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The hammer throw is a track-and-field event in which a 7.30 kg 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 eventually returns to the ground some distance away. The world record for the horizontal distance is 86.75 m, achieved in 1986 by Yurly Sedykh. Ignore air resistance and the fact that the ball was released above the ground rather than at ground level. Furthermore, assume that the ball is whirled around a circle that has a radius of 2.66 m and that its velocity at the instant of release is directed 22.5∘ above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release. Number _____________ Units __________
The hammer throw is a track-and-field event in which a 7.30 kg 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 eventually returns to the ground some distance away. The world record for the horizontal distance is 86.75 m, achieved in 1986 by Yurly Sedykh. Ignore air resistance and the fact that the ball was released above the ground rather than at ground level. Furthermore, assume that the ball is whirled around a circle that has a radius of 2.66 m and that its velocity at the instant of release is directed 22.5∘ above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release. Number _____________ Units __________
Final answer:The magnitude of the centripetal force acting on a 7.30 kg hammer, whirled in a circle with a radius of 2.66 m and traveling at a velocity of 7 m/s, is approximately 134.46 Newtons.Explanation:To find the magnitude of the centripetal force on the hammer at the instant before its release, we must first calculate the centripetal acceleration. The formula for centripetal force is Fc = m * ac, where m is the mass of the object, and ac is the centripetal acceleration. Centripetal acceleration can be found using the formula ac = v2 / r, where v is the velocity of the object and r is the radius of the circular path.Given a velocity of 7 m/s and a radius of 2.66 m, the centripetal acceleration is:ac = v2 / r = 72 / 2.66≈ 18.42 m/s2With the centripetal acceleration and the mass of the hammer, 7.30 kg, the centripetal force is:Fc = m * ac = 7.30 kg * 18.42 m/s2≈ 134.46 NTherefore, the magnitude of the centripetal force acting on the hammer just prior to the moment of release is approximately 134.46 Newtons....