/*! 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 85 A golf ball is launched at an an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A golf ball is launched at an angle of \(20^{\circ}\) to the horizontal, with a speed of \(60 \mathrm{~m} / \mathrm{s}\) and a rotation rate of \(90 \mathrm{rad} / \mathrm{s}\). Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

Short Answer

Expert verified
The golf ball makes approximately 30 revolutions by the time it reaches maximum height.

Step by step solution

01

Calculate Time to Reach Maximum Height

The time to reach maximum height can be calculated using the vertical component of the initial velocity. First, find this component using the formula: \( v_{y} = v \sin(\theta) \), where \( v = 60 \mathrm{~m/s} \) and \( \theta = 20^{\circ} \). This gives: \( v_{y} = 60 \sin(20^{\circ}) \approx 20.52 \mathrm{~m/s} \). The formula to find the time is: \( t = \frac{v_{y}}{g} \) where \( g = 9.8 \mathrm{~m/s^2} \). Thus, \( t \approx \frac{20.52}{9.8} \approx 2.09 \mathrm{~s} \).
02

Calculate Number of Revolutions

The ball rotates at a rate of \(90 \mathrm{~rad/s}\). To find the number of revolutions, calculate the total radians the ball rotates by multiplying the rotation rate by the time: \( \text{Total Radians} = 90 \times 2.09 \approx 188.1 \mathrm{~rad} \). Since there are \(2\pi\) radians in one full revolution, the number of revolutions is: \( \text{Revolutions} = \frac{188.1}{2\pi} \approx 29.94 \).
03

Round to Nearest Whole Number

Since the number of revolutions is usually expressed as a whole number, we round 29.94 to the nearest whole number, resulting in 30 revolutions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Motion
Angular motion describes the movement of an object around a point or axis. Imagine tracing a path around a circle; when something rotates, it moves in angular motion. In this exercise, the golf ball spins as it travels upward, showcasing its angular motion.
The speed of this motion is measured in radians per second. A radian is a measure of angular distance, much like a degree, but it's based on the radius of a circle. The ball's rotation rate is given as 90 radians per second, meaning the ball turns this amount every second.
When exploring angular motion, it's essential to distinguish between linear speed (straight-line motion) and angular speed (rotational movement). Even though both involve speed, they're measured differently – linear in meters per second and angular in radians per second.
  • Angular Speed: Measure of how fast something spins.
  • Radial Measure: Is based on the ratio between the length of an arc and its radius.
Rotational Dynamics
Rotational Dynamics is the study of forces and torques in situations where an object undergoes rotation. For a rotating object, factors like mass and how it's distributed affect how easily it spins or stops. In our exercise, while we don’t delve into these forces, it’s important to understand that they influence the motion.
The golf ball’s ability to keep spinning relies on its constant rotational force, denoted by its rotation rate. This rotation, if unimpeded, will continue as long as there isn’t an external force stopping it – similar to how once a swing gets going, it won't just stop on its own unless something intervenes.
  • Moment of Inertia: This is a concept within rotational dynamics indicating how hard or easy it is to change an object's rotation.
  • Torque: A force that causes an object to rotate.
While basic rotational dynamics include these concepts, this exercise focuses more on the ball's continuous spin till it reaches maximum height without external interferences.
Kinematics
Kinematics is a branch of physics that explores motion without considering its causes. It informs us about an object’s trajectory, speed, and time in motion. Every object in motion has a path - and kinematics helps predict where and when an object is at a certain point.
In this exercise, we broke down the ball's motion into components - horizontal and vertical. To solve for time to maximum height, which is a kinematic problem, we focused on the vertical component of motion because that's directly impacted by gravity.
  • Vertical Velocity: This is calculated using the sine component of the angle with which the ball is launched.
  • Time Calculation: Determined using the formula for constant acceleration under gravity.
Gravity
Gravity is an omnipresent force that draws objects towards the center of Earth. It plays a pivotal role in projectile motion, defining how an object travels when hurled into the air. In our case, gravity impacts the ball by influencing its upward motion until it reaches a peak and begins descending.
The force of gravity is constant at approximately 9.8 m/s² near the surface of the Earth. When solving the exercise, we used this value to figure out when the ball reaches its maximum height by dividing the initial vertical speed by the gravitational constant.
  • Acceleration due to Gravity: Always acts downward, slowing the upward part of any projectile’s journey.
  • Impact on Vertical Motion: Gravity leads to a decrease in the vertical component of the velocity over time until it becomes zero at peak.
Trigonometry
Trigonometry is a branch of mathematics dealing with the relationships between the angles and sides of triangles. It comes into play in physics when analyzing angles of direction or components of motion, just like with the golf ball being launched at an angle.
In this problem, we used trigonometry to find the vertical component of the initial velocity by utilizing the sine function. Since the ball is launched at a 20-degree angle to the horizontal, trigonometric functions help separate this motion into horizontal and vertical parts, essential for further calculations.
  • Sine Function: Gives us the ratio of the opposite side (vertical component) to the hypotenuse (initial velocity) in a right triangle.
  • Trigonometric Identities: Useful in resolving vector components of angles.
Understanding these concepts allows us to unravel complex projectile motion problems by breaking them into manageable pieces.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the \(20 \mathrm{~cm}\) mark. (Treat the stick as a thin rod.)

A wheel, starting from rest, rotates with a constant angular acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). During a certain \(3.00 \mathrm{~s}\) interval, it turns through \(90.0\) rad. (a) What is the angular velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval? (b) How long has the wheel been turning before the start of the \(3.00\) s interval?

What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second.

A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

At \(t=0\), a flywheel has an angular velocity of \(4.7 \mathrm{rad} / \mathrm{s}, \mathrm{a}\) constant angular acceleration of \(-0.25 \mathrm{rad} / \mathrm{s}^{2}\), and a reference line at \(\theta_{0}=0 .\) (a) Through what maximum angle \(\theta_{\max }\) will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at \(\theta=\frac{1}{2} \theta_{\max } ?\) At what (d) negative time and (e) positive time will the reference line be at \(\theta=10.5\) rad? (f) Graph \(\theta\) versus \(t\), and indicate the answers to (a) through (e) on the graph.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.