PHY183-Lecture25pre - Example: Hammer Throw Throw the...

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1 March 1, 2012 Physics for Scientists&Engineers 1 1 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 25 Angular Acceleration Examples (with a little help from MC Hammer) March 1, 2012 Physics for Scientists&Engineers 1 2 Example: Hammer Throw Throw the “hammer”, a 12 cm diameter ball attached to a grip by a steel cable, a maximum distance The hammer’s total length is 121.5 cm, and its total weight is 7.26 kg (4 kg for women) The athlete has to accomplish the throw while not leaving a circle of radius 7 feet (= 2.135 m) March 1, 2012 Physics for Scientists&Engineers 1 3 Example: Hammer Throw Yipsi Moreno (Cuba) 73.33m, Paris ‘03 Sergei Litvinov (USSR) 84.80m, Seoul ‘88 March 1, 2012 Physics for Scientists&Engineers 1 4 Hammer Throw (2) Litvinov took seven turns before releasing hammer. These took 1.52 s, 1.08 s, 0.72 s, 0.56 s, 0.44 s, 0.40 s, and 0.36 s Frequency increased linearly with each turn => constant angular acc. Question: what is value of α ? Answer: total time = 1.52 s+…+ 0.36 s = 5.08 s; total angle: Constant angular acceleration: θ all = 7 2 π = 14 44.0 = 1 2 t 2 = 2 t 2 = 2 44.0 (5.08 s) 2 = 3.41 s -2 March 1, 2012 Physics for Scientists&Engineers 1 5 Example: Hammer Throw (4) Question: Assuming that the radius of the circle on which the hammer moves is 1.67 m (= length of hammer + arms of the athlete), what is the linear speed with which the hammer gets released? Answer: Under constant angular acceleration from rest for a period of 5.08 s, the final angular velocity reached is Velocity: ω = t = (3.41 s -2 ) (5.08 s)=17.3 s -1 v = r = (1.67 m) (17.3 s -1 ) = 28.9 m/s March 1, 2012 Physics for Scientists&Engineers 1 6 Example: Hammer Throw (5) Question: What is the centripetal acceleration and centripetal force that the hammer thrower has to exert on the hammer right before the hammer gets released? Answer: The centripetal acceleration right before release is given by: With a mass of 7.26 kg for the hammer, the centripetal force required is then Same force as weight of 370 kg object! a c = v = (28.9 m/s) (17.3 s -1 ) = 500. m/s 2 F c = ma c = (7.26 kg) (500 m/s 2 ) = 3630 N
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2 March 1, 2012 Physics for Scientists&Engineers 1 7 Cutting Curves Cars move through U-turn shown at constant speed. Coefficient of friction between the tires and the road is ߤ s = 0.9. Inner radius of curve r = 8.3 m, outer radius R = 22.2 m. How much time will it take
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This note was uploaded on 04/02/2012 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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PHY183-Lecture25pre - Example: Hammer Throw Throw the...

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