SPRING 2002 FINAL EXAMINATION

PHYSICS 220A #85077

1. The blocks of m1 = 15kg and m2 = 20kg shown in the figure are connected by a string of negligible mass passing over a pulley of radius R = 0.250m. The block on the table moves to the right with a constant acceleration of magnitude a = 2.00m/s2. The coefficient of kinetic friction between the 15-kg block and the table is 0.240. Find the moment of inertia of the the pulley.

Solution.

 

 We write SF = ma for m1:

             x- component:  FT1 - Ffr = m1a;

             y- component:  FN -  m1g = 0.      

 We have Ff = mkFN = mkm1g ,

Then we get FT1 = mkm1g + m1a.

We write SF = ma for m2:

             y- component: m2g - FT2 = m2a, which gives FT2 =  m2g - m2a.

We write St = Ia for the pulley:

             FT2r -  FT1r= I(a/r);

            (m2g - m2a)r - (mkm1g + m1a)r = I(a/r);

            [(20kg)(9.80m/s2) - (20kg)(2.00m/s2)](0.250m) - [(0.240)(15kg)(9.80m/s2) + (15kg)(2.00m/s2)](0.250m)

             = I(2.00m/s2)/(0.250m),

which gives I = 5.67kg·m2.

2. A uniform rod of mass m and length 2l stands vertically on a rough horizontal floor and is allowed to fall. Assuming that slipping has not occurred. Find (a) the angular velocity of the rod when it makes an angle q  with the vertical, and (b) the normal force exerted by the floor on the rod in this position, and the coefficient of static friction involved if slipping occurs when q  = 30°.

Solution.

 

 The forces acting on the rod are the weight mg, normal force FN and the frictional force Ffr.

From the conservation of energy we have

              Ei = Ef; 

              Ei = mgl .  

              Ef = mglcosq  + 1/2Iw2 = mglcosq  + 1/2[1/3m(2l)2]w2.

              mgl = mglcosq  + 1/2Iw2, which gives

              w2 = 3g/2l(1 - cosq ), and w = [3g/2l(1 - cosq )]1/2.

We find the angular acceleration from differentiating w2 = 3g/2l(1 - cosq )

              2wdw/dt = 3g/2lsinq dq/dt, where w/dt = a, and dq/dt = w.

then we have a = (3g/4l)sinq .                     

We write SF = ma for the rod:

 y-comment: mg - FN = may; 

but we have  ay = asinq = lasinq  = (3g/4)sin2q .

then we have FN = mg - m(3g/4)sin2q = (mg/4)(4 - 3sin2q). 

and  

x-component: Ffrmax = msFN = max = mla cosq = (3/4)mgsinq cosq .

At q  = 30°, We have

             ms = [(3/4)mgsinq cosq ]/[(mg/4)(4 - 3sin2q)]

                  = (3sinqcosq)/(4 - 3sin2q)

                  = 0.400.

                  

 

 

3. A wire of length L, Young's modulus Y, and cross-sectional area A is stretched elastically by an amount DL. By Hooke's Law, the restoring force is -kDL. Find (a) k, and (b) the work done in stretching the wire by DL.

Solution.

 

  (a) According to Hook's law |F| = kDL.

Young's modulus is defined as

                  Y = (F/A)/(DL/L)

                     = k(L/A),

which give k = YA/L.

(b) We find the work done from:

                  W = -ò0DL Fdx = -ò0DL-kxdx = ò0DLYA/Lxdx

                      = 1/2YA/L(DL)2.

4. The seal over a circular hole of diameter 1cm in the side of an aquarium tank ruptures. The water level is 1m above the hole and the tank is standing on a smooth plastic surface. What force must an attendant apply to the tank to prevent it from being set into motion?

Solution.

 

 We choose the rupture level as zero level.

Applying Bernoulli's equation, we find the speed with which the water comes from the rupture:

              P1 + 1/2rv12 + rgy1 = P2 + 1/2rv22 + rgy2;

              Pa + 0 + rgh =  Pa + 1/2rv22 + 0;

                           (9.80m/s2)(1m) = 1/2v22, which gives v2 = Ö2gh.

From Newton's second law, we have:

               F = dp/dt = d(mv)/dt = vdm/dt

                  = v2Av2 = (rpd2/4)v22

                  = (1000kg/m3)p(0.01m)2/4(2)(9.80m/s2)(1m)

                  = 1.54N.

OPTIONAL PROBLEMS:

1. A particle of mass m is shot with an initial velocity v0 making an angle q with the horizontal. The particle moves in the gravitational field of the Earth. Find the angular momentum of the particle about the origin when the particle is (a) at the origin, (b) at the highest point of its trajectory, and (c) just before it hits the ground. (d) what torque causes its angular momentum to change?

Solution.

 

 L = r x mv 

(a) Since r = 0, L0 = 0.

(b) At the highest point of the trajectory,

          L = [(v02sin2q /2g)i + (v02sin2q /2g)j] x mvx0i

              = -m(v02sin2q /2g)(v0cosq)k

              = -mv03sin2qcosq /2gk.

(c) L = Ri x mv

         = m[Ri x (v0cosq i - v0sinq j]

         = -mRv0sinq k 

         = (-mv03sin2q sinq /2g)k.

(d) The downward force of  gravity exerts a torque in the -z direction.

 

2. A flywheel of mass 12kg and radius of gyration 20cm is mounted on a light horizontal axle of radius 5cm which rotates on frictionless bearings. A string wound round the axle has attached to its free end a hanging mass of 4kg, and the system is allowed to start from rest. If the string leaves the axle after the mass has descended 3m, what torque must be applied to the flywheel to bring it to rest in 5 revs?

Solution.

 

 We write write SF = ma for the mass:

             mg - FT = ma. 

We have a = ra,  then mg - FT = mra.

For the flywheel, we write St = Ia:

              FTr = Ia = MR2a.

Combining the equations, we have:

              mgr = a(mr2 + MR2), 

             (4kg)(9.80m/s2)(0.05m) = a[(4kg)(0.05m)2 + (12kg)(0.2m)2], 

which gives a = 4 rad/s2.

From conservation of energy for the system, we have:

              Ei = Ef.

              Ei = mgh;

              Ef = 1/2mv + 1/2mIw2

                         = 1/2mr2w2 + 1/2MR2w2.

             mgh = 1/2mr2w2 + 1/2MR2w2;

             w2 = 2mgh/(mr2 + MR2) = 480rad2/s2.

In the final stage, the work done by the retarding torque in the 5 rev = 5(2p) rads must equal the kinetic energy possessed by the flywheel before the torque is applied. Thus

              - 10pt = 1/2Iw2;

               - 10pt = 1/2MR2w2;

               - 10pt =  1/2(12kg)(0.2m)2(480rad2/s2), which gives

               t = - 3.67N·m.