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Exercises in Classical Mechanics 1 Moments of Inertia 2 Half-Cylinder

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Exercises in Classical Mechanics 1 Moments of Inertia 2 Half-Cylinder Exercises in Classical Mechanics CUNY GC, Prof. D. Garanin No.1 Solution ||||||||||||||||||||||||||||||||||||||||| 1 Moments of inertia (5 points) Calculate tensors of inertia with respect to the principal axes of the following bodies: a) Hollow sphere of mass M and radius R: b) Cone of the height h and radius of the base R; both with respect to the apex and to the center of mass. c) Body of a box shape with sides a; b; and c: Solution: a) By symmetry for the sphere I®¯ = I±®¯: (1) One can ¯nd I easily noticing that for the hollow sphere is 1 1 X ³ ´ I = (I + I + I ) = m y2 + z2 + z2 + x2 + x2 + y2 3 xx yy zz 3 i i i i i i i i 2 X 2 X 2 = m r2 = R2 m = MR2: (2) 3 i i 3 i 3 i i b) and c): Standard solutions 2 Half-cylinder ϕ (10 points) Consider a half-cylinder of mass M and radius R on a horizontal plane. a) Find the position of its center of mass (CM) and the moment of inertia with respect to CM. b) Write down the Lagrange function in terms of the angle ' (see Fig.) c) Find the frequency of cylinder's oscillations in the linear regime, ' ¿ 1. Solution: (a) First we ¯nd the distance a between the CM and the geometrical center of the cylinder. With σ being the density for the cross-sectional surface, so that ¼R2 M = σ ; (3) 2 1 one obtains Z Z 2 2σ R p σ R q a = dx x R2 ¡ x2 = dy R2 ¡ y M 0 M 0 ¯ 2 ³ ´3=2¯R σ 2 2 ¯ σ 2 3 4 = ¡ R ¡ y ¯ = R = R: (4) M 3 0 M 3 3¼ The moment of inertia of the half-cylinder with respect to the geometrical center is the same as that of the cylinder, 1 I0 = MR2: (5) 2 The moment of inertia with respect to the CM I can be found from the relation I0 = I + Ma2 (6) that yields " µ ¶ # " # 1 4 2 1 32 I = I0 ¡ Ma2 = MR2 ¡ = MR2 1 ¡ ' 0:3199MR2: (7) 2 3¼ 2 (3¼)2 (b) The Lagrange function L is given by L('; '_ ) = T ('; '_ ) ¡ U('): (8) The potential energy U of the half-cylinder is due to the elevation of its CM resulting from the deviation of ' from zero. Taking into account that the geometrical center remains at the same height, one obtains the height of the CM y = a(1 ¡ cos '); (9) so that U(') = Mgy = Mga(1 ¡ cos '): (10) Kinetic energy of the half-cylinder depends on whether it freely slides on the surface (zero friction) or it rolls on the surface (large enough friction). In the sliding case the horizontal position of the CM remains the same and only its height y changes. Thus one obtains 1 1 1 1 1 ³ ´ T = I'_ 2 + My_2 = I'_ 2 + Ma2 sin2 ' '_ 2 = I + Ma2 sin2 ' '_ 2 (11) 2 2 2 2 2 and 1 ³ ´ L = I + Ma2 sin2 ' '_ 2 ¡ Mga(1 ¡ cos '): (12) 2 In the rolling case the horizontal displacement of the geometrical center is given by xc = ¡R'; (13) where we have directed the x axis to the right. Thus the horizontal displacement of the CM is x = xc + a sin ' = ¡R' + a sin ': (14) The kinetic energy is thus 1 1 ³ ´ 1 1 h i T = I'_ 2 + M x_ 2 +y _2 = I'_ 2 + M (¡R'_ + a cos ' '_ )2 + a2 sin2 ' '_ 2 2 2 2 2 1 h ³ ´i = I + M R2 ¡ 2Ra cos ' + a2 '_ 2 (15) 2 and the Lagrange function becomes 1 h ³ ´i L = I + M R2 ¡ 2Ra cos ' + a2 '_ 2 ¡ Mga(1 ¡ cos '): (16) 2 2 c) For ' ¿ 1 the Lagrange function simpli¯es to 1 1 L =» I'_ 2 ¡ Mga'2 (17) 2 2 in the sliding case and to 1 h i 1 L =» I + M (R ¡ a)2 '_ 2 ¡ Mga'2 2 2 1 1 = I00'_ 2 ¡ Mga'2 (18) 2 2 in the rolling case. With the help of Eq. (4) and (7) one obtains " # µ ¶ 1 32 4 2 I00 = I + M (R ¡ a)2 = MR2 1 ¡ + M R ¡ R 2 (3¼)2 3¼ " # 1 16 8 16 = MR2 ¡ + 1 ¡ + 2 (3¼)2 3¼ (3¼)2 · ¸ 3 8 = MR2 ¡ ' 0:6512MR2: (19) 2 3¼ These results are easy to interpret. For ' ¿ 1;in the sliding case the half-cylinder approximately rotates around its center of mass, whereas in the rolling case the half-cylinder approximately rotates around its contact point with the surface, R ¡ a being the distance between the CM and the contact point. Since in the latter case the moment of inertia is larger, the frequency of oscillations is smaller. The Lagrange equation in our case has the form d @L @L ¡ = 0 (20) dt @'_ @' that in the rolling case becomes I00'Ä + Mga' = 0 (21) or 2 'Ä + !0' = 0 (22) with the oscillation frequency v s u r u 4 Mga t g 3¼ g !0 = 00 = 3 8 ' 0:807 3 (23) I R 2 ¡ 3¼ R p that is close to that of a simple pendulum g=R: In the sliding case one obtains v s u v r u 4 u 2 Mga u g 3¼ u g 3¼ g !0 = = ³ ´ = t ' 1:1519 : (24) t 1 32 32 I R 1 ¡ R 1 ¡ 2 R 2 (3¼)2 (3¼) 3 Self-rotation (15 points) How can a cat manage always to land on her feet? How can a system with zero angular momentum set itself into rotation? Consider a person standing on a rotating platform without friction, so that its angular momentum is conserved and is zero, Lz = 0: The person having together with the platform a moment of inertia Izz ´ I moves a point mass m by (massless) hand around a closed contour in the x-y plane, de¯ned in the frame of the platform. By which angle the person on the platform rotates as the mass m makes a full turn? a) Write down the condition Lz = 0 in terms of the projections of the point-mass position and velocity on the axes of the body (platform) frame. 3 b) Change to the polar coordinates (r; ') for the point mass and obtain a relation between d' and the in¯nitesimal rotation of the platform dθ. Let ¢θ be the angle of rotation of the platform corresponding to one full turn of the point mass. What do you expect for ¢θ in the limits I ! 1 and I ! 0? c) Consider a particular case of rotation of the point mass around a circle with radius R and the center at the distance l > R from the center of the platform and calculate ¢θ: What is the condition for ¢θ to be maximal? Solution: a) The angular momentum of the system is given by Ã! Ã! L = I ¢!+ [r £ p] = I ¢!+m [r £ v] ; (25) where ! is the angular velocity of the body and plarform and v is the velocity of the point mass m: The latter consists of two contributions, one due to the rotation of the platform and the other, u; due to the motion with respect to the platform, v = [! £ r] + u: (26) Plugging Eq. (26) into Eq. (25) yields Ã! L = I ¢!+mr2! ¡ m (! ¢ r) r+m [r £ u] : (27) We are interested in the z component of the above equation taking into account Lz = 0: Also we take into account that !kez: Since the motion of the point mass is de¯ned in the body frame, it is convenient to project r and u onto the rotating body-frame axes: (®) (®) r =r®e ; u =u®e : (28) Since for the motion in the x-y plane one has ! ¢ r = 0; this yields ³ ´ 2 0 = Lz = I + mr ! + m (rxuy ¡ ryux) (29) and thus m (r u ¡ r u ) m (r r_ ¡ r r_ ) ! = ¡ x y y x = ¡ x y y x : (30) I + mr2 I + mr2 This formula gives the back reaction of the moving mass m onto the body+platform subsystem. b) Now we change to the polar coordinates rx = r cos '; ry = r sin ': (31) It is clear that the rhs of Eq. (30) should be sensitive to' _ but not tor: _ Indeed, using r_x =r _ cos ' ¡ r sin ' ';_ r_y =r _ sin ' + r cos ' '_ (32) one can simplify Eq. (30) to mr2 ! ´ θ_ = ¡ ';_ (33) I + mr2 or, in the di®erential form, mr2 dθ = ¡ d': (34) I + mr2 One can see that in the limit I ! 1 the platform does not rotate, dθ = 0. In the limit I ! 0 one has dθ = ¡d', and the situation depends on whether the center of rotation of the platform is inside or outside the contour circomscribed by the mass m: If it is inside, then the full rotation of m corresponds to ¢' = 2¼: In this case ¢θ = ¡2¼: If, however, the the contour does not encircle the axis of the platform rotation, then the full turn corresponds to ¢' = 0 and thus one obtains ¢θ = 0: One can see that in the latter case ¢θ = 0 in both limits I ! 1 and I ! 0; so that there should exist some relation between the parameters for which the person can rotate itself most e±ciently. 4 c) In the particular case of the contour being a circle of radius R with the center at l from the axis of the platform rotation, it is convenient to parametrize the motion of m with the polar angle à with respect to the circle R: The relations between (r; ') and à are the following rx = r cos ' = l + R cos à ry = r sin ' = R sin Ã: (35) First, one obtains from here 2 2 2 2 2 r = rx + ry = l + 2lR cos à + R : (36) Then d' can be found from drx = cos ' dr ¡ r sin ' d' = ¡R sin à dà dry = sin ' dr + r cos ' d' = R cos à dÃ: (37) Elimination of dr yields rd' = R cos ' cos à dà + R sin ' sin à dà (38) or r2d' = R (l + R cos Ã) cos à dà + R2 sin2 à dà = R (R + l cos Ã) dÃ: (39) Inserting this and Eq.
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