a) An iron ring is to fit snugly on a cylindrical iron rod.At 20 o C, the inner
ID: 1682480 • Letter: A
Question
a) An iron ring is to fit snugly on a cylindrical iron rod.At 20oC, the inner diameter of the ring is 6.420 cm. Toslip over the rod, the ring must be larger than the rod diameter,equal to 6.420 cm, by about 0.008 cm. To what temperature must thering be brought if its hole is to be large enough so it will slipover the rod? Is this temperature lower or higher than the roomtemperature? The coefficient of linear expansion of iron is12×106 1/Co. b) A mass attached to the end ofa string revolves in a circle on a frictionless tabletop. The otherend of the string passes through a hole in the table. Initially,the mass revolves with the speed of 2.4 m/s in a circle of radius0.70 m. The string is then pulled slowly through the hole so thatthe radius is reduced to 0.38 m. What is the speed of the mass now?What is the work done by the puller of the string during theprocess?
Explanation / Answer
L1= original diameter of ring= 6.420 cm =0.06420m L2= New diameter of ring = 0.06420 + 0.00008 m =0.06428m = coefficient of linear expansion = 12 x10-6/C, T1= 20.C, T2= new temperature L2= L1( 1+ {T2-T1}) 0.06428 = 0.06420 (1+ 12x 10-6{T2-20}) 6428/6420 = (1 - 0.00024 + 12x 10-6T Solving for T2= 123.8.C That's the answer to the first part b) For solving IInd part . we use conservation of angularmomentum Initial angularmomentum = I Where I = initial moment of inertia As the mass is rotating around a hole, it takes the shape ofring I (ring) = mr2= m (0.70)2= 0.49m ( wherem = mass is mass of block) = v/r = 2.4/0.70= 3.43 rad/second Initial angularmomentum = 0.49(m)(3.43) Final angular momentum = I'' I' = m(0.38)2= m(0.1444) I= I'' 0.49(m)(3.43) = m(0.1444)' cancelling m on both sides, we get ' = final angular velocity = 11.64 rad/second speed of mass ='r'= 11.64x0.38 = 4.42 m/s approx c)Work done by puller of spring = change in K.E. of mass Initila K.E. of mass = 1/2 I2= 1/2(0.49m)(3.43)2= 2.88 m J apprx Final K.E. = 1/2 I''2= 1/2m(0.1444)(11.64)2= 9.78 m J approx Work done = 9.78 m -2.88 m = 6.9 m J approx To get the exact work done, we need the mass of block c)Work done by puller of spring = change in K.E. of mass Initila K.E. of mass = 1/2 I2= 1/2(0.49m)(3.43)2= 2.88 m J apprx Final K.E. = 1/2 I''2= 1/2m(0.1444)(11.64)2= 9.78 m J approx Work done = 9.78 m -2.88 m = 6.9 m J approx To get the exact work done, we need the mass of blockRelated Questions
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