PHYS 150

Dr. Johnny B. Holmes

Introduction
Rotation of a Rigid Body About a Fixed Axis
Angular Momentum
Oscillatory Motion
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In this last quarter of the course we consider: (a) the description and cause of rotations which will include rotational energy and rotational analogues to force and momentum; and (b) the description and causes of oscillations.

In the first part we consider the description and cause of rotations.

In the second part we consider the concept of ANGULAR MOMENTUM. We further develop this idea to formulate the conservation of angular momentum. This is an important concept which is especially useful in studying any orbiting system including the solar system and the atomic system.

In the last part we consider the basis of oscillations and their description.

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Outline
Supplementary Homework Problems
Answers to Supplementary Problems
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1. Review of circular motion
1. angles - definition and units [recall that q (in rads) = s/r.]
2. circular motion at constant speed [recall that r = constant; v_r=0; v_t=w r; a_r=w²r=v²/r; a_t=0.]
3. Newton's 2nd Law and circular motion [SF_r = ma_r]
4. satellites [F_r = F_gravity = Gm₁m₂/r²; a_r = w²r leads to Gm_earthm_sat/r² = m_satw²r.]
2. Rotational vectors
1. angle as vector
2. angular velocity vector: w = dq /dt (where w is not necessarily constant)
3. angular acceleration vector: a = dw/dt = d²q /dt².
4. tangential and radial components
   r = constant; q = q(t); x(t) = r cos[q(t)] , y(t) = sin[q(t)]
   v_x = dx/dt = (dx/dq)*(dq /dt) = -w r sin[q(t)]; v_y = w r cos[q(t)]; v = [v_x²+v_y²]^1/2 = wr; q_v = q ± 90^o.
   a_x = dv_x/dt = -w² r cos[q(t)] - a r sin[q(t)]; a_y = dv_y/dt = -w² r sin[q(t)] + a r cos[q(t)];
   let a_r = -w²r, and let a_t = a r; then a = [a_r² + a_t²]^1/2.
3. Torque and rotation
1. torque - rotational analogue of force
2. torque as vector: cross product: t = r ´ F; magnitude: t = r F sin(q_rF); direction: right hand rule.
3. moment of inertia (rotational analogue of mass): I = S m_i r_i²
4. Torque and Newton's 2nd Law: St = Ia
4. Rotational energy
1. kinetic energy for rotating particle: KE = (1/2)mv² = (1/2)m(wr)²
2. kinetic energy for rotating RIGID body: KE = (1/2)Iw²
3. work through torque: W = ò F · ds leads to W = ò t dq
4. power through torque: P = dW/dt = F · v leads to P = t w .
5. translation plus rotation: KE = (1/2)mv² + (1/2)Iw² .

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42. The diameter of Jupiter is 144,000 km and at its nearest, it is 6.26 x 10⁸ km from the earth. What is the angle that Jupiter makes at the eye a) in radians? b) in degrees? c) in seconds of angle?

43. a) How fast are you going right now due to the fact that the earth is rotating about its axis? (Memphis is at about 35° N lattitude, and the radius of the earth is 6,400 km.) b) How fast are you going right now due to the fact that the earth is orbiting the sun? (The radius of the earth's orbit is 1.49 x 10⁸ km.)

44. A space station is built in the shape of a spoked wheel with the rim having a square cross section. When the wheel is set spinning, artificial gravity is set up. a) Inside the rim, which of the four "walls" would be considered the floor? b) If the outside "wall" is 300 meters from the center of the wheel, how fast (in rev/sec) would the wheel have to spin to "create" an artificial gravity of (1/2)g (where g is acceleration due to gravity on earth's surface)? c) In one of the spokes, which direction would be considered up? d) What would be the magnitude of the effective gravity at a point 200 meters from the center?

45. A man is at the earth's equator. In terms of N, E, S, W, up, and down, what is the direction of the following due to the earth's rotation about its axis: a) w ? b) a ? c) v? d) a?

46. A car accelerates uniformly from rest to a speed of 15 m/s in a time of 20 seconds on wheels of radius 30 cm. Find the angular acceleration of one of its wheels a) at t = 0 sec, and b) at t = 20 sec. c) Find the number of revolutions turned by the wheel in the process. d) What is the angular speed of the wheel at t = 0 sec; and e) at t = 20 sec? f) What is the tangential acceleration (due to spinning only) of a point on the outside part of the wheel at t = 0 sec; g) at t = 20 sec? h) What is the radial acceleration of a point on the outside part of the wheel at t = 0 sec? i) at t = 20 sec?

47. a) What is the moment of inertia about the axis of a solid aluminum (density of 2.7 gm/cm³ ) cylinder of length 5 cm and radius of 0.8 cm? b) What is the moment of inertia of a hollow aluminum cylinder of length 5 cm, inner radius 6 mm, and outer radius 8 mm? c) What is the moment of inertia of a solid cylinder of length 5 cm and radius 6 mm? d) What torque is necessary to accelerate the cylinder of part-a from 0 cps to 7 cps in 5 seconds? e) What torque is necessary to accelerate the cylinder of part (b) from 0 cps to 7 cps in 5 sec?

48. A mass, M, of 200 grams hangs from a string wound around an aluminum (density of 2.7 gm/cm³ ) disc of radius of 10 cm and thickness 2 cm. a) What is the moment of inertia of the disc? b) If the weight is allowed to fall one meter to the floor, what will its speed be as it hits the floor assuming negligible friction? c) What will the rotational speed of the disc be as the weight hits the floor? d) What is the acceleration of the block? e) What is the tension in the string? f) What is the torque acting on the disc? g) What is the angular acceleration of the disc?

49. What is the power output of an electric motor that supplies a torque of 1 Nt m to a wheel that rotates at 60 cycles per second?

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42. a) 2.3 x 10^-4 rad; b) .013°; c) 47.4 seconds.

43. a) 1,370 km/hr or 850 mph; b) 106,800 km/hr or 66,750 mph.

44. a) one farthest from the center of wheel; b) .0203 rev/s or 1.22 rpm; c) toward center of the wheel; d) 3.27 m/s²= .33*gearth.

45. a) N; b) none since a is zero; c) E; d) down.

46. a) 2.5 rad/s²; b) 2.5 rad/s²; c) 500 rad = 79.6 rev; d) 0 rad/s e) 50 rad/s; f) .75 m/s²; g) .75 m/s²; h) 0 m/s²; i) 750 m/s².

47. a) 8.69 gm cm² or 8.69 x 10^-7 kg m²; b) 5.94 x 10^-7 kg m²; c) 2.75 x 10^-7 kg m²; d) 7.64 x 10^-6 Nt m; e) 5.22 x 10^-6 Nt m.

48. a) 8.48 x 10^-3 kg m²; b) 1.93 m/s; c) 19.3 rad/s; d) 1.87 m/s²; e) 1.59 Nt; f) .159 Nt m; g) 18.7 rad/s² .

49. 377 Watts.

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Outline
Supplementary Homework Problems
Answers to Supplementary Homework Problems
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1. Definition and analogy with linear momentum: p = mv becomes L = Iw ; also t = r ´ F leads to L = r ´ p.
2. Conservation of angular momentum: SL_initial = SL_final .
3. Spinning tops and gyroscopes

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50. An ice skater spins around with her arms extended. Use the approximation that she is a uniform cylinder of mass 40 kg and radius 25 cm plus arms that have a mass of 2.5 kg each, and neglect friction. If she rotates at a speed of 2 rev/sec., with her arms extended (radius 50 cm), and then suddenly she pulls her arms in (radius 25 cm) how fast will she be going after she pulls her arms in?

51. The sun (mass of 2 x 10³⁰ kg) rotates about its axis with a period of 25 days. Lets assume the sun is a sphere of radius 7 x 10⁵ km and uniform density (a bad assumption). a) Assuming the planets were not formed and the sun collapsed from a uniform spherical gas cloud of radius 10,000 times its present radius, what was the period of rotation of that original gas cloud? b) Assuming the planet Jupiter (mass of 2 x 10²⁷ kg, distance to sun of 7.75 x 10⁸ km, rotational period about the sun of 12 years) was formed but none of the other planets were with the sun, what would the period of rotation of the original gas cloud have been? c) What is the relation between answers to parts a & b?

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50. 3.21 rev/sec.

51. a) 2.5 x 10⁹ days = 6.84 million years; b) 1.35 x 10⁸ days = 370,000 yrs; c) b is 18.5 times faster than a.

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Outline
Supplementary Homework Problems
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1. Oscillations using springs
1. review: numerical
2. review: energy
3. analytical: S F = ma (with F = -kx, Newton's Second Law becomes a differential equation)
4. with friction
5. with friction and a driving force
6. resonance
2. Other cases

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SUPPLEMENTARY HOMEWORK PROBLEMS (S-):

52. (This is similar to problem 25 except that it includes friction.) Given that an object of mass 3 kg starts from the origin (xo = 0) with a speed of 5 m/s (vxo = 5m/s), and given that it is attached to a spring that exerts a force Fx-sp(x) = -kx where k = 24 Nt/m [the spring constant], and given that there is a frictional force Fx-fr = -bv where b = .4 Nt s/m, find: a) the acceleration at t = 0 when the object starts; b) assuming that the object experiences this acceleration for 0.1 seconds, find [by numerical methods] its approximate new speed and new position at t = 0.1 sec; c) do this process nine more times to find the approximate speed and position at the end of the first full second; d) graph the acceleration versus time, the velocity versus time, and the position versus time all between t=0 and t=1 second. *e) [extra credit] Try using a computer program (or program your calculator) to do the above numerical approximation, and follow the motion for several seconds. See what happens to the motion as your vary the mass and/or the spring constant and/or the frictional constant.

53. For the same problem as the one above, add in an applied force that behaves as: Fx-applied = 2 Nt sin([0.7 rad/sec] t) and see what happens in each of the cases.

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