This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PHY 2053 Recitation Week # 13 Ch. 9 # 52, 72. Ch. 10 # 22, 24, 27 9-52. A helicopter has two blades as shown, each of which has a mass of 240 kg and can be approximated as a thin rod of length 6.7 m. The blades are rotating at an angular speed of 44 rad/s. (a) What is the total moment of inertia of the two blades about the axis of rotation? (b) Determine the rotational kinetic energy of the spinning blades. 52. REASONING Each blade can be approximated as a thin rod rotating about an axis perpendicular to the rod and passing through one end. The moment of inertia of a blade is given in Table 9.1 as 2 1 3 ML , where M is the mass of the blade and L is its length. The total moment of inertia I of the two blades is just twice that of a single blade. The rotational kinetic energy KE R of the blades is given by Equation 9.9 as 2 1 R 2 KE I ϖ = , where ω is the angular speed of the blades. SOLUTION a. The total moment of inertia of the two blades is ( 29 ( 29 2 2 2 2 2 1 1 2 2 3 3 3 3 240 kg 6.7 m 7200 kg m I ML ML ML = + = = = ⋅ b. The rotational kinetic energy is ( 29 ( 29 2 2 2 6 1 1 R 2 2 KE 7200 kg m 44 rad/s 7.0 10 J I ϖ = = ⋅ = × 9-72. The drawing shows a person (weight W = 584 N, L 1 = 0.840 m, L 2 = 0.410 m) doing push- ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. 1 72. REASONING The drawing shows the forces acting on the person. It also shows the lever arms for a rotational axis perpendicular to the plane of the paper at the place where the person’s toes touch the floor. Since the person is in equilibrium, the sum of the forces must be zero. is in equilibrium, the sum of the forces must be zero....
View Full Document
This note was uploaded on 09/21/2011 for the course PHY 1500 taught by Professor Staff during the Spring '09 term at University of Central Florida.
- Spring '09