This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Problem Set 3: Problem 1.
Problem: A hammer thrower is swinging a hammer whose head A has mass M . The head rotates at constant speed, v, in a horizontal circle of radius . The angle of wire BC relative to the horizontal plane is as shown. (a) Solve for the tension in wire BC, T , and the speed v. Express your answers in terms of M , , and gravitational acceleration, g. (b) Compute T and v for M = 7 kg, = 1 m and = 33o . Solution: The following figure shows the forces acting on the hammer head. Specifically, in the vertical direction, we have the weight of the head, M g, where g is gravitational acceleration. In the direction of Cable BC, we have the tension force, T .
.. . . . . . . . . . . . ..... .. . . . . . . . . . . ...... .. . . . .. ..... . . .. ... . . . ... .. . . ... ... . . . ... . ... . ... . . ..... . ... .. . ... . ... ...... .. . . . .... . ..... ... . ..... .. . .... .. . . .. . ..... . . . . . . .. . . . . .. . .. . .. . .. . .. . ... . .. .... . . .. ....... .. . . . . . . . . . . . . . . . ... ... .. . . .. . . T cos T T sin Mg (a) Newton's Second Law tells us that in the vertical direction Fz = T sin  M g = 0 In the radial direction, we have Fn = T cos = M Substituting for T from above yields v 2 = g cot v2 = v2 = T cos M = T = Mg sin ...
View
Full
Document
This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.
 Spring '06
 Shiflett

Click to edit the document details