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pnwokolo

Joined: 11 Mar 2010
Posts: 11
Location: Toledo,OH

 Posted: Fri Mar 12, 2010 7:37 pm    Post subject: Calculus-based physics-Please help solve this problem! In testing a design for a yo-yo, an engineer begins by constructing a simple prototype– a string wound about the rim of a wooden disk. She puts an axle riding on nearly frictionless ball bearings through the axis of the wooden disk and fixes the ends of the axle. In order to measure the moment of inertial of the disk, she attaches a weight of mass m to the string and measures how long it takes to fall a givne distance. (a) Assuming the moment of inertia of the disk is given by I, and the radius of the disk is R, find the time for the mass to fall a distance h, starting from rest. (b) She doesn’t have a very accurate stopwatch, but she still wants to get a measurement good to a few percent. She decides a fall time of 2 seconds would work. How big a mass should she use? Imagine you were setting up this experiment and make reasonable estimates of the parameters you need.
hepcj

Joined: 23 Jun 2007
Posts: 125

 Posted: Fri Mar 12, 2010 10:02 pm    Post subject: Look. While I enjoy helping students that are genuinely stuck with physics problems. I am not here to do your physics homework for you. Rather than just posting questions. Please, make an attempt to answer the question before posting, explaining why you are stuck. KE = (1/2)I omega^2 As there is no friction and the rod does not move in its holder, this is equal to the potential energy of the weight falling. PE = mgh Since the mass is under the influence of gravity it is a constant acceleration therefore the average velocity of the mass will be the difference between the starting and final velocities; if the mass starts at zero velocity, this will just be v_f/2. If the mass is timed for a time t then it will fall a height h = v_f/2 *t or v_f = 2h/t The string is wound around the rim of the disc, therefore it rotates with an angular velocity of omega of omega = v_f/R = 2h/(t*R) t = 2h/(omega*R) and mgh = 0.5 I omega^2 omega = (2mgh /I)^(1/2) t = 2h/[(2mgh/I)^(1/2)] t^2 = 4h^2*I/[(2mgh)] = 2h I/(mg) t = [(2hI)/(mg)]^0.5 b) for a 2 second time then you need to use the above equation setting t =2 and using estimates for the unknown parameters, then transpose the equation to make m the subject.
pnwokolo

Joined: 11 Mar 2010
Posts: 11
Location: Toledo,OH

 Posted: Sat Mar 13, 2010 6:52 am    Post subject: Site Admin, Thanks very much for your help. I am very much grateful and will continue to participate in the splung.com physics forum. As you said, I will always do my best to do the problem first before posting it in your site. Once again, thanks very much. Patrick.
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