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EP 222: Classical Mechanics Tutorial Sheet 5: Solution This Tutorial Sheet Contains Problems Related to Angular Momentum, Inertia Tensor, and Rigid Body Motion

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EP 222: Classical Mechanics Tutorial Sheet 5: Solution This Tutorial Sheet Contains Problems Related to Angular Momentum, Inertia Tensor, and Rigid Body Motion EP 222: Classical Mechanics Tutorial Sheet 5: Solution This tutorial sheet contains problems related to angular momentum, inertia tensor, and rigid body motion. 1. Three equal point masses m are located at (a; 0; 0), (0; a; 2a), and (0; 2a; a). Find the principal moments of inertia about the origin and a set of principal axes. Soln: By using the formulas for various components of inertia tensor, one can easily calculate it to be 0 10ma2 0 0 1 I = @ 0 6ma2 −4ma2 A 0 −4ma2 6ma2 on setting up the characteristic polynomial det(I − λI) = 0; leads to eigenvalues λ = 2ma2; 10ma2; 10ma2, last two of which are degenerate. One 0 0 1 0 1 1 0 0 1 can show that the corresponding eigenvectors are p1 1 , 0 , and p1 1 2 @ A @ A 2 @ A 1 0 −1 leading to principle axes directions u^ = p1 ^j + k^ , u^ = ^i, and u^ = p1 ^j − k^ 1 2 2 3 2 2. Obtain the inertia tensor of a system, consisting of four identical particles of mass m each, arranged on the vertices of a square of sides of length 2a, with the coordinates of the four particles given by (±a; ±a; 0). Soln: Consider four identical particles of mass m, arranged on the vertices of a square, with sides of length 2a, as shown Y m m (−a,a) (a,a) O X (−a,−a) (a,−a) m m Noting that for all particles, the z coordinate is zero (zi = 0), we obtain 4 X 2 2 2 Ixx = mi(yi + zi ) = 4ma i=1 4 X 2 2 2 Iyy = mi(xi + zi ) = 4ma i=1 4 X 2 2 2 Izz = mi(xi + yi ) = 8ma i=1 1 Now 4 X 2 2 2 2 Ixy = − mixiyi = −m a − a − a + a = 0 = Iyx i=1 4 X Ixz = − mixizi = −m(0 + 0 + 0 + 0) = 0 = Izx i=1 4 X Iyz = − miyizi = −m(0 + 0 + 0 + 0) = 0 = Izy i=1 Thus, tensor of inertia is diagonal here, due to the high symmetry of the problem 0 4ma2 0 0 1 I = @ 0 4ma2 0 A 0 0 8ma2 3. A rigid body consists of three point masses of 2 kg, 1 kg, and 4 kg, connected by massless rods. These masses are located at coordinates (1, -1,1), (2,0,2), and (-1,1,0) in meters, respectively. Compute the inertia tensor of this system. What is the angular momentum vector of this body, if it is rotating with an angular veloctiy ! = 3^i−2^j+4k^? Soln: We have X 2 2 Ixx = mi(yi + zi ) = 2(1 + 1) + 1(0 + 4) + 4(1 + 0) = 12 i X Ixy = − mixiyi = −2(−1) − 1(0) − 4(−1) = 6 = Iyx i X Ixz = − mixizi = −2(1) − 1(4) − 4(0) = −6 = Izx i X 2 2 Iyy = mi(xi + zi ) = 2(2) + 1(8) + 4(1) = 16 i X Iyz = − miyizi = −2(−1) − 1(0) − 4(0) = 2 = Izy i X 2 2 Izz = mi(xi + yi ) = 2(2) + 1(4) + 4(2) = 16 i Therefore 0 12 6 −6 1 I = @ 6 16 2 A −6 2 16 Given angular velocity can be expressed as 0 3 1 ! = @ −2 A 4 2 We know that, in the matrix form, one can write 0 1 0 1 0 1 Lx Ixx Ixy Ixz !x @ Ly A = @ Iyx Iyy Iyz A @ !y A Lz Izx Izy Izz !z Thus, for this case 0 1 0 1 0 1 0 1 Lx 12 6 −6 3 0 @ Ly A = @ 6 16 2 A @ −2 A = @ −6 A Lz −6 2 16 4 42 or L = −6^j + 42k^ 4. Obtain the moment of inertia tensor of a thin uniform rod of length l, and mass M, assuming that the origin of the coordinate system is at the center of mass of the rod. Soln: We assume that the rod is lying in the xy plane, making an angle θ with respect to the x axis, as shown in the figure below Y P r dr θ X O Because the rod lies in the xy plane with z = 0, therefore, all its off-diagonal compo- nents involving z coordinate will be zero. Nonzero components are Z Z 2 2 2 Ixx = dm(y + z ) = dmy Z Z 2 2 2 Iyy = dm(x + z ) = dmx Z 2 2 Izz = dm(x + y ) Z Ixy = − dmxy m We define a linear density λ = l , so that dm = λdr, x = r cos θ, and y = r sin θ. Note 3 that here r varies from −l=2 to l=2, while θ is constant. Therefore, l Z 2 2 2 2 ml 2 Ixx = λ sin θ r dr = sin θ l 12 − 2 l Z 2 2 2 2 ml 2 Iyy = λ cos θ r dr = cos θ l 12 − 2 l Z 2 2 2 ml Izz = λ r dr = l 12 − 2 l Z 2 2 2 ml Ixy = −λ cos θ sin θ r dr = − sin θ cos θ l 12 − 2 So that 0 ml2 2 ml2 1 12 sin θ − 12 sin θ cos θ 0 ml2 ml2 2 I = @ − 12 sin θ cos θ 12 cos θ 0 A ml2 0 0 12 5. Obtain the moment of inertia tensor of a thin uniform ring of radius R, and mass M, with the origin of the coordinate system placed at the center of the ring, and the ring lying in the xy plane. Soln: Consider a ring of mass m and radius R, as shown Y R dθ dθ R θ X Let us compute I for this zz Z 2 2 Izz = dm(x + y ) Mass is distributed uniformly on the ring, therefore, one can define a linear mass density λ M λ = 2πR Mass dm of a infinitesimal segment of ring which subtends angle dθ on the center is M 2 2 2 dm = λRdθ = 2π dθ. This element is on a circle of radius R, therefore, x + y = R . Thus Z 2 Z 2π M 2 MR 2 Izz = R dθ = dθ = MR 2π 2π 0 4 Let us calculate other elements Z 2 2 Ixx = dm(y + z ) Ring is in the xy plane, so z = 0, and using plane polar coordinates y = R sin θ, so that 2 Z 2π 2 Z 2π 2 MR 2 MR MR Ixx = sin θdθ = (1 − cos 2θ)dθ = 2π 0 4π 0 2 Similarly, we can show MR2 I = = I yy 2 xx All the off-diagonal elements in this case are zero, for example Z M Z I = − dmxy = − (R cos θ)(R sin θ)dθ xy 2π MR2 Z 2π = − sin 2θdθ = 0 4π 0 Ixz = Iyz = 0 trivially, because everywhere on the ring z = 0. Thus, the inertia tensor of a uniform ring of mass M and radius R lying in the xy plane, with its center treated as origin 0 MR2 1 2 0 0 I = MR2 : @ 0 2 0 A 0 0 MR2 Because the inertia tensor is diagonal, therefore, x; y; and z axes are its principal axes. 6. Obtain the moment of inertia tensor of a thin uniform disk of radius R, and mass M, with the origin of the coordinate system placed at the center of the disk, and the disk lying in the xy plane. Soln: Consider a uniform circular disk of mass M and radius R, as shown A disk can be divided into a large number of rings with radii 0 ≤ ρ ≤ R Because mass M is distributed uniformly, we can define a surface mass density (mass 5 per unit area) σ M σ = πR2 Therefore, mass dm of the ring of radius ρ, and width dρ is 2M dm = σ2πρdρ = ρdρ R2 Assuming that the disk lies in the xy plane, with origin at its center, so that x2+y2 = ρ2 Z Z R 4 2 2 2 2M 3 2M R MR Izz = dm(x + y ) = 2 ρ dρ = 2 = R 0 R 4 2 Other diagonal elements, keeping in mind that z = 0, on the disk Z Z 2 2 2 Iyy = dm(x + z ) = dmx Now we consider a area element located at (ρ, θ) as shown So that M dm = σρdρdθ = ρdρdθ πR2 Given that x = ρ cos θ, we have Z Z R Z 2π 2 2 M 3 2 MR Iyy = dmx = 2 ρ cos θdθ = πR ρ=0 θ=0 4 MR2 Similarly we can prove Ixx = Iyy = 4 , and all off-diagonal elements are zero, so that the inertia tensor is diagonal 0 MR2 1 4 0 0 MR2 I = @ 0 4 0 A : MR2 0 0 2 Thus the chosen coordinate axes are also the principal axis of the disk. 7. Obtain the moment of inertia tensor of a uniform solid sphere of radius R, and mass M, with the origin of the coordinate system placed at the center of the sphere. Note that this problem can be done by dividing the sphere into a large number of infinites- imally thin disks. Soln: Because of high symmetry, any three mutually perpendicular axes passing 6 through the center will be its principle axes, and moment of inertia about any of those axes will be the same. Let us calculate the moment of inertia about one such axis We divide the sphere into a large number of disks of thickness dx, and integrate their contribution Moment of inertia of the the disk shown, with respect to the axis 1 dI = dmr2 2 dm is the mass of the given disk, given by M 3M dm = ρdV = πr2dx = r2dx; 4 3 3 3 πR 4R So that, using r2 = R2 − x2, we have 3M 3M dI = r4dx = (R2 − x2)2dx 8R3 8R3 On integrating over x 2 [−R; R], we obtain the required moment of inertia Z R Z R 3M 2 2 2 3M 2 2 2 I = 3 (R − x ) dx = 3 (R − x ) dx 8R −R 4R 0 Z R 3M 4 2 2 4 = 3 (R − 2R x + x )dx 4R 0 3M 2R5 R5 3M 8R3 2 = (R5 − + ) = = MR2 4R3 3 5 4R3 15 5 So that its inertia tensor is 0 2MR2 1 5 0 0 2MR2 I = @ 0 5 0 A : 2MR2 0 0 5 7 8.
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