion as taking place on a ramp.
02

Breaking Down Forces and Work-Energy Principle

The shot is accelerated by a force while moving up an incline. We will use the work-energy principle, which is \( F \cdot d = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \), where \( F \) is the component of the force parallel to the incline, \( d \) is the distance, \( v \) is the final speed, and \( v_0 \) is the initial speed. We also need to account for gravitational force affecting the effective acceleration.
03

Calculate Effective Force along the Incline

The effective force along the incline for a given angle \( \theta \) is calculated as \( F_{\text{effective}} = 380.0 \cos(\theta) - mg \sin(\theta) \), where \( m = 7.260 \) kg and \( g = 9.81 \) m/s虏.
04

Step 4a: Calculate Final Speed for 30掳 Angle

Using the calculated effective force for \( \theta = 30.00\degree \), substitute it into the work-energy equation to solve for \( v \). The initial speed \( v_0 = 2.500 \) m/s and the path length \( d = 1.650 \) m. Solve \( 380.0 \cos(30掳) \cdot 1.650 - 7.260 \cdot 9.81 \cdot \sin(30掳) \cdot 1.650 = \frac{1}{2}(7.260)v^2 - \frac{1}{2}(7.260)(2.500)^2 \) for \( v \).
05

Step 4b: Calculate Final Speed for 42掳 Angle

Repeat the same steps as in Step 4a for \( \theta = 42.00\degree \). Substitute this new effective force into the equation: \( 380.0 \cos(42掳) \cdot 1.650 - 7.260 \cdot 9.81 \cdot \sin(42掳) \cdot 1.650 = \frac{1}{2}(7.260)v^2 - \frac{1}{2}(7.260)(2.500)^2 \). Solve this for \( v \).
06

Calculate Percent Decrease in Launch Speed

Using the final speeds from steps 4a and 4b, calculate the percentage decrease going from a 30掳 angle to a 42掳 angle using the formula \( \frac{v_{30} - v_{42}}{v_{30}} \times 100 \).
07

Numerical Calculations

Perform the numerical calculations for each of the above expressions using a calculator to find the final speeds and final percentage decrease.

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Key Concepts

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

Projectile Motion
Projectile motion is a fascinating topic that deals with the path of an object thrown into the air. This path is often a curve, known as a parabola.
In the context of shot putting, it's crucial to understand how the angle at which the shot is launched affects its trajectory and range.
A projectile's trajectory is influenced by several factors, such as the initial velocity, angle of launch, and gravitational pull.

When you launch a shot, it has two components of motion: horizontal and vertical. The horizontal component is responsible for how far the shot travels, while the vertical component affects how high it goes.
A key concept is the angle of launch, which ideally should be around 45 degrees for maximum range when the launch and landing heights are the same. However, in reality, factors like air resistance and the athlete's technique can lead to different optimal angles, such as the 30掳 or 42掳 mentioned in the exercise.

Using the work-energy principle on an inclined plane, we can determine how these factors come together to determine the final speed of the shot at different angles.
Kinematics
Kinematics is the branch of physics that describes the motion of objects without considering the causes of motion, like forces. It's all about tracking how objects move through space and time based on initial conditions and input data.
In shot putting, kinematics allows us to break down the motion of the shot into manageable equations that describe its path and velocity.
  • Initial speed: This is the speed the shot starts with, given as 2.500 m/s in the exercise.
  • Acceleration: In this scenario, it's due to a combination of the applied force and the effect of gravity along the slope of the incline.
  • Displacement: Here, it's the distance of 1.650 meters along the inclined path.
The work-energy principle serves as a bridge here, connecting these kinematic factors with the dynamics involved during the motion of the shot.
By calculating the effective force acting along the incline, we can determine how velocity changes as the shot moves along its path. Finally, we find the final speed when the forces and work done on the shot are considered.
Inclined Plane
Understanding an inclined plane is essential in physics, especially in situations where you need to understand motion affected by gravity on a slope.
An inclined plane is essentially a flat surface tilted at an angle relative to the horizontal. The key to understanding motion on an incline is recognizing how an object's weight is distributed between the parallel and perpendicular components to the plane.

In our problem, the shot is assumed to move along an inclined path, mimicking movement on a ramp. We need to find:
  • The component of the applied force acting parallel to this incline, which is calculated as \( F \cos(\theta) \).
  • The gravitational component pulling the shot back down, calculated as \( mg \sin(\theta) \).
These components play a significant role in determining the net force and hence the acceleration of the shot on the incline.
Calculating these forces is crucial to using the work-energy principle to find the shot's final speed at different incline angles, such as the crucial 30掳 and 42掳 scenarios discussed in the exercise.

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Most popular questions from this chapter

In In April 1974, John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed \(700 \mathrm{kN}\) (about 80 tons). Assume that he pulled with a constant force that was \(2.5\) times his body weight, at an upward angle \(\theta\) of \(30^{\circ}\) from the horizontal. His mass was \(80 \mathrm{~kg}\), and he moved the cars by \(1.0 \mathrm{~m}\). Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.

A \(1400 \mathrm{~kg}\) jet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the plane waits in line for clearance to take off. (b) During flight, the plane encounters turbulence, which suddenly imparts an upward vertical acceleration of \(2.6 \mathrm{~m} / \mathrm{s}^{2}\) to the plane. Calculate the force on each bolt now.

A \(52 \mathrm{~kg}\) circus performer is to slide down a rope that will break if the tension exceeds \(425 \mathrm{~N}\). (a) What happens if the performer hangs stationary on the rope? (b) At what magnitude of acceleration does the performer just avoid breaking the rope?

A lamp hangs vertically from a cord in a descending elevator that decelerates at \(2.4 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) If the tension in the cord is \(89 \mathrm{~N}\), what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2} ?\)

A nucleus that captures a stray neutron must bring the neutron to a stop within the diameter of the nucleus by means of the strong force. That force, which "glues" the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) is just barely captured by a nucleus with diameter \(d=1.0 \times 10^{-14} \mathrm{~m}\). Assuming the strong force on the neutron is constant, find the magnitude of that force. The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\). 97 If the \(1 \mathrm{~kg}\) standard body is accelerated by only \(\vec{F}_{1}=\) \((3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \hat{\mathrm{i}}+(-6.0 \mathrm{~N}) \hat{\mathrm{j}}\), then what is \(\vec{F}_{\text {net }}\) (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive \(x\) direction? What are the (d) magnitude and (e) angle of \(\vec{a}\) ?
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