Science Student Posted February 5, 2014 Share Posted February 5, 2014 2 metal disks are welded together. One has a radius of R1 = 0.025m with a mass of m1 = 0.80kg and the other has a radius of R2 = 0.05m with a mass of m2 = 1.60kg. A light string is wrapped around the edge of the smaller disk. At the end of the string is a block with mass mb = 1.50kg. The block is released y = 2.00m from the ground. What is the speed of the block just before it strikes the ground? U(1) + K(1) = U(2) + K(2 U(1) + 0 = 0 + K(2) U(1) = mbgy K(2) = (1/2)mb(v^2) + (1/2)(I)(w^2), where I = (1/2)(m1) (R1)^2 + (1/2)(m2) (R2)^2 = 0.0025kgm^2, and w^2 = (v^2)/(R1^2). mbgy = (1/2)mb(v^2) + (1/2)(I)(w^2) 2mbgy = mb(v^2) + (I)(v^2)/(R1^2) v^2 = (2mbgy)/(mb + (I/(R1^2))) v = 3.27m/s But the answer in the textbook is 3.40m/s. The textbook seems to never be wrong. Does anyone see anything wrong with my work? Link to comment Share on other sites More sharing options...
swansont Posted February 5, 2014 Share Posted February 5, 2014 I get 0.00225 for the moment of inertia calculation Link to comment Share on other sites More sharing options...
Science Student Posted February 5, 2014 Author Share Posted February 5, 2014 I get 0.00225 for the moment of inertia calculation Oh thank-you so much!!!! I checked my work about 15 times and could not see that. Link to comment Share on other sites More sharing options...
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