Physics I Chapter H Solutions

Chapter H - Problems Blinn College - Physics 2425 - Terry Honan Problem H.1 A wheel rotates from rest to 12 rad ê s in 3 s. Assume the angular acceleration is constant. (a) What is the magnitude of the wheel's angular acceleration? (b) Through what angle (in radians) does the wheel rotate? Solution to H.1 HaL w = w0 + a t ï 12 = 0 + a 3 ï a = 4 HbL Dq = rad s 1 1 Hw0 + wL t = H0 + 12L 3 = 18 rad 2 2 Problem H.2 (a) What is the angular velocity of the Earth in its orbit about the Sun. (b) What is the angular velocity of the Moon in its orbit about the Earth. Solution to H.2 A constant angular velocity is related to the period T by w = 2p . T HaL T = 1 year = 365.24 µ 24 µ 3600 = 3.1557 µ 107 s ï w = 1.99 µ 10-7 HbL T = 27.32 days = 27.32 µ 24 µ 3600 = 2.36 rad s 05 µ 106 s ï w = 2.66 µ 10-6 rad s Problem H.3 A grinding wheel rotating at 100 rev ê min is turned off. It slows with a constant angular acceleration of 2 rad ê s. (a) How long does it take the wheel to stop? (b) Through what angle, in radians, does it turn while slowing? Solution to H.3 w = 0 and w0 = 100 rev 2 p rad 1 min rad µ µ = 10.472 min rev 60 s s HaL w = w0 + a t ï 0 = 10.472 - 2 t ï 5.24 s HbL w2 - w20 = 2 a Dq ï Dq = 0-10.4722 = 27.4 rad 2µ2 Problem H.4 A car races around a circular track of radius 250 m at a constant speed of 45 m ê s. (a) What is its angular velocity? (b) What is the magnitude and direction of its acceleration? 2 Chapter H - Problems Solution to H.4 R = 250 m and v = 45 m HaL v = r w ï w = HbL ac = w2 r = v rad = 0.18 R s v2 m = 8.1 R s2 The direction of the acceleration is centripetal, meaning that is directed toward the center. Problem H.5 By rotating through 1.25 rev of a 1 m radius arc, a discus thrower uniformly accelerates a discus from rest to 25 m ê s. (a) What is the final angular velocity of the discus? (b) What is its angular acceleration? (c) What is the total time of acceleration? Solution to H.5 HaL v = r w ï w = v 25 rad = = 25 r 1 s HbL w2 -w20 = 2 a Dq ï 252 - 0 = 2 a H2 p µ 1.25L ï a = 39.8 HcL Dq = rad s 1 1 Hw0 + wL t ï 2 p µ 1.25 = H0 + 25L t ï t = 0.628 s 2 2 Problem H.6 Assume that a baseball is a uniform sphere of radius 3.80 cm. If it moves at a speed of 38 m ê s (the speed of the center of mass) and rotates at 125 rad ê s then what is the ratio of the rotational kinetic energy to the translational kinetic energy. Solution to H.6 For a uniform solid sphere I = Krot = 2 M R2 5 1 1 2 1 I w2 = M R2 w2 = M R2 w2 2 2 5 5 1 M R2 w2 Krot 2 R2 w2 = 5 = 1 Ktrans 5 v2 M v2 2 Insert the numbers: R = 0.038 m , v = 38 Krot m rad and w = 125 giving = 0.00625 . s s Ktrans Problem H.7 Three masses glued onto a light rigid sheet in the xy plane. The masses and positions are: 4 kg at H-3 m, 5 mL, 6 kg at H4 m, 0L and 8 kg at H0, -3 mL. (a) Suppose this sheet is rotated about the y axis at an angular velocity of 12 rad ê s. What is its total rotational kinetic energy? (b) Suppose it rotates about the z axis at 12 rad/s. What is its rotational kinetic energy? Chapter H - Problems 3 Solution to H.7 (a) For a discrete distribution of mass: I = ⁄i mi ri2 where ri is the perpendicular distance of the mass mi from the axis. If the axis is the y axis then ri = †xi § I = ⁄i mi ri2 = ⁄i mi x2i = 4 µ 32 + 6 µ 42 + 8 µ 0 = 132 km ÿ m2 The kinetic energy is K= (b) If the axis is the z-axis then ri = 1 1 I w2 = 132 µ 122 = 9500 J. 2 2 x2i + y2i . I = ⁄i mi ri2 = ⁄i mi Ix2i + y2i M = 4 µ I32 + 52 M + 6 µ I42 + 0M + 8 µ I0 + 32 M = 304 km ÿ m2 K= 1 1 I w2 = 304 µ 122 = 21 900 J. 2 2 Problem H.8 Two identical thin rods of length L and mass M are perpendicular and joined at their center. What is the moment of inertia about an axis at the end of one rod that is: (a) perpendicular to the plane, (b) parallel to the other rod. HaL HbL axis axis Solution to H.8 The rod connected to the axis has, in both cases, a moment of I1 = 1 M L2 . 3 The other rod's moment varies. (a) Use the parallel axis theorem to find the other moment. The distance from the center of mass to the axis is D = L ê 2. I2 = Icm + M D2 = 1 L 2 1 M L2 + M J N = M L2 . 12 2 3 The total moment is the sum. I = I 1 + I2 = 2 M L2 . 3 (b) In this case all the mass of the second rod is a distance L ê 2 rom the axis. It's moment is L 2 2 I2 = M J N = 1 1 1 7 M L 2 . ï I = I1 + I2 = J + N M L 2 = M L2 4 3 4 12 4 Chapter H - Problems Problem H.9 What are the moments of inertia of the following flat uniform bodies of mass M about the axis mentioned? (a) an L µ L square about a diagonal (b) a circle of radius R about a diameter (c) a circle of radius R about a tangent HaL HbL axis HcL axis axis Solution to H.9 (a) The moment of a flat a µ b rectangle of mass M about a perpendicular axis through the center is I = 1 M Ia2 + b2 M. This 12 follows from the perpendicular axis theorem. (For a flat body in xy-plane, Iz = Ix + Iy ) Here take the x and y axes to axes through the center that are parallel to the sides. For Ix and Iy the moment is equivalent to a uniform rod about a perpendicular axis through the center. It follows that for a square Iz = 1 M L2 . Now take the x and y axes to be the diagonals. By symmetry Ix = Iy = I and by the 6 perpendicular axis theorem Iz = I x + I y ï I = (b) Iz = 1 1 Iz = M L2 2 12 1 M R2 . Now take the x and y axes to be the perpendicular diameters.. By symmetry Ix = Iy = I and by the perpen2 dicular axis theorem Iz = I x + I y ï I = 1 1 Iz = M R2 2 4 (c) Use the parallel axis theorem to find this moment in terms of the previous. . The distance from the center of mass to the axis is D = R. The axis parallel to the tangent through the center is the diameter, so Icm is the answer to part (b). I = Icm + M D2 = 1 5 M R2 + M R2 = M L2 . 4 4 Problem H.10 (a) A uniform solid cylinder of mass m and radius R rotates about an axis at its rim parallel to the central axis. What is its moment of inertia. (b) What is the moment of inertia of a uniform solid sphere of mass m and radius R about an axis tangent to its surface. (c) What is the moment of inertia of a uniform thin-shelled hollow sphere of mass m and radius R about an axis tangent to its surface. Solution to H.10 For all three cases we use the parallel axis theorem. Chapter H - Problems (a) I = Icm + M D2 = 1 3 m R2 + m R2 = m R2 . 2 2 (b) I = Icm + M D2 = 2 7 m R2 + m R2 = m R2 . 5 5 (c) I = Icm + M D2 = 2 5 m R2 + m R2 = m R2 . 3 3 5