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The hammer throw is a track-and-field event in which a \(7.3-\mathrm{kg}\) 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 the Known Quantities

The mass of the ball (hammer) is given as \( m = 7.3 \) kg. The path of the ball forms a circle with a radius \( r = 1.8 \) m. The release angle is \( \theta = 41^{\circ} \). The ball is released, eventually achieving a horizontal distance (range) of \( d = 86.75 \) m.

02

Calculate the Initial Velocity

Using the projectile motion formula for range, we know:\[d = \frac{v^2 \sin(2\theta)}{g}\]Plug in the known values to solve for the initial velocity \( v \):\[86.75 = \frac{v^2 \sin(82^{\circ})}{9.8}\]Solving for \( v \),\[v^2 = \frac{86.75 \times 9.8}{0.9903}\]\[v \approx 29.47 \, \text{m/s}\]

03

Apply the Formula for Centripetal Force

The centripetal force \( F_c \) can be calculated using:\[F_c = \frac{mv^2}{r}\]Substitute the given values:\[F_c = \frac{7.3 \times (29.47)^2}{1.8}\]This yields:\[F_c \approx 3508.92 \, \text{N}\]

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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 concept that describes the trajectory of an object in motion under the influence of gravity. It involves two components: horizontal and vertical motion. In the context of the hammer throw:

• The hammer is released at an angle, combining both the forward and upward movements.
• As it rises, gravity acts on the hammer pulling it downwards, creating a curved path known as a parabolic trajectory.
• The horizontal distance covered, or range, depends on the initial speed and angle of release.

Projectile motion explains how the hammer can travel large distances before landing. By considering only gravity and neglecting air resistance, calculations become more straightforward allowing for predictions of its landing spot.

Circular Motion

Circular motion is key to understanding the initial phase of the hammer throw, as it is spun in a circle before release. This motion has distinct characteristics:

• An object moving in a circle experiences acceleration toward the center of the circle, called centripetal acceleration.
• This requires a force to keep the object moving in its circular path, known as centripetal force.
• For the hammer, the athlete provides this force through the tension in the chain, directed towards the center of the circle.

In our problem, the radius of this circular path is given, and the centripetal force just before release helps us understand the force exerted by the athlete on the hammer.

Initial Velocity Calculation

The initial velocity is a crucial factor in determining the range of the projectile in the hammer throw. It defines how fast the hammer is traveling at the moment it leaves the circular path.

• It's determined using the formula for projectile range: \(d = \frac{v^2 \sin(2\theta)}{g}\).
• This formula relates the distance traveled with initial velocity, angle of release, and gravitational acceleration \(g\).
• By rearranging the equation, the required initial velocity can be calculated given the distance and angle.

In the solution, this calculation provided an initial velocity of approximately \(29.47 \, \text{m/s}\), which was essential for further computations of the centripetal force.