# Applying translational equilibrium summationdisplay f

• Test Prep
• GrandIceLion2233
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Applying translational equilibrium, summationdisplay F x = - F x,wall + T x F x,wall = T cos θ = 2 m g sin θ cos θ = 2(1 . 53 kg)(9 . 8 m / s 2 ) cos θ sin θ = 46 . 7111 N .
Version 069 – Test 4 – swinney – (57305) 8 013 10.0points A constant torque of 20 N · m acts on a wheel rotating about a fixed axis. The applied force acts for 6 . 6 s, during which time the angu- lar speed of the wheel increases from 0 to 14 rad / s. What is the moment of inertia of the wheel? 1. 2 5 015 10.0points
= (20 N · m)(6 . 6 s) 14 rad / s = 9 . 42857 kg · m 2 . 014 10.0points A solid sphere with moment of inertia about its center of mass I cm = 2 5 M r 2 rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy? Two objects of masses 25 kg and 45 kg are connected to the ends of a rigid rod (of negli- gible mass) that is 70 cm long and has marks every 10 cm, as shown. 25 kg 45 kg A B C D E F G H J 10 20 30 40 50 60 Which point represents the center of mass of the sphere-rod combination?
Version 069 – Test 4 – swinney – (57305) 9 2. C 3. A 4. F 5. G correct 6. D 7. J 8. E 9. H Explanation: A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass 2 m and a length . A piece of clay with mass m and velocity v 0 hits one end of the rod, gets stuck and causes the clay-rod system to spin about the pivot point O at the center of the rod in a horizontal plane. Viewed from above the scheme is ω ω v 0 m 2 m (a) (b) (c) Before During After With respect to the pivot point O , what is the magnitude of the initial angular mo- mentum L i of the piece of clay and the final moment of inertia I f of the clay-rod system? After the collisions the clay-rod system has an angular velocity ω about the pivot.