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Angular VelocityHello! I've been stuck on this question all night and was wondering if anyone had any ideas. In the mens hammer throw field event, athletes compete to throw a “Hammer” as far as possible. A “Hammer” consists of a ball of mass 7.257 kg attached to a cable of length 1.215 meters. Athletes typically spin the hammer 4 times before releasing. The world record for a hammer throw is 86.74 meters by Yuriy Syedikh of the Soviet Union in 1986.So far, I've managed to calculate: • the speed of the ball when released at 45º (neglecting air resistance) = 3.45m/s • tension in cable just before release = 101 (is N correct units?) • min. coefficient of static friction needed by shoes to prevent slipping = 0.290 (assuming Yuri is 100kg) • total distance ball moves assuming 4 revolutions completed = 30.5 metres I've been asked to calculate the final angular velocity of the hammer just before it's released. I understand that the angular velocity is the measurement of the angular displacement per unit time and have found a few relationships to show this, such as v=w/r , w=2πf (where f = frequency in revolutions). However, considering I can't work out the frequency, I'm unsure as to how I can calculate w and therefore calculate v?Secondly, how to calculate the torque Yuri supplied to the ball (assuming constant angular acceleration, and that all mass of the hammer is located in the ball)? I have the relationships: torque = r x F[perpendicular] angular acceleration = torque ÷ (m*r^2) angular acceleration = ∆w/∆t angular acceleration = torque/moment of intertia I found that if angular acceleration is constant, torque is also constant (is this correct?). So: I = mr^2 I = 7/257 x 1.215^2 I = 10.7 kgm^2 However because I haven't been given an angular acceleration value, how can I find torque?? Any help on either questions would be greatly appreciated! Thanks :) |

Quote:
The tension (measured in N because it is a force) is equal to the centripetal force on the hammer. Thus T = Fc = mv^2 / r = m(omega)^2 * r -Dan Edit: Oh yes, the torque. You know that the hammer was thrown upward at an angle of 45 degrees. Sketch a quick Free Body Diagram. You've got the weight acting downward, and the force applied to the hammer at 45 degrees upward, and the centripetal force (acting in the direction of the horizontal component of the applied force at 45 degrees.) The upward component and weight balance, so you can find the upward component of the applied force. The horizontal component of the applied force is what's causing the centripetal force so that does not contribute to the torque. So the only part of the force applied to the hammer that contributes to the torque is the upward component of the applied force. |

Thanks so much! One quick question re torque (and I probably should have mentioned) that the ball is being thrown horizontally. Will that affect this at all? Thanks again!! |

Quote:
Then it's a matter of (torque) = I (alpha). -Dan |

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