hw 25 - homework 25 HETTIGER CHRISTOF Due 4:00 am 2F...

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homework 25 – HETTIGER, CHRISTOF – Due: Mar 29 2008, 4:00 am 1 Question 1, chap 13, sect 1. part 1 of 1 10 points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force vector F is applied at the other end, at an angle θ to the rod. L m F θ If vector F were to be applied perpendicular to the rod, at what distance d from the axis of rotation should it be applied in order to produce the same torque vector τ ? 1. d = L sin θ correct 2. d = L cos θ 3. d = L 4. d = L tan θ 5. d = 2 L Explanation: The torque the force generates is τ = F L sin θ . Thus the distance in question should be L sin θ . Question 2, chap 13, sect 1. part 1 of 1 10 points A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and per- pendicular to the page. Four forces are ex- erted tangentially to the rims of the wheels, as shown below. F 2 F F F 2 R 3 R What is the magnitude of the net torque on the system about the axis? 1. τ = 5 F R 2. τ = 2 F R correct 3. τ = 0 4. τ = F R 5. τ = 14 F R Explanation: The three forces F give counter-clockwise torques while the other force 2 F gives a clock- wise torque. So the total torque is τ = summationdisplay F i R i = 2 F 3 R + F 3 R + F 3 R + F 2 R = 2 F R . Question 3, chap 13, sect 1. part 1 of 1 10 points A disk of mass M and radius R rotates about its axis. A string is wrapped around the disk and exerts a constant force F = 2 M g at an angle θ , as shown in the diagram.
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homework 25 – HETTIGER, CHRISTOF – Due: Mar 29 2008, 4:00 am 2 F θ R M Find the magnitude of the tangential ac- celeration of a point on the rim of the disk, given that the moment of inertia of the disk is I = 1 2 M R 2 . 1. bardbl vectora t bardbl = 4 g sin θ 2. bardbl vectora t bardbl = 2 g tan θ 3. bardbl vectora t bardbl = 4 g cos θ 4. bardbl vectora t bardbl = 2 g cos θ 5. bardbl vectora t bardbl = 4 g tan θ 6. bardbl vectora t bardbl = 4 g tan θ 7. bardbl vectora t bardbl = 4 g correct 8. bardbl vectora t bardbl = 2 g 9. bardbl vectora t bardbl = 2 g sin θ 10. bardbl vectora t bardbl = 2 g tan θ Explanation: Let : I = 1 2
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