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The hammer throw is a track-and-field event in which a 7.3-kg ball (the "hammer") is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion and eventually returns to earth some distance away. The world record for this distance is \(86.75 \mathrm{m},\) achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball is released above the ground rather than at ground level. Furthermore, assume that the ball is whirled on a circle that has a radius of \(1.8 \mathrm{m}\) and that its velocity at the instant of release is directed \(41^{\circ}\) above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.

Step by step solution

01

Identify Known Values

We know the mass of the hammer, which is 7.3 kg. The radius of the circle is 1.8 m, and the angle of release is 41 degrees above the horizontal.

02

Determine Velocity at Release

Since the hammer follows projectile motion after release, we can use the horizontal distance formula.The horizontal range at the maximum point can be calculated using the formula:\[R = \frac{v^2 \sin(2\theta)}{g}\]where R is the range (86.75 m), \(\theta\) is the angle (41°), and g is the acceleration due to gravity (9.81 m/s²).Rearranging the formula to solve for v:\[v = \sqrt{\frac{R \cdot g}{\sin(2\theta)}}\]

03

Calculate Velocity

Substitute the known values into the rearranged formula:\[v = \sqrt{\frac{86.75 \times 9.81}{\sin(82°)}}\]Calculate v using a calculator.

04

Determine Centripetal Force Formula

The centripetal force \(F_c\) required to keep the ball moving in a circle can be expressed as:\[F_c = \frac{m \cdot v^2}{r}\]where m is the mass (7.3 kg), v is the velocity calculated in the previous step, and r is the radius (1.8 m).

05

Calculate Centripetal Force

Substitute the calculated velocity and known values into the centripetal force formula:\[F_c = \frac{7.3 \cdot v^2}{1.8}\]Perform the calculation to find \(F_c\).

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Key Concepts

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

Projectile Motion

When we talk about projectile motion, we're referring to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. This type of motion happens in two dimensions and involves both a horizontal and a vertical component. In our hammer throw scenario, after the hammer is released, it travels in a projectile motion.

• The horizontal component of the velocity is constant since there is no air resistance considered.
• The vertical component is influenced by gravity and changes over time.
• The path followed by the hammer is a parabola, a classic footprint of projectile motion.

Understanding the separate horizontal and vertical motions allows us to predict the path of the projectile.

Hammer Throw

A hammer throw is a captivating event where an athlete spins a heavy ball attached to a wire or grip in circular motions and hurls it as far as possible. During the rotations, centripetal force is vital, as it keeps the hammer moving in a circle. This event tests not only the athlete's strength but also their technique and timing of release. In our exercise, this event is simplified by ignoring air resistance and considering only the gravitational forces acting once released.

• The hammer's path follows the principles of projectile motion after release, curving upward before descending.
• Proper release, precisely at the right speed and angle, maximizes the hammer's flight distance.

Learning to calculate projectile paths and the forces involved helps in understanding the physics governing such sports.

Horizontal Range

The horizontal range of a projectile is the distance it covers in the horizontal direction. Calculating the horizontal range helps you understand how far an object can move before returning to the ground. In our problem, the hammer's horizontal range is given as the world record distance of 86.75 meters.

• The range is influenced by the initial velocity and angle of projection, as seen in the formula: \[R = \frac{v^2 \sin(2\theta)}{g}\]
• Maximizing range involves understanding the angle of release, as shown where the optimal angle is typically 45°, but specifics like angle and initial speed matter too.

These calculations allow athletes to optimize their throws and improve their chances of success in competition.

Velocity Calculation

Calculating the velocity of a projectile at the point of release is crucial in predicting where it will land. This velocity is a combination of both the hammer's speed and the angle of release, which together determine the energy with which it moves forward and upward. Use the rearranged horizontal range formula \[v = \sqrt{\frac{R \cdot g}{\sin(2\theta)}}\] to find the velocity.

• This calculation provides a precise velocity that directly impacts how far the hammer can travel.
• By understanding each component's effect, you can better predict and control the outcomes of the projectile throw.

The velocity at release is not only a key part of sports science but also a fundamental concept in physics, crucial for understanding motion.