Problem 1:
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 2:
Near the surface of the earth a uniform disk of mass M1 and radius R is pivoted on a frictionless horizontal axle through its center. A small mass M2 is attached to the disk at radius R/2, at the same height as the axle. If this system is released from rest.
| (a) What is the angular acceleration of the disk immediately after it is released? |
| (b) What will be the magnitude of the maximum angular velocity that the disk will reach? |
Solution:
| Concepts, principles, relations
that apply to the problem: The moment of inertia, Newton's 2nd law applied to rotational motion, energy conservation |
| Why do they apply? The system is constrained to rotate about a fixed axis, gravity is responsible for a torque about this axis, the force of gravity is a conservative force. |
| How do they apply? (a) The moment of inertia of the disk about its center is Idisk = ½M1R2. The total moment of inertia of the system about the center of the disk is I = Idisk + M2(R/2)2 = (2M1 + M2)R2/4. When the system is released the torque is t = Ia = M2gR/2. The angular acceleration a when the system is released therefore is a = M2gR/2I, = 2M2g/[(2M1 + M2)R]. (b) Energy conservation. ½Iwmax2 = M2gR/2. wmax2 = M2g R/I = 4M2g/[(2M1 + M2)R]. |
| Details of the calculation: None |
Problem 3:
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 4:
A homogeneous ladder of length L leans against a wall (see figure). The coefficient of friction with the ground is µ1, the coefficient of friction with the wall is µ2. Find the minimal angle q, below which the ladder will slip.
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? In equilibrium: Fy = 0: N1 + f2 = mg (1) Fx = 0: N2 = f1 (2) The torque about the CM is zero: N2sinq + f2cosq + f1sinq = N1cosq (3) (N2 + f1)sinq = (N1- f2)cosq (rewriting (3)) 2f1sinq = (mg - 2f2)cosq (eliminating N1 and N2 from (3) using (1) and (2)) tanq = (mg - 2f2)/(2f1) The angle q is smallest when both f1 and f2 have their maximum value. tanq = (mg- 2m2N2)/(2m1N1) = (N1 + m2m1N1- 2m1m2N1)/(2m1N1) = (1 - m2m1)/(2m1). For smaller angles the ladder will slip. |
| Details of the calculation: None |
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 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 7:
A perfectly uniform billiard ball at rest is struck by a cue at a height h. This causes it to have an initial velocity v0 and an initial angular velocity w0. How far will it go on a surface with coefficient of friction m before pure rolling sets in?
Solution:
| Concepts, principles, relations
that apply to the problem: Sliding friction: F = mN, N = normal force. Friction opposes the relative motion. Rolling: A definite relationship between linear and angular velocity is needed, and therefore a definite relationship between linear and angular momentum. Newton's second law: F = dP/dt, t = dL/dt. |
| Why do they apply? Friction acting at a point away from the CM results in a torque about the CM, thus changing linear as well as angular momentum. |
| How do they apply? Let v0 = v0i and w0 = w0k. The direction of the frictional force depends on v0 and w0. If they are both positive, then F = -mNi = -mmgi. Let us assume that this is the case. The friction results in Dp = -Dpi, DL = -DLk. We need to find the time interval Dt after which the condition for rolling is satisfied and rolling sets in. The condition for rolling is v(t) = -aw(t), or p(t)/m = -aL(t)/I, with v = vi and w = wk and p = pi and L = Lk. |
| Details of the calculation: F = -mmgi = m(dv/dt)i. dv/dt = -mg, v(t) = v0 - mgt. t = dL/dt = I(dw/dt)k = -mmgak. dw/dt = -mmga/I, w(t) = w0 - mmgat/I. Now find the time t when v(t) = -aw(t). v0 - mgt = -aw0 + mmga2t/I. For a sphere I = (2/5)ma2. Therefore v0 - mgt = -aw0 + (5/2)mgt, (7/2)mgt = v0 + aw0. t = (v0 + aw0)/((7/2)mg) is the time when rolling sets in. d = v0t - (1/2)mgt2 is the distance the ball has traveled before this time.
Think
about:
|
Problem 8:
A sphere of radius R, mass M, and radius of gyration k = (I/M)1/2 about any axis through its center rolls with linear velocity v on a horizontal plane. The direction of motion is perpendicular to a vertical step of height h, where h < R. The sphere and step are perfectly rough and inelastic. Show that the sphere will surmount the step upon collision if v2 > 2ghR2(R2 + k2)/(R2 - hR + k2)2.
Solution:
| Concepts, principles, relations
that apply to the problem: DP = FDt, DL = tDt, relationship between linear and angular momentum for rolling motion |
| Why do they apply? The sphere makes an inelastic collision with the step. It starts rotating about the pivot point. Its momentum and angular momentum change. The change in momentum and angular momentum are not independent of each other but are connected by the fact that the pivot point is fixed. |
| How do they apply? Just after the sphere has contacted the pivot point it is rotating about the pivot point. Its kinetic energy is T = (1/2)MR2(da/dt)2 + (1/2)Mk2(da/dt)2. KE of motion of CM + KE of motion about CM For the sphere to surmount the step we need T > Mgh just after the sphere has contacted the pivot point. At that moment, the magnitude of the momentum is MR(da/dt) = Mvsina - FDt = Mv(R-h)/R - FDt. Here Mvsina is the momentum component perpendicular to the direction from the pivot point to the center of the sphere from before the collision and FDt is the impulse due to the tangential force F. At the same time the magnitude of the angular momentum about the CM is I(da/dt) = Iv/R + FRDt. Here Iv/R is the angular momentum before the collision and FRDt is the change in angular momentum because of the impulse. We therefore can write FDt = Mv(R-h)/R - MR(da/dt) = (I/R)(da/dt) - Iv/R2. |
| Details of the calculation: Solving for (da/dt) we have: MR(da/dt) = Mv(R-h)/R -(I/R)(da/dt) + Iv/R2. (R2 + k2)(da/dt) = v(R - h + k2/R), (da/dt) = v(R2 - hR + k2)/[(R2 + k2)R]. T = (1/2)M(R2 + k2)(da/dt)2 > Mgh yields (1/2)v2(R2 - hR + k2)2/[(R2 + k2)R2] > gh, v2 > 2ghR2(R2 + k2)/(R2 - hR + k2)2. |
Problem 9:
A uniform spherical planet of radius a revolves around the sun in a circular orbit of radius r0 and angular velocity w0. It rotates about its axis with angular velocity W0 (period T0) normal to the plane of the orbit. Due to tides raised on the planet by the sun, its angular velocity of rotation is decreasing. Find an expression which gives the orbital radius r as a function of the angular velocity W of rotation and the parameters r0 and T0 at any later or earlier time.
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of angular momentum, Newton's law of gravitation | ||||
| Why do they apply? No external forces act on the sun-planet system, there are no external torques. Therefore the total angular momentum of the system is conserved. Newton's law of gravitation yield a relationship between the period and the orbital radius of circular orbits for planets orbiting a sun (Kepler's third law). | ||||
| How do they apply? The total angular momentum of the system is the sum of the angular momenta of the revolution and the rotation. Let the z-direction be the direction perpendicular to the plane of the orbit. The direction of all angular momenta is the z-direction.
Conservation of angular momentum: | ||||
| Details of the calculation: (2/5)a2W0 + r02[GMs/r03]1/2 = (2/5)a2W + r2[GMs/r3]1/2. r = (1/GMs)[(2/5)a2[(2p/T0) - W] + [r0GMs]1/2]2. |
Problem 10:
A 60 kg woman stands at the rim of a horizontal turntable having a moment of
inertia of 500 kgm2, and a radius of 2 m. The turntable is initially
at rest and is free to rotate about a frictionless vertical axis through its
center. The woman then starts walking around the rim clockwise (as viewed from
above the system) at a constant speed of 1.5 m/s relative to the Earth.
(a) In what direction and with what angular speed does the turntable rotate?
(b) How much work does the women do to set herself and the turntable in motion?
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of angular momentum |
| Why do they apply? The system consists of the woman and the turntable. No external torques act on the system, so the total angular momentum of the system is conserved. It is zero before the woman starts to walk, and it is zero afterwards. |
| How do they apply? (a) When the women walks with angular velocity w = -(v/r)k = -((1.5 m/s)/(2 m))k = -(0.75/s)k, her angular momentum is L = Iw = (-60 kg 4 m2 0.75/s)k = -(180 kgm2/s)k. The angular momentum of the turntable will be L = (180 kgm2/s)k, its angular velocity w = (180 kgm2/s)k/(500 kgm2) = (0.36/s)k. The turntable rotates counterclockwise. (b) The kinetic energy of the women is (1/2)Iw2 = (1/2)240 kgm2(0.75 s)2 = 67.5 J. The kinetic energy of the turntable is (1/2)Iw2 = (1/2)500 kgm2(0.36 s)2 = 32.4 J. The work done by the women is W = (67.5 + 32.4) J = 99.9 J. |
| Details of the calculation: None |
