Problem 1:
A uniform massive rod AB of length L rests at the inclination ±10o with respect to the horizontal with end A in contact with a vertical wall. The two configurations are shown in figures a and b. End B of the rod is supported by a massless cord BC, which is of length 1.2L and is attached at C, a point on the wall above A. What coefficient of static friction between A and the wall is necessary to achieve each configuration?
Solution:
 | Concepts, principles, relations
that apply to the problem: Static friction: Fmax = mN, N = normal force Static equilibrium: F = 0, t = 0 | ||||||
| Why do they apply? In static equilibrium the total force and the total torque about the CM must be zero. | ||||||
| How do they apply? The diagram shows the forces on the rod. The forces acting at the ends of the rod can exert a torque about the CM.
Note: Ff could be positive or negative. | ||||||
| Details of the calculation: We have 3 equations and 3 unknowns. We are, however, only interested in the ratio Ff/FR. Inserting (ii) into (iii) we have Tsin(f - q) - Ffcosq - Tcosfsinq = 0, Ff = [Tsin(f - q) - Tcosfsinq]/cosq. m = |Ff|/N = |Ff|/FR = Ff/(Tcosf) = [sin(f - q) - cosfsinq]/[cosfcosq]. cosf = L cos10o/(1.2L) = 0.82067. f = 34.848o.
|
Problem 2:
Three particles are connected by rigid rods of negligible mass lying along the y-axis as shown.
If the system rotates about the x-axis with angular speed of 2 rad/s, find
(a) the moment of inertia about the x-axis and the total rotational
kinetic energy evaluated from (1/2)Iw2,
and
(b) the linear speed of each particle and the total kinetic energy evaluated
from å(1/2)mivi2.
Solution:
| Concepts, principles, relations
that apply to the problem: The moment of inertia of a system about an axis |
| Why do they apply? We are asked to find the moment of inertia and then use it in a simple calculation. |
| How do they apply? (a) The moment of inertia is I = åmiri2. Here ri is the perpendicular distance of particle i from the x-axis. I = 4 kg 9 m2 + 2 kg 4 m2 + 3 kg 16 m2 = 92 kgm2. The rotational kinetic energy is Trot = (1/2)Iw2 = 46´4/s2 = 184 J. (b) The linear speed of particle i is vi = wri. The linear speed of the 4 kg mass is v = 6 m/s, and its kinetic energy is (1/2)mv2 = 72 J. The linear speed of the 2 kg mass is v = 4 m/s, and its kinetic energy is (1/2)mv2 = 16 J. The linear speed of the 3 kg mass is v = 8 m/s, and its kinetic energy is (1/2)mv2 = 96 J. The sum of the kinetic energies of the three particles is 184 J. |
| Details of the calculation: None |
Problem 3:
A uniform carpenter's square has the shape of an L, as shown in the figure. Locate the center of mass relative to the origin of the coordinate system.
Solution:
| Concepts, principles, relations that apply to the
problem: The center of mass |
| Why do they apply? We are asked to find the CM of an object. |
| How do they apply? We can think of the system as being made up of two subsystems, as shown in the figure.
The CM of the left subsystem lies at xCM = 2 cm, yCM = 9 cm. The CM of the right subsystem lies at xCM = 8 cm, yCM = 2 cm. If the mass of a 1 cm by 1 cm square is 1 unit, then the mass of the left subsystem is 72 units and the mass of the right subsystem is 32 units. We find the CM of the system by treating each subsystem as a separate particle, with all its mass concentrated at its center of mass. Then we have for the whole system xCM = (72 units 2 cm + 32 units 8 cm)/(104 units) = 3.85 cm, yCM = (72 units 9 cm + 32 units 2 cm)/(104units) = 6.85 cm. The CM of the system lies outside of the system. For irregular-shaped objects it is quite common for the CM to lie outside the system. This special point outside the system responds to external forces as if the total mass of the system were concentrated there. |
| Details of the calculation: None |
Problem 4:
Three stars with masses m1, m2, and m3 are forming a peculiar triple-star system, where each of the stars is situated in the corners of an equilateral triangle with a side length d. The stars are attracting each other with gravitational forces. Determine the direction and magnitude of the rotational velocity w which will leave the relative position of the three stars unchanged.
Solution:
| Concepts, principles, relations that apply to the
problem: Orbiting, uniform circular motion, the center of mass (CM) |
| Why do they apply? For the relative positions of the stars to remain unchanged all stars must have circular orbits lying in the plane of the triangle about a common point. The force on each star must be directed towards this point, the center of the circular orbits. We show that this point is the CM. |
| How do they apply? Place the triangle into the xy plane. Place the CM of the system at the origin. M = m1 + m2 + m3, MR = m1r1 + m2r2 + m3r3 = 0. Star 1: F1 = F12 + F13 = Gm1m2(r2 - r1)/d3 + Gm1m3(r3 - r1)/d3. F1 = (Gm1/d3)[m2r2 + m3r3- m2r1 - m3r1 + m1r1 - m1r1] = (Gm1/d3)[MR - Mr1] = -(GMm1/d3)r1. The direction of F1 is towards the origin, towards the CM. For the star to move in a circle of radius r1 we need F1 = mv12/r1, v1 = r1w. (GMm1/d3)r1 = mr1w2, w = (GM/d3)1/2. |
| Details of the calculation: Focusing on star 2 and star 3 we will get the same result for w. w = ±(GM/d3)1/2k. |
Problem 5:
Galileo wants to demonstrate that unequal masses fall with
a uniform acceleration by dropping two spherically symmetric masses
simultaneously from the same height z. Ideally we know that they should hit the
ground at the same time, t = (2z/g)1/2, where g is the acceleration
due to gravity.
Unfortunately for Galileo, the
surface of the first sphere is slightly irregular, so the otherwise negligible
air drag induces a rotation with an angular frequency proportional to the
velocity, w =
Îv1. The second sphere does
not rotate. Assume that Î is constant
and ignore possible heating of the air by friction.
Find the ratio of impact times t1/t2
of the two spheres. The answer should involve only the coefficient
Î and the mass m1 and moment
of inertia I1 of the first sphere.
Solution:
| Concepts, principles, relations
that apply to the problem: Energy conservation |
| Why do they apply? Gravitational potential energy is converted into kinetic energy. The rotating sphere has translational as well as rotational kinetic energy. |
| How do they apply? For the non-rotating sphere we have as a function of the distance z’ below the point of release mgz’ = ½mv2, v2(z’) = 2gz’. This yields t(z’) = (2z’/g)1/2. For the rotating sphere we have as a function of the distance z’ below the point of release m1gz’ = ½m1v12 + ½I1w2 = ½m1v12 + ½I1Î2v12 = ½(m1 + Î2I1)v12, v12 = 2g[m1/(m1 + Î2I1)]z'. This yields t1(z') = (2z'/g’)1/2, with g’ = g[m1/(m1 + Î2I1)]. Therefore t1/t2 = (g/g’)1/2 = [(m1 + Î2I1)/m1)]1/2. |
| Details of the calculation: None |
Problem 6:
A solid sphere toy globe of mass M and radius R rotates freely without friction with an initial angular velocity w0. A bug of mass m starts at one pole N and travels with constant speed v to the other pole S along a meridian in time T. The axis of rotation of the globe is held fixed. Show that during the time the bug is traveling the globe rotates through an angle
Useful
integral:
Solution:
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Concepts, principles, relations
that apply to the problem: |
|
Why do they apply? |
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How do they apply? |
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Details of the calculation: |
Problem 7:
A solid iron cylinder (density = 7.87 g/cm3) of
radius r = 5 cm and length l = 20 cm rolls down a ramp which has an incline of
20o (no sliding). The initial height is 3m above ground.
(a) What is the magnitude of the linear acceleration at half the height?
(b) The cylinder arrives at the bottom of the ramp. What is the angular
momentum of the cylinder about its central axis if it suddenly lifted up from
the ground at the two ends of this axis?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's second law, energy conservation, rolling |
| Why do they apply? We use Newton's 2nd law to find the linear acceleration. The rolling constraint is used to eliminate the frictional force in the equation. |
| How do they apply? (a) mgsin20o – f = ma, rf = Ia, a = a/r (rolling). Here a is the acceleration of the CM, a is the angular acceleration about the symmetry axis passing through the CM. mgsin20o = (m + I/r2)a. a = mgsin20o/(m + I/r2) at any position on the ramp. I = ½ mr2, m = rpr2l = 7.87 g/cm3* p*25 cm2*20 cm = 12.36 kg. a = 9.8m/s2 * sin20o/1.5 = 2.23 m/s2. |
| Details of the calculation: (b) Rolling: w = v/R. Energy conservation: ½ mv2 + ½ Iw2 = mgh, (3/4)mv2 = mgh. v = (4gh/3) = 4*9.8 m/s = 39.2 m/s. L = Iw = (1/2)mrv is the angular momentum about the symmetry axis at the bottom of the ramp. Lifting the cylinder at the two ends does not exert a torque about the axis, therefore L stays constant. L = (1/2)12.36 kg*0.05m* 39.2 m/s = 12.1 kgm2/s. |
Problem 8:
A uniform disk of radius R and mass M is spinning about its diameter with angular velocity w, as shown below. Located on the rim of the disk, at an angle q from the spin axis, is point P, and P is moving with speed vp. Point P is now suddenly fixed. Show that the subsequent linear speed vc of the center of the disk is vc = vp/5.
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's second law: DP = FDt, DL = tDt |
| Why do they apply? To reduce the velocity of point P to zero, an impulse must be delivered. The force acts at a point away from the CM. This results in a torque and an angular impulse about the CM. |
| How do they apply? The impulse changes the momentum of the CM and the angular momentum about the CM in such a way that point P is at rest. This determines the magnitude of DP. The velocity of the CM is given by vc = DP/M. |
| Details of the calculation: The disk rotates about the z-axis. Assume that when it is aligned with the yz-plane the impulse is delivered to the point P on the rim, a distance y = Rsinq from the z-axis and z = Rcosq from the y-axis. DP = FDt i, DL = r ´ DP = -F Rsinq Dt k + F Rcosq Dt j. Just after the impulse, the momentum of the CM is DP and the angular momentum about the CM is L = Izwk - F Rsinq Dtk + F Rcosq Dtj. The velocity of the point P after the impulse is (FDt/M)i - (1/Iz)(Izw - F Rsinq Dt)Rsinq i + (1/Iy)(F Rcosq Dt)Rcosq i = 0. For a disk of radius R we have:   Therefore: FDt/M - wRsinq - FR2sin2q Dt/(MR2/4) + FR2cos2q Dt/(MR2/4) = 0. FDt = (M/5)wRsinq . The speed of point P before the impulse is vp = wRsinq. Therefore FDt = (M/5)vp, vc = FDt/M = vp/5. |
Problem 9:
A homogeneous thin rod of mass m and length 2a slides on a smooth,
horizontal table, one end being constrained to slide without friction in a fixed
straight line. It is initially at rest, with its extension normal to the line,
when it is struck at the free end with an impulse Q parallel to the line.
(a) Determine the initial motion of the rod.
(b) Show that the force exerted by the line on the rod is given by
Q2sinq/[ma((4/3)
-sin2q)2], where
q is
the angle between the rod and the line.
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's second law: F = dP/dt, t = dL/dt. The linear and the angular motion are coupled by a constraint. |
| Why do they apply? The force delivering the impulse acts at a point away from the CM. This results in a torque about the CM. The force of constraint also acts away from the CM and therefore can produce a torque. |
| How do they apply? The impulse is in the x-direction. It will result in momentum P of the CM and angular momentum L about the CM. The force of constraint is in the y-direction. It can change the y-component of P and it can change L. The y coordinate of the CM and the angular coordinate q are linked by constraints. |
| Details of the calculation: (a) initial motion of CM: P = Qi, V = (Q/m)i, initial motion about CM: L = -Qak, w = -(Qa/I)k. moment of inertia of the rod about the CM:  .(b) subsequent motion: There is no force in the x-direction, Vx = Q/m = constant. There is a force in the y-direction which produces linear acceleration of the CM and angular acceleration about the CM. d2y/dt2 = -F/m, dL/dt = F cosq ak = I(d2q/dt2)k. Therefore d2y/dt2 = -I(d2q/dt2)/(m cosq a). (y is the y-coordinate of the CM.) But y and q are linked by constraints. y = a sinq, d2y/dt2 = - a sinq (dq/dt)2 + a cosq (d2q/dt2). Therefore - a sinq (dq/dt)2 + a cosq (d2q/dt2) = -I(d2q/dt2)/(m cosq a), d2q/dt2 = cosq sinq (dq/dt)2/[(4/3) - sin2q] We are supposed to show that F = Q2sinq/[ma((4/3) - sin2q)2]. Using F cosq ak = I(d2q/dt2)k, we are supposed to show that d2q/dt2 = (Q2/I)cosqsinq/[ma((4/3) -sin2q)2]. [To solve this differential equation guess (dq/dt) = A/((4/3) - sin2q)1/2 and verify the solution. This is an obvious guess, since the solution for F is given. A is determined from the initial conditions. At q = 90o, dq/dt = Qa/I = 3Q/ma = 31/2A. A = 31/2Q/ma.] F = I(d2q/dt2)/[cosq a] = [ma/(3cosq)][A2sinq cosq/((4/3) -sin2q)2] = Q2sinq/[ma((4/3) -sin2q)2] |
Problem 10:
Two identical uniform cylinders of radius R each are placed on top of each other next to a wall as shown. After a disturbance, the bottom cylinder slightly moves to the right and the system comes into motion. Find the maximum subsequent speed of the bottom cylinder. Neglect friction between all surfaces.
Solution:
| Concepts, principles, relations
that apply to the problem: Energy conservation |
| Why do they apply? No friction à no torque The cylinders will not rotate. Assume a constraint. The cylinders will stay in contact. Calculate the velocity of the bottom cylinder as a function of X, the horizontal position of its center, using energy conservation. Calculate its acceleration a function of X. At the horizontal position X for which dvx/dt = 0, the acceleration changes sign. The force of constraint now must pull on the cylinders together. This constraint cannot be met. The cylinders will not stay in contact. The lower cylinder will slide to the right with the horizontal velocity it has at the horizontal position X for which dvx/dt = 0. Because its acceleration and its velocity have both pointed in the positive x-direction before the cylinder reached that position X, the speed the lower cylinder has at that position is its maximum speed. |
| How do they apply?
Let M be the mass of each cylinder and R the radius.
Assume the cylinders stay in contact. |
|
Details of the calculation: |
