The hammer throw is a track-and fuid event in which 3730 -gg ball (the hammer) is whirled around in a circle several times and released. It then moves upward on the familiar curved path of projectile motion and eventvally returns to the zround some distanca away, The world record for the horizontal distance is 86.75 m , achieyed in 1986 by Xuriy Sodym. Ignore air resigtance and the fact that the ball was released above the ground rather than at ground leyel, furthermore, assume that the ball is whinhed around a circle that has radius of 277 m and that its velocity at the instant of release is directed 56.59 above the horizontal. Find the magritude of the centripetal force acting of the ball just prior to the moment of release, Number \( \square \) Units \( \square \)

To find the magnitude of the centripetal force acting on the hammer just prior to the moment of release, we can use the formula for centripetal force: \[ F_c = \frac{m v^2}{r} \] where: - \( F_c \) is the centripetal force, - \( m \) is the mass of the hammer, - \( v \) is the velocity of the hammer at the moment of release, - \( r \) is the radius of the circular path. ### Step 1: Extract Known Conditions - Mass of the hammer, \( m = 3730 \) g = \( 3.73 \) kg (since \( 1 \) kg = \( 1000 \) g). - Radius of the circular path, \( r = 277 \) m. - The horizontal distance (range) is not needed for this calculation. - The angle of release is not needed for centripetal force calculation. ### Step 2: Calculate the Velocity To find the velocity \( v \) of the hammer at the moment of release, we can use the range formula for projectile motion. The horizontal distance \( R \) is given by: \[ R = \frac{v^2 \sin(2\theta)}{g} \] where: - \( R = 86.75 \) m, - \( \theta = 56.59^\circ \), - \( g = 9.81 \, \text{m/s}^2 \). We can rearrange this formula to solve for \( v \): \[ v^2 = \frac{R g}{\sin(2\theta)} \] ### Step 3: Calculate \( v \) Now, we will calculate \( v \) using the known values. 1. Calculate \( \sin(2\theta) \): \[ 2\theta = 2 \times 56.59^\circ = 113.18^\circ \] \[ \sin(113.18^\circ) \] 2. Substitute into the equation for \( v^2 \): \[ v^2 = \frac{86.75 \times 9.81}{\sin(113.18^\circ)} \] ### Step 4: Calculate the Centripetal Force Once we have \( v \), we can substitute it back into the centripetal force formula. Let's perform these calculations. Calculate the value by following steps: - step0: Calculate: \(86.75\times 9.81\) - step1: Multiply the numbers: \(851.0175\) Calculate or simplify the expression \( \sin(113.18 * \pi / 180) \). Calculate the value by following steps: - step0: Calculate: \(\sin\left(\frac{113.18\pi }{180}\right)\) Calculate or simplify the expression \( 851.0175 / 0.919273 \). Calculate the value by following steps: - step0: Calculate: \(\frac{851.0175}{0.919273}\) - step1: Convert the expressions: \(\frac{\frac{340407}{400}}{\frac{919273}{1000000}}\) - step2: Multiply by the reciprocal: \(\frac{340407}{400}\times \frac{1000000}{919273}\) - step3: Reduce the numbers: \(340407\times \frac{2500}{919273}\) - step4: Multiply: \(\frac{340407\times 2500}{919273}\) - step5: Multiply: \(\frac{851017500}{919273}\) Calculate or simplify the expression \( 3.73 * (925.750566^2) / 277 \). Calculate the value by following steps: - step0: Calculate: \(\frac{3.73\times 925.750566^{2}}{277}\) - step1: Convert the expressions: \(\frac{3.73\left(\frac{462875283}{500000}\right)^{2}}{277}\) - step2: Multiply the numbers: \(\frac{\frac{373\times 462875283^{2}}{5000000^{2}}}{277}\) - step3: Multiply by the reciprocal: \(\frac{373\times 462875283^{2}}{5000000^{2}}\times \frac{1}{277}\) - step4: Multiply the fractions: \(\frac{373\times 462875283^{2}}{5000000^{2}\times 277}\) ### Step 1: Calculate \( \sin(113.18^\circ) \) We found that: \[ \sin(113.18^\circ) \approx 0.919273 \] ### Step 2: Calculate \( v^2 \) Using the range formula: \[ v^2 = \frac{86.75 \times 9.81}{\sin(113.18^\circ)} \approx \frac{851.0175}{0.919273} \approx 925.750566 \] ### Step 3: Calculate the Centripetal Force Now, substituting \( v^2 \) into the centripetal force formula: \[ F_c = \frac{m v^2}{r} = \frac{3.73 \times 925.750566}{277} \] Calculating this gives: \[ F_c \approx 11540.29831 \, \text{N} \] ### Final Answer The magnitude of the centripetal force acting on the hammer just prior to the moment of release is approximately: \[ \boxed{11540.30} \, \text{N} \]