APM6/03//08 Tutorial letter 03//08 Mechanics APM6 Semester Department of Mathematical Sciences IMPORTANT INFORMATION: This tutorial letter contains solutions to assignment 3, Sem. BARCODE Define tomorrow. university of south africa
ASSIGNMENT 03 Solution Total Marks: 00 UNIQUE ASSIGNMENT NUMBER: 870 Question : Marks Find the following moments of inertia: (a) A rod AB of length a and mass M, rotating about an axis perpendicular to the rod, through end point A. (b) A rod AB of length a and mass M, with a particle of mass M attached at its centre and a particle of mass M attached at end B, rotating about an axis perpendicular to the rod, through end point A. (c) A rod AB of length a and and negligible mass, with a particle of mass M attached at its centre and a particle of mass M attached at end B, rotating about an axis perpendicular to the rod, through end point A. Solution (a) This is a simple application of the parallel axis theorem: The moment of inertia for rotation about an axis perpendicular to the rod, through the centre of the rod is I G = 3 Ma, so the moment of inertia for rotation about a parallel axis, through point A, is (since the distance between points A and G is a) is given by See Example 0.8! I A = I G + Ma = 4 3 Ma. (b) The rod and the axis here is the same as the one in (a), but now two particles are attached to the rod: Particle with mass M attached at the centre G, and Particle with mass M attached at end point B. To find the moment of inertia for the whole object (rod and the two particles), we will add up the moments of inertia of the three components: I A = I rod A + I particle A + I particle A. Here, I rod A = 4 3 Ma from (a). For the particles, the moment of inertia is calculated using I = mr,
APM6/03//08 where m is the mass of the particle and r the distance from the particle to the axis of rotation in this question, therefore, from the particle to point A. So, we get I A = 4 3 Ma + Ma + (M) (a) = 0 3 Ma. (c) This question differs from (b) in that the rod is now considered to be massless. It follows that I rod A = 0, and I A = Ma + (M) (a) = 9Ma. Question : 0 Marks Find the moment of inertia of a thick hollow sphere, with an inner radius R and an outer radius R and mass M, about an axis through its centre. Solution We will give two alternative methods for calculating the moment of inertia. Method : by integration The thick sphere can be sliced into concentric hollow spheres, each specified by its radius r where r ranges from R to R. If the thickness of one such hollow sphere is dr, then its mass is given by dm = ρ 4πr dr (density times area of the hollow sphere times its thickness) and according to Example.6, the moment of inertia of the hollow sphere is di = 3 (dm)r. For the moment of inertia of the entire object, we need to integrate over the values of r from R to R : I = di () = () R R R 3 (dm)r = R 3 (ρ 4πr dr)r (3) = ρ 8 3 π R = ρ 8 3 π R r 4 dr () ( ) R R. (7) (4) (6) 3
Finally, we wish to express this in terms of the mass of the thick sphere. The volume of such a thick sphere is the volume of a sphere with radius R, minus the volume of a sphere with radius R, that is, volume = 4 3 πr3 4 3 πr3 = 4 3 π ( R 3 R 3 ). On the other hand, the mass M must be equal to density times volume. Therefore, we must have M = ρ 4 3 π ( ) R 3 R 3 ρ = 3M 4π ( R 3 R3 ). When we substitute this into (*), we get I = 3M 4π ( R 3 ) 8 R3 3 π ( R R) ) = ( R M R R 3 R3 Method : An object with a part removed If we combine the object in this question and the missing part, a sphere with radius R, we get a solid sphere with radius R. It follows that the moment of inertia of the object here must equal the moment of inertia of a solid sphere with radius R, minus the moment of inertia of a solid sphere with radius R. We know the moments of inertia of the latter objects, so this should be fairly easy. However, we must make sure we get the masses of the solid spheres right: the mass of the object in this question must be M, so what should be the masses of the corresponding big sphere and small sphere? The masses of these should be determined by their volumes, and the density. The small sphere and the big sphere will have the volumes V = 4 3 πr3, V = 4 3 πr3 respectively, and the object in this question will therefore have the volume V = 4 3 πr3 4 3 πr3. So if it has the mass M, then the density of the material must be such that M = ρv ρ = M V = M 4 3 πr3 4 3 πr3 It follows that the masses of the small and big spheres are M = ρv = M = ρv =. 3M = 4π ( R 3 R3 3M 4π ( R 3 R3 3M 4π ( R 3 R3 ) 4 3 πr3, ) 4 3 πr3. ). 4
APM6/03//08 For the moment of inertia for the whole object we therefore get I = I I = M R M R = 3M 4π ( R 3 ) 4 R3 3 πr3 R 3M 4π ( R 3 ) 4 R3 3 πr3 R = M R R R 3. R3 Note that both methods do give the same answer, as they should! Question 3: Marks Find the moment of inertia when the lamina bounded by the function y = + x and the X-axis, between x = 0 and x = 3, rotates about the X-axis. (Hint: slice the lamina into thin rods, perpendicular to the X-axis!) Solution The lamina looks like this: If we divide it into thin slices, then the one at position x on the X axis is approximately a thin rod of length y = + x and mass dm = ρ ( + x ) dx. Its moment of inertia, when it rotates about the X axis, is therefore given by di = 4 3 dm ( + x ) = 4 3 ρ ( + x ) ( + x dx = 3 ρ ( + x ) 3 dx. )
To find the total amount of inertia, we integrate over all the thin rods: I = di = 3 0 3 ρ ( + x ) 3 dx = 3 ρ ( 8x + 4x 3 + 6 x + 7 x 7 )] 3 0 = 887 3 ρ. Note that the moment of inertia depends on density ρ; if we alternatively wish to express it in terms of the mass M of the object, we can calculate the connection between the mass and the density by finding the area of the object: area = 3 0 ( ) + x dx = 3 0 (x + 3 ) x 3 =, hence and δ = M area = M I = 887 M 6.36 M. 3 Question 4: Marks Find the moment of inertia of an A A A cube, rotating about an axis which goes through the midpoints of two opposing sides. Solution We can slice the cube into thin squares, perpendicular to the axis of rotation; we can assume that each such slice has mass m i where the sums of the masses m i give the mass of the whole cube: M = m i. 6
APM6/03//08 Then the total moment of inertia is the sum of the moments of inertia of the thin squares. To find their moments of inertia, we can use the perpendicular axis theorem: I Z = I X + I Y, where I X and I Y are moments of inertia for rotation about an axis parallel to the square. From Example 0. of the study guide we get I X = I Y = ( ) A 3 (M) = MA, therefore I Z = I X + I Y = 6 MA and thus the total moment of inertia is I i = I Z = 6 m ia, I = mi A = mi A = 6 6 6 MA. Alternatively, we can use the result in Example.9: The cube here is a special case of solid block there, with a = b = c = A. As derived there, the moment of inertia of the cube for rotation about an axis along one of the edges between two sides would be I edge = ( A + A ) M = 3 3 MA, so using the parallel axis theorem, the moment of inertial for an axis parallel to this, but going through the middle of two opposing sides would be I = I edge Md where ( ) A d = = A. [Note the minus sign: it is there because we are now calculating the value of I for an axis through the centre of mass, from the value of I about a parallel axis not through the centre of mass!] This gives again I = 3 MA M A = 6 MA. 7
Question : 0 Marks A pendulum is made of two disks (each with a mass M and radius R) which are separated by a massless rod. One of the disks is pivoted through its centre by a small pin. The disks hang on the same plane and their centres are a distance l apart. Find the period for small oscillations. Solution The object looks as shown below. We can imagine the discs as being attached to a rod AB of mass zero and length l. We will need to find the moment of inertia of the object for rotation about an axis perpendicular to the object, through point A. But this will be equal to I A = I(disc at A about A) + I(disc at B about A) The moment of inertia of disc A about the axis through A is just the moment of inertia of a disc rotating about its centre; while by applying the parallel axis theorem, the moment of inertia of disc B about A equals the moment of inertia of disc B rotating about its centre, plus Ml : I A = MR + MR + Ml = M ( R + l ). We will also be needing the centre of mass of the object; but due to symmetry, this is easily seen to lie in the middle of the rod, the distance l/ from point A. Now, when the object is set to oscillate as a pendulum as described, a sketch of the situation could be as follows: We are showing the situation where the rod makes the counterclockwise angle θ with a vertical line down from point A. The only force acting on the rod which does not act at point A is gravity Mg, acting at point G (the centre of mass). Therefore, the equation of motion for rotation about point A is (see (0.3)): I A θk = r Mg where r is the vector from A to G. The direction of vector r will depend on the angle of the rod. we choose i and j to be unit vectors as in the picture below, If 8
APM6/03//08 then r = l i Mg = Mg ( cos θi sin θj ) and r mg = i ( cos θi sin θj ) = lmg sin θk. Therefore, the equation of motion becomes I A θk = lmg sin θk θ = lmg I A sin θ. For small oscillations (θ small) we can use the approximation sin θ θ. Then the equation of motion is approximately θ = lmg θ, I A which describes harmonic motion with period T = π I A lmg. [Here is how this comes about: The solution to the differential equation is given by ẍ = A x x (t) = cos (At) + sin (At), as can be seen by differentiating this function twice with respect to time t. But x (t) = cos (At) + sin (At) 9
clearly describes periodic motion; and since the period of cos (t) and sin (t) is π, (meaning that they return back to their original values after that time), the period of cos (At) and sin (At) is π.] A So, after substituting the value of I A = M ( R + l ) into this, we see that the period of the small oscillations in this question is M (R T = π + l ). lmg Question 6: Marks Two men, each with a mass M, are standing at the centre of a uniform, horizontal beam with a mass m which is rotating at a uniform angular velocity ω about a vertical axis through its centre. Then the two men walk out to the ends of the beam and ω is then the angular velocity. Find an expression for ω in terms of ω, M and m. (Include all the calculations.) Solution The angular momentum of the whole object formed by the beam and the two men must be the same initially and afterwards; and in both cases the angular momentum is given by L = I θ where I is the moment of inertia of the object, and θ the angular velocity. INITIALLY, we have a rod of mass m and length a (let s say) with both men at the centre of the rod, so the moment of inertia for rotation about the centre of the rod is I = 3 ma [the men stand at the axis of rotation so they do not contribute to the moment of inertia]. And we are told that the angular velocity is θ = ω. So, the initial angular momentum is L in. = 3 ma ω. FINALLY, after the men have moved to the ends of the beam, the moment of inertia equals I = 3 ma + Ma + Ma = 3 ma + Ma. (this is the moment of inertia of the rod, as before, plus the moments of inertia of the two men with masses M at the ends of the beam, the distance a from the centre). The angular velocity is now θ = ω. Therefore, the final moment of inertia is ( ) L fin. = 3 ma + Ma ω. Now, the values of L in. and L fin. will be the same, since no no moments of forces exist to change the angular momentum. We get L in. = L fin. 3 ma ω = ω = ( ) 3 ma + Ma ω 3 ma ( 3 ma + Ma )ω ω = m (m + 6M) ω. [Note that the value of a, determining the length of the beam, will cancel out everywhere here!] 0