/*! 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 53 (III) A hammer thrower accelerat... [FREE SOLUTION] | 91Ó°ÊÓ

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(III) A hammer thrower accelerates the hammer (mass \(=7.30 \mathrm{kg}\) ) from rest within four full turns (revolutions) and releases it at a speed of 26.5 \(\mathrm{m} / \mathrm{s}\) . Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius \(1.20 \mathrm{m},\) calculate \((a)\) the angular acceleration, \((b)\) the (linear) tangential acceleration, \((c)\) the centripetal acceleration just before release, \((d)\) the net force being exerted on the hammer by the athlete just before release, and \((e)\) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

Short Answer

Expert verified
Angular acceleration: 9.69 rad/s², Tangential acceleration: 11.63 m/s², Centripetal acceleration: 585.42 m/s², Force: 4275.83 N, Angle: 1.14°.

Step by step solution

01

Convert Linear Speed to Angular Speed

The final linear speed, \( v = 26.5 \ \mathrm{m/s} \). To find the angular speed \( \omega \), use the relation: \( v = r\omega \). Here, \( r = 1.20 \ \mathrm{m} \).\[ \omega = \frac{v}{r} = \frac{26.5}{1.20} \approx 22.08 \ \mathrm{rad/s} \]
02

Calculate Total Angular Displacement

In four full turns, the angular displacement \( \theta \) is \( 4 \times 2\pi \) radians.\[ \theta = 8\pi \ \mathrm{rad} \]
03

Calculate Angular Acceleration (a)

Using the equation for angular motion, \( \omega^2 = \omega_0^2 + 2\alpha\theta \), and knowing the initial angular speed \( \omega_0 = 0 \), solve for \( \alpha \).\[ \alpha = \frac{\omega^2}{2\theta} = \frac{(22.08)^2}{2 \times 8\pi} \approx 9.69 \ \mathrm{rad/s^2} \]
04

Determine Linear Tangential Acceleration (b)

The linear tangential acceleration \( a_t \) is related to angular acceleration \( \alpha \) by \( a_t = r\alpha \).\[ a_t = 1.20 \times 9.69 \approx 11.63 \ \mathrm{m/s^2} \]
05

Calculate Centripetal Acceleration (c)

Centripetal acceleration \( a_c \) is given by \( a_c = v^2 / r \).\[ a_c = \frac{(26.5)^2}{1.20} \approx 585.42 \ \mathrm{m/s^2} \]
06

Calculate Net Force (d) Being Exerted

The net force \( F \) is mass \( m \) times the net acceleration, which includes both tangential and centripetal components acting perpendicularly:\[ F = m \sqrt{a_t^2 + a_c^2} \]\[ F = 7.30 \sqrt{(11.63)^2 + (585.42)^2} \approx 4275.83 \ \mathrm{N} \]
07

Find the Angle (e) with Respect to Radius

The angle \( \theta_F \) of the force with respect to the radius can be calculated using:\[ \tan \theta_F = \frac{a_t}{a_c} \]\[ \theta_F = \arctan \left(\frac{11.63}{585.42}\right) \approx 1.14^\circ \]

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Key Concepts

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

Angular Acceleration
Angular acceleration is a key concept when discussing rotational motion. It describes the rate at which an object's angular velocity changes over time.
In the case of the hammer thrower, the hammer accelerates uniformly, and we use the formula for angular acceleration, \( \alpha \), which is given by:
  • \( \alpha = \frac{\omega^2}{2\theta} \)
Here, \( \omega \) represents the final angular speed, and \( \theta \) is the angular displacement. Angular acceleration has units of radians per second squared (rad/s²).
Practically, this means the hammer is speeding up at a constant rate as it moves around its circular path. Calculating \( \alpha \) involves knowing the total angular displacement and the final angular speed, as shown in the exercise solution.
This uniform acceleration is crucial as it ensures the hammer achieves the needed speed for a successful throw.
Centripetal Acceleration
Centripetal acceleration is a fundamental concept that comes into play in any circular motion. It is the acceleration directed towards the center of the circle, keeping the object moving in a circular path.
For the hammer thrower, the centripetal acceleration \( a_c \) is calculated by the formula:
  • \( a_c = \frac{v^2}{r} \)
The variables \( v \) and \( r \) represent the linear speed and radius of the circular path, respectively. The units are meters per second squared (m/s²).
As the hammer's speed increases, the centripetal acceleration grows significantly, emphasizing its dependence on the square of the velocity. Just before the release of the hammer, this type of acceleration reaches a peak value, ensuring the hammer remains in its path and doesn't fly off tangent to the circle.
Tangential Acceleration
Tangential acceleration is another important concept necessary to understand motions involving rotation. This type of acceleration is tangent to the circle and reflects the change in the linear speed along the circle's path.
In relation to angular acceleration, the tangential acceleration \( a_t \) is expressed as:
  • \( a_t = r\alpha \)
Where \( r \) is the radial distance from the center of the circle, and \( \alpha \) is the angular acceleration. The units are meters per second squared (m/s²).
This acceleration is responsible for increasing the hammer's speed as it continues along its circular path. In the context of the hammer throw, it is the component of the acceleration that changes the speed of the hammer along the circular path. Without it, there would be no increase in the hammer's speed, which is critical for achieving the desired release velocity.

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