PHYSICS 220A FINAL EXAMINATION SPRING 2001

1. An Atwood's machine consists of two masses, m1 and m2, which are connected by a massless inelastic cord that passes over a pulley, Fig. If the pulley has radius R and moment of inertia I about its axle, determine the acceleration of the masses m1 and m2. Assume m2 > m1.

 Solution. We assume that m1 > m2 and choose the coordinates show on the force diagram. Note that we take the positive direction in the direction of the acceleration for each object. Because the linear acceleration of the masses is the tangential acceleration of the rim of the pulley, we have

                  a = a = aR0.

               We write SFy = may for m2,

                  m2g - FT2 = m2a.

               We write SFy = may for m1,

                  FT1 -m1g = m1a.

               We write St = Ia for the pulley about its axle:

                  FT2R0 - FT1R0 = Ia = Ia/R0, or    F - F = Ia/R02.

               If we add the two force equations, we get

                  FT1 - FT2 = (m1 + m2)a + (m1 - m2)g.

               When we add these two equations, we get

                  a = (m2 - m1)g/(m1 + m2 + I/R02).

        

          

  

2. A light rigid rod 1.00m in length rotates in the xy plane about a pivot through the rod's center. Two particles of mass 4.00kg and 3.00kg are connected to its ends (Fig. 2) Determine the angular momentum of the system about the origin at the instant the seed of each particle is 5.00m/s.

Solution. The moment of inertia of the compound object its center is:

             I = S mr2 =(4.00kg)(0.500m)2 + (3.00kg)(0.500m)2

               = 1.75kg.m2

              Its angular speed is

                w = v/r = (5.00m.s)/(1/2)(1.00m) = 10.0rad/s

              By the right-hand rule, we find that this is directed out of the plane.

             So its angular momentum is 

                L = Iw = (1.75kg.m2)(10.0rad/s)

                   = 17.5kg.m2/s out of the plane.

            We could also compute this from:

                L = S mr x v

                   = (4.00kg)(0.500m)(5.00m/s)     out of plane

                       + (3.00kg)(0.500m)(5.00m/s)  out of plane

                   = 17.5kg.m2/s  out of plane.

  

  3. A 15-m uniform ladder weighing 500N rests against a frictionless wall. the ladder makes a 60.0° angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter is 4.00m from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00m up, what is the coefficient of static friction between ladder and ground?

Solution. Refer to the free-body diagram.

       (a) SFx = f - nw  = 0    

            SFy = ng -800N - 500N = 0

         Taking torques about an axis at the foot of the ladder,

             -(800n)(4.00m)sin30° - (500N)(7.5m)sin30° + nw(15.0m)cos30° = 0

        solving the torque equation for nw,

              nw = [(4.00m)(800N) + (7.5m)(500N)]tan30°/(15.0m) = 267.5N

        Next substitute this value into the Fx equation to find 

               f = nw = 268N in the positive x direction.

         Solving the equation SFy = 0 gives

             ng = 1300N in the positive y direction.

         (b) In this case, the torque equation St = 0 gives

              -(9.00m)(800N)sin30° - (7.50m)sin30° + (15.0m)(nw)sin60° = 0

          or      nw = 421N.    

                Since f = n = 421N and f = fmax = mng,    we find

                          m = fmax/ng = 421N/1300N = 0.324.

  

                              Fig 3

 4. A large storage tank filled with water develops a small hole in its side at a point 16m below the water level. If the rate flow from the leak is 2.5 x 10-3m3/min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.

Solution. Assuming the top is open the the atmosphere then P1 = Pa. 

      (a) P1 + 1/2rv12 + rgy1 = P2 + 1/2rv22 + rgy2

           A1 >> A2, so v1 << v2.

          Assuming v1 = 0 and P1 = P2 = P0,

                 v2 = Ö(2gy1) = Ö[2(9.80m/s2)(16m)] = 1.77m/s.

      (b) Flow rate = ( 2.5 x 10-3m3/min)(60min/s) = 4.167 x 10-5m3/s

           Flow rate = A2v2 = (pd2/4)(17.7m/s) = 4.167 x 10-5m3/s

          Thus,           d = 1.73 x 10-3m = 1.73mm.          

  

5. (Optional) Two blocks m1 = 15kg and m2 = 20kg are connected by a string of negligible mass passing over a pulley of radius R = 0.025m and mass M = 2kg.   The coefficient of kinetic friction between the block m1 and the horizontal surface is 0.015. Calculate the speed of the blocks when the block m1 moves 2m along the horizontal surface from rest. Assume the pulley is a uniform disk with frictionless bearing, and the string is nonslipping on the pulley. Use conservation of energy to solve this problem.

Solution. From conservation of  energy we have

             Ki + Ui = Kf + Uf. + Wfr

            Ki = 0, Ui = m2gh,  where h = 2m.

            Kf = 1/2(m1 + m2)v2 + 1/2Iw2 = 1/2(m + m + I/R2)v2,        Uf = 0

            where  w = v/R, I = 1/2MR2.

           Wfr = Ffrd = mkm1gd

          Solving the equations, we have

            v2 = (m2gh - mkm1gd)/(1/2)(m1 +m2 + 1/2M)

                =  [(20kg)(9.80m/s2)(2m) - (0.015)(15kg)(9.80m/s2)(2m)]

                    /[(1/2)(15kg + 20kg  +(1/2)(2kg)]  

               = 21.53(m/s)2

                    v = 4.6m/s