Physics 325 Discussion 2 Solution2/2/2015Problem 1+ˆz+ˆrθTmgThe first thing we need to do is draw a free-body diagram and writethe forces in the cylindrical coordinate system (see right). The tensionneeds to be broken down into ˆr and ˆz components. Summing the forcesin each coordinate direction:Fr= −T sin θ (1.1)Fz= T cos θ − mg (1.2)Fφ= 0 (1.3)Now, we can use these expressions in the given form of Newton’s secondlaw, respectively:−T sin θ = m¨r − r˙φ2(1.4)T cos θ − mg = m¨z (1.5)0 = mr¨φ + 2 ˙r˙φ(1.6)These equations can be simplified by noting the mass moves in a horizontal circle, such that ˙z = ¨z = 0,and since the mass is fixed to a string, this also gives ˙r = ¨r = 0. Cancelling these terms, note thenthat Equation (1.4) is reminiscent of the equation for the centrifugal force under uniform circular motion:Fc= mrω2= mv2/r, we will use this relationship shortly. From the diagram, we can write r = L sin θ.From Equation (1.5), we can just write the tension as:T =mgcos θ(1.7)From Equation (1.6),¨φ = 0, which means˙φ is a constant. Because the mass is moving in a circle in uniformcircular motion (i.e.˙φ is a constant), then we can write˙φ = v/r (see the note above). Now, we can writeEquation (1.4) as:T sin θ =mv2r(1.8)⇒ v2=rT sin θm=rg sin θcos θ=gL sin2θcos θ(1.9)⇒ v =sgL sin2θcos θ(1.10)Problem 2The forces given in this problem are the linear drag force and the normal force:Fr= −c ˙r (1.11)Fφ= N (1.12)1We can then use the form of Newton’s second law given in the statement of the problem:Fr= −c ˙r = m¨r − r˙φ2(1.13)Fφ= N = mr¨φ + 2 ˙r˙φ(1.14)The equation relevant to this problem is Equation (1.13), where Equation (1.14) gives the normal force. Asnoted in the problem, the rod rotates at a constant angular velocity:φ(t) = ωt ⇒˙φ = ω (1.15)We can solve the differential equation given by Equation (1.13) by guessing the form of the solution asr(t) = eλt. At this point, we need to solve for the value of λ that solves the differential equation for thisfor the guessed form of r. Using this, we can write ˙r = λeλt= λr and ¨r = λ2eλt= λ2r, such that we canrewrite Equation (1.13) as:−cλr = mλ2r − mrω2φ2(1.16)0 =mλ2+ cλ − mω2 r (1.17)We see that we can solve for λ as:λ =−c ±√c2+ 4m2ω22m(1.18)As expected the second order differential equation gives two solutions for r. Thus, in general, we mustwrite r as a linear combination of the two solutions:r(t) = aeλ1t+ beλ2t(1.19)For any given set of boundary conditions, we can then find a and b that gives the expression for r(t). Giveninitial conditions, r = r0and dr/dt = 0 at t = 0, we can solve by evaluating Equation (1.19) at the givenpoint:r(0) = a + b (1.20)drdtt=0=aλ1eλ1t+ bλ2eλ2t t=0= aλ1+ bλ2(1.21)Then solving:a + b = r0, aλ1+ bλ2= 0 (1.22)for a and b gives:a =r0λ2λ2− λ1(1.23)b =r0λ1λ1−
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