QUIZ 12 - CH 10 - ROTATIONAL DYNAMICS : TORQUE, ROTATIONAL AND TRANSLATIONAL KINETIC ENERGY, ANGULAR MOMENTUM, NEWTON'S SECOND LAW OF ROTATION, CONSERVATION OF ANGULAR MOMENTUM * means try out only but still on exam |
1**, 2**, 4**, 8**, 16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95** |
WE WISH TO GROUP THESE PROBLEMS INTO IMPORTANT SEVERAL
CATEGORIES:
1. Rotating Pulley, String, and a Single Translating Object
connected to it. WE WILL CLASSIFY PROBLEMS BELOW ACCORDING TO THIS SCHEME WHERE POSSIBLE. For example # 14 would be classified as 1 a or 1 b since you can either use Conservation of Energy or Newton's Law's of Translation and Rotation. THIS IS A WORK IN PROGRESS : CHECK BACK SOON. |
DISCUSSIONS |
1**, 2**, 4**, 8**, 16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95** |
1. ** DO These exercises on at a time, an connect the dots mentally. Note the torque in (f) is ZERO since the force goes through the axis . THIS IS IMPORTANT AND WILL REPEATEDLY BE ENCOUNTERED when you selectively choose which force to use in CHAPTER 11, STATICS. |
2.** I will make an interesting comment only . If I had chosen the axis a the right end of the beam, the force of magnitude F_{1} would exert NO TORQUE. |
4.** We mentioned the force of magnitude 11.9 N exerts no torque about the axis through the center. The 14.6 N-force twists disk clockwise---R*sin 40 is the moment arm for that force)---and the 8.50 N-force rotates counter clockwise. |
8.** CLOCKWISE ROTATION, TORQUE POINTS OUT. Right fingers rotate counter clockwise and your right thumb points OUT. Moment arm = L*sin 37. |
16.** GROUP 1a or 1b (a) Use I*alpha = R*T or I*(a/R^{2}) = T and ma = mg - T. Solve these simultaneously for T and a. (b) Use v^{2} = 2*a*h from Ch. 2. You can also use conservation of energy: mgh = (1/2)*m*v^{2} + (1/2)*I*w^{2} where w = v/R. (c) Use Ch. 2 methods. (d) This is critical for ENGINEERING 36 "Engineering Mechanics -Statics" . On the Pulley the vertical forces sum to ZERO: 0 = N - M*g - T , where N is the magnitude of the normal force on the pulley from the axle. Note: Rotational Inertia I restricts the translational acceleration downward such that a < g. |
17.** GROUP 2b Labels: A, block on table; B, falling mass and P, pulley. (I) M_{A}*a = T_{A}, (II) I*(a/R^{2}) = T_{B} - T_{A}, (III) M_{B}*a = M_{B}*g - T_{B} . (a) and (b) : Solve these three equations simultaneously for a and two tensions. TIP: Add the equations to cancel the tensions, go back and substitute. (c) ENGINEERING 36 "Engineering Mechanics -Statics" : Vertically: 0 = F_{v} - Mg - T_{B} . Horizontally: 0 = F_{h} - T_{A}. |
70.** GROUP 1b M*a = Mg*sin 36.9 - T - u_{k}*Mgcos 36.9, where u_{k} is the coefficient of kinetic friction. I*(a/R^{2}) = T Note: Rotational Inertia I restricts the translational acceleration down the incline such that a < g*sin theta, the result for a block sliding down an inclined plane without being connected to a string wrapped around a pulley and without friction. |
1**, 2**, 4**, 8**, 16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95** |
20.** GROUP 3a mgh = (1/2)*m*v^{2} + (1/2)*I*w^{2} where w = v/R and I = m*R^{2}, useful for simplifying purposes. |
21.** GROUP 4a KE_{total} = (1/2)*m*v^{2} + (1/2)*I*w^{2} , where w = v/R and I is given by formulae for several objects. Find [(1/2)*I*w^{2} ]/KE_{total} in each case. Any guesses on which is the largest ? smallest? What happens ion the limit I goes to infinity? SEE EXAMPLE 10.5 FOR AN APPLICATION OF CONSERVATION OF ENERGY TO AN INCLINE |
68**. GROUP 1b PULLEY: I*(a/R) = r*F -RT, where F is the magnitude of applied force and T is rope tension magnitude. Note tension acts at distance R = 0.25 m from axis whereas the applied force acts 0.12 m away. Also m*a = T - mg, the equation for the rising mass attached to rope. |
22.** GROUP 4b m*a = mg*sin theta - f_{s}. I*(a/R^{2}) = f_{s}. Solve simultaneously for a and f_{s} . f_{smax} = u_{s}*N, where u_{s} is the coefficient of static friction. Evaluate this from what you know from chapter 5. Check results for doubled mass. |
34.** Mathematically, Group 1and b---Rotating Pulley (wheel),
Applied Torque portion of principle. I*alpha = L*F allows you to get alpha, from the entire range of qu4eesiotn are answered. . We skipped lecture discussion of power. Let us go over this now. differential work = dW = (torque)*d theta Integrating both sides we get W = (1/2)*I*w^{2} = rotational KE , using torque = T*(alpha) and a couple other calculus tricks; consult the textbook Note: W = Torque*total theta Instantaneous Power = dW/dt = (1/2)*2*I*w*dw/dt = I*alpha*w = torque*w. (a) and (b) Use kinematics to get alpha, which you know from above discussion and w (c) W = (1/2)*I*w^{2 }. Note: W = Torque*total theta^{ }(d) Average power = (change in W)/time = (1/2)*I*w^{2}/time (e) Instantaneous Power = torque*w. |
33.** Mathematically, Group 1and b---Rotating Pulley
(Propeller), Applied Torque portion of principle. I*alpha = L*F allows you to get alpha, from the entire range of qu4eesiotn are answered. . We skipped lecture discussion of power. Let us go over this now. differential work = dW = (torque)*d theta Integrating both sides we get W = (1/2)*I*w^{2} = rotational KE, using torque = T*(alpha) and a couple other calculus tricks; consult the textbook. W = Torque*total theta Instantaneous Power = dW/dt = (1/2)*2*I*w*dw/dt = I*alpha*w = torque*w. (a) Find alpha kinematically (Ch. 9) then use it to complete the question (b) Pure kinematics Ch. 9. (c) W = Torque*total theta (d) (1/2)*I*w^{2} = rotational KE; compare with part (c). |
1**, 2**, 4**, 8**, 16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95** |
ANGULAR MOMENTUM AND ITS CONSERVATION |
36.** Reviewed in class, but still as important as
ever. For a single particle I = mr^{2}, where r is the
perpendicular distance from the axis of rotation to the particle
that is rotating about it. In general L_{z} =
I*w. If the point particle was rotating by itself, then it has L_{z} = mr^{2}*w If the point particle was rigidly attached to a rotating solid cylinder about an axis through the center and perpendicular to the cylinder's face, then I_{total} = I_{cylinder} + mr^{2} and L_{z} = = I_{total}*w |
37.** (a) (R*sin 36.9)*p = | L_{z} | (b) (Rcos36.9)*mg =| torque| = |dL_{z}/dt | |
43.** More complex than you think: I includes a central
cylinder (hands and arms wrapped) and a long, slender
cylindrical rod (hands and arms outstretched). I_{i}
= 0.40 kg*m^{2} + I_{rod }and I_{f}
= 0.40 kg*m^{2} + I_{cylinder}
. See table 9.2. L_{iz}_{ }= L_{fz} .I_{i}*w_{i} = I_{f}*w_{f}. |
45.** L_{iz }= L_{fz} . I_{i}*w_{i} = I_{f}*w_{f}. I_{i} = I_{cylinder} . I_{f} = I_{person } + I_{cylinder} . Note: For a single particle, representing the person, I = mr^{2},
where r is the perpendicular distance from the axis of rotation to
the particle that is rotating about it. In general L_{z}
= I*w. |
THE FOLLOWING PROBLEMS CAN BE GROUPED TOGETHER---45***( in hindsight), 46**, 48**, 49**, 95**---VIA THE IDEA OF CONSERVATION OF ANGULAR MOMENTUM INVOLVING A POINT PARTICLE COLLIDING WITH AN EXTENDED RIGID BODY. THE AXIS OF ROTATION RUNS THROUGH OR ALONG THE RIGID BODY. |
46.*** See example 10-12** and #95.** L_{iz }= L_{fz} . L_{mudi} + L_{doori }= L_{mudf} + L_{doorf}. D_{*}p_{i} + I_{door}*w_{doori} = D*p_{f }+ I_{door}*w_{doorf} D = width of door/2 = Width/2 = perpendicular distance from the axis to the line of action of the mud's linear momentum; p_{i} = mv_{i}, where v_{i}= 12.0 m/s and m is the mud mass. p_{f }= mv_{f}, where v_{f} = D*w_{doorf} , where w_{doorf} is the final angular speed of the door. Note: D*p_{f }+ I_{door}*w_{doorf } = mD^{2}*w_{doorf } + I_{door}*w_{doorf }= (I_{mud}+ I_{door})*w_{doorf} See Table 9.2*** (d) for I_{door} in terms of Width. Note: w_{doori} = 0. |
48.*** Let R = Earth
radius, m = mass of asteroid, M = Earth mas. L_{iz }= L_{fz} . L_{asteroid_i} + L_{earth_i }= L_{asteroid_f} + L_{earth_f}_{ . }0 + I_{ earth }*w_{i }= R*p_{f} + I_{ earth }*w_{f}_{ } . I_{ earth }*w_{i }= R*mv_{f} + I_{ earth }*w_{f } . I_{ earth }*w_{i }= R*m*R*w_{f} + I_{ earth }*w_{f}_{ } . I_{ earth }*w_{i }= m*R^{2}*w_{f} + I_{ earth }*w_{f } . I_{ earth }*w_{i }=I_{asteroid} *w_{f} + I_{ earth }*w_{f } . I_{ earth }*w_{i }= (I_{asteroid} + I_{ earth })*w_{f } . Prove if the day is 25 % longer then w_{f } = (4/5)w_{ i} . Solve for I_{asteroid} and then m in terms of M, the Earth's mass. See table 9.2 for I_{sphere} = I_{earth}. = (2/5)*MR^{2}. |
49.** See example 10-12*** and #95.*** L_{iz }= L_{fz} . L_{ball_i} + L_{bar_i }= L_{ball_f} + L_{bar_f}. D_{*}p_{i} + I_{bar}*w_{bar_i} = D*p_{f }+ I_{bar}*w_{bar_f} D =1.50 m = perpendicular distance from the axis to the line of action of the ball's linear momentum; p_{i} = mv_{i}, where v_{i}= 10.0 m/s and m is the ball's mass. p_{f }= mv_{f}, where v_{f} = - 6.00 m/s since it rebounds in the negative direction, whether left of right in your diagram. . Note: w_{bar_i} = 0. Find w_{bar_f. }(b) * Linear momentum is not conserved if you consider the bar and ball to be the system. Some of the linear momentum gets transferred to the Earth through the pivot which exerts an extra NORMAL force on the Earth during the collision. See #28***, Ch. 8 . That NORMAL force has the same direction as the initial momentum of the ball and is applied at the pivot during impact. Thus linear momentum is NOT conserved within the ball-bar system. That system is "OPEN" regarding linear translations during the small time period just before and after collision. * On the other hand, even though the pivot may be rigidly attached to the Earth, the rotation process at the pivot is considered frictionless; there are no friction forces between the axle's circular outer surface and the same shaped inner pivot surface rubbing against it angularly as bar swings. Thus no angular momentum get transferred to Earth since the pivot experiences NO torque and the pivot's outer surface exerts NO torque on Earth at the Earth-pivot interface. Thus angular momentum is conserved within the ball-bar system which is "CLOSED" during the small time period just before and after collision. |