Dynamics_Part18

Dynamics_Part18 - Problem Set 3: Problem 1. Problem: A...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online