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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.

Step by step solution

01

Identify Components

First, we need to identify and break down the given components of the problem. The golf ball is launched at a speed of \( v = 60 \, \text{m/s} \) and at an angle \( \theta = 20^{\circ} \). It has a rotation rate of \( \omega = 90 \, \text{rad/s} \).
02

Calculate Time to Maximum Height

To find the time to reach maximum height, we need the initial vertical velocity \( v_{y_i} = v \sin\theta \). The maximum height is reached when the vertical velocity is 0. Using the equation \( v_{y_f} = v_{y_i} - gt \), where \( g = 9.81 \, \text{m/s}^2 \) and \( v_{y_f} = 0 \), solve for \( t \): \[t = \frac{v \sin\theta}{g} = \frac{60 \sin 20^{\circ}}{9.81}\].
03

Solve for Revolutions

The number of revolutions the ball makes is determined by the rotational motion and time. Calculate \( t \) from the previous step, then compute the number of revolutions using \( \omega \), which is in radians per second: \[ \text{Number of Revolutions} = \frac{\omega t}{2\pi} \].
04

Perform the Calculations

1. Calculate \( v_{y_i} = 60 \sin 20^{\circ} \approx 20.52 \, \text{m/s} \).2. Calculate the time, \( t = \frac{20.52}{9.81} \approx 2.09 \, \text{s} \).3. Use this time to find revolutions: \[ \text{Revolutions} = \frac{90 \times 2.09}{2\pi} \approx 30 \].
05

Conclude the Solution

The calculations show that the golf ball makes approximately 30 full revolutions by the time it reaches its maximum height.

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Key Concepts

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

Rotational Motion
Rotational motion is a key concept when studying objects that spin while they move. Imagine a spinning top or a rotating planet; these objects exhibit rotational motion. In our golf ball scenario, the ball spins around its axis while traveling through the air. The rate of spin is described in terms of angular velocity, denoted by the Greek letter \( \omega \).
  • Angular velocity measures how quickly an object spins, and it is often expressed in radians per second (rad/s).
  • One revolution of a full circle corresponds to \( 2\pi \) radians.
Understanding rotational motion helps us calculate how many full spins the golf ball makes before it stops ascending. The process involves determining the time it spends rotating and its speed in terms of angular displacement.
To find the number of revolutions, we examine the relationship between angular velocity and time: \[ \text{Revolutions} = \frac{\omega t}{2\pi} \] Here, \( \omega \) is the angular velocity and \( t \) is the time determined from the ball's projectile motion.
Vertical Velocity
Vertical velocity is crucial in projectile motion as it dictates the object's rise and fall in the vertical direction. When dealing with angled launches, like our golf ball, we separate the initial velocity into horizontal and vertical components.
  • The initial vertical velocity \( v_{y_i} \) is calculated using trigonometry: \[ v_{y_i} = v \sin\theta \]
  • A golf ball launched at an angle means part of its speed propels it upward.

Next, we use the kinematic equations to find out when this vertical motion ceases—when the vertical velocity decreases to zero. The equation is: \[ v_{y_f} = v_{y_i} - gt \] With \( v_{y_f} = 0 \), we can solve for the time \( t \) it takes for the ball to reach its maximum height. This aids in calculating both the ball's peak altitude and the duration of its rotational motion.
Kinematic Equations
Kinematic equations allow us to analyze motion mathematically, specifically the motion of objects subject to gravity. These equations describe how things move with constant acceleration, like our golf ball, which is subject to gravity.
  • They enable us to calculate displacement, final velocity, and time during an object's motion.
  • One main equation used is: \[ v_f = v_i + at \]

In the golf ball's case, vertical motion falls under constant gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \). We calculated the time to reach the maximum height using:\[ v_{y_f} = v_{y_i} - gt \]Setting \( v_{y_f} \) to zero lets us solve for time \( t \). This calculated \( t \) helps us determine the total number of rotations during the ascent. Understanding these equations gives insights into how different motion components influence the ball's trajectory.
Trigonometry in Physics
Trigonometry in physics often helps break down vectors into components, especially in projectile motion problems. In our example, the initial speed of the golf ball needs to be resolved into its horizontal and vertical components.
  • This process uses trigonometric functions like sine and cosine.
  • For the vertical component, the sine function is crucial: \[ v_{y_i} = v \sin\theta \]

With the launch angle \( \theta \) and initial speed \( v \), we calculate how fast the ball travels upwards. This information, in conjunction with gravity, is used to find how long the ball takes to reach its peak.Trigonometry aids in visualizing motion and helps solve projectile problems by simplifying complex motions into manageable calculations. Embracing trigonometry aids in predictive modeling of how objects will behave in real-world scenarios, like predicting where our golf ball will be in its path at any point in time.

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Most popular questions from this chapter

A diver makes \(2.5\) revolutions on the way from a \(10-\mathrm{m}\) -high platform to the water. Assuming zero initial vertical velocity, find the average angular velocity during the dive.

Two uniform solid cylinders, each rotating about its central (longitudinal) axis at \(235 \mathrm{rad} / \mathrm{s}\), have the same mass of \(1.25 \mathrm{~kg}\) but differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius \(0.25 \mathrm{~m}\), and \((\mathrm{b})\) the larger cylinder, of radius \(0.75 \mathrm{~m} ?\)

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by \(\theta=0.40 e^{2 t}\), where \(\theta\) is in radians and \(t\) is in seconds. Consider a point on the object that is \(4.0 \mathrm{~cm}\) from the axis of rotation. At \(t=0\), what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?

In Fig. 10-31, wheel \(A\) of radius \(r_{A}=10 \mathrm{~cm}\) is coupled by belt \(B\) to wheel \(C\) of radius \(r_{C}=25 \mathrm{~cm}\). The angular speed of wheel \(A\) is increased from rest at a constant rate of \(1.6 \mathrm{rad} / \mathrm{s}^{2}\). Find the time needed for wheel \(C\) to reach an angular speed of 100 rev/min, assuming the belt does not slip. (Hint: If the belt does not slip, the linear speeds at the two rims must be equal.)

Cheetahs running at top speed have been reported at an astounding \(114 \mathrm{~km} / \mathrm{h}\) (about \(71 \mathrm{mi} / \mathrm{h})\) by observers driving alongside the animals. Imagine trying to measure a cheetah's speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering \(114 \mathrm{~km} / \mathrm{h}\). You keep the vehicle a constant \(8.0 \mathrm{~m}\) from the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius \(92 \mathrm{~m}\). Thus, you travel along a circular path of radius \(100 \mathrm{~m} .\) (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah's speed is \(114 \mathrm{~km} / \mathrm{h}\), and that type of error was apparently made in the published reports)

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