For the AP Physics Kids....

Hey, guys. I'm not quite sure why the images aren't coming through; I'll take a look and see what I can do. Best of luck tomorrow!

~Doc

Answers - Angular Motion 1. (a) For constant linear velocity, the ratio of the distance x moved to the time t, x/t = constant = v or x = vt. (b) For constant angular velocity, the ratio of the angle Θ moved to the time t, Θ/t = constant = ω or Θ= ωt. 2. (a) Constant linear acceleration a equals the change in linear velocity divided by the time to change the velocity a = (v - vo)/(t - 0) or v = vo + at. (b) Constant angular acceleration α equals the change in angular velocity divided by the time to change the velocity α = (ω - ωo)/(t - 0) or v = vo + at. 3. dx/dt = v = vo + at, or dΘ/dt = ω = wo + at, or 4. (a) ω(t) = ωo + αt. ω(4.0 s) = π s-1 + (4 π s-2)(4.0 s) =17 π s-1 (b) Θ(t) = Θo + ωot + 1/2 αt2 Θ(4.0s) = 0 + ( π s-1)(4.0s) + 1/2 (4 πs-2)(16s2) = 36 π In one rotation, the wheel turns through an angle of 2 π radians. The number of turns made by the wheel = 36 π/2 π = 18. 5. (a) ω(t) = dΘ/dt = d(b + ct + et2)/dt = c + 2et. ω(t1) = c + 2et1 (b) α(t) = dω/dt = d(c + 2et)/dt= 2e = constant = α(t1) (c) v(t1) = ω(t1)r = (c + 2et1)r (d) tangential acceleration, a(t1) = α(t1)r = 2er (e) centripetal or radial acceleration(t1) ={ v(t1)}2/r = (c + 2et1)2r2/r = (c + 2et1)2r 6. The constant angular velocity ω = (Θ - 0)/(t - 0) = Θ/t. In time t, both points 1 and 2 rotate through angle Θ so both points have the same angular velocity. For point 1, v1 = s1/t, and for point 2, v2 = s2/t. Since s2 > s1, v2 > v1. This is also shown from v1 = ωr1 and v2 = ωr2. Again since r2 > r1, v2 > v1. 7. Angular acceleration α = (ω10s - ωo)/(3 s - 0) =(6.0 rad/s - 0)/(3 s) = 2 rad/s2. Angle turned through from t = 0 to t = 10 s = ωot + 1/2 αt2 = 0 + 1/2(2 rad/s2)(100 s2) = 100 rad. ω(10 s) = ωo + αt = 0 + (2.0 rad/s2)(10s) = 20 rad/s. After t = 10 s, α = 0. Angle turned through from t = 0 to t = 20 s Θ (20 s) = Θ10s + ω10s + 1/2 αt2 = 100 rad + 20 rad/s(10 s) + 0 = 300 rad. 8. For one particle of mass m and velocity v, the kinetic energy K =1/2 mv2. For all of the particles that make up the disk from i = 1 to i = N, K = 1/2 m1v12 + 1/2 m2v22 + 1/2 m3v32 + . . . .1/2 mNvN2 = 1/2 mivi2 = 1/2 mi (ωri)2 = 1/2 miω2ri2 = 1/2 ω2 miri2 = 1/2 ω2I =1/2 Iω2, where I is the moment of inertia of the rigid body. 9. (a) I = mL2 + m(4L2) = 5mL2. (b) In general, K = 1/2 Iω2. For object closest to the axis O, K = 1/2 mL2ω2. For object farthest from O, K = 1/2 4mL2ω2. Total K = 1/2 L2ω2 (1 + 4) = 5/2(L2ω2). 10. τ = r x F τ1 = r1F1 sin 300 = 2.0 m(1.5 N)(0.5) = 1.5 N-m Point the fingers of your right hand in the direction of r1 and rotate it into F1 and your thumb points into the page. τ1 is into the page. τ2 = r2F2 sin 300 = 1.0 m(1.0 N)(0.8) = 0.5 N-m Point the fingers of your right hand in the direction of r2 and rotate it into F2 and your thumb points out of the page. τ2 is out of the page. Resultant torque τ= (1.5 - 0.8)N-m = 0.7 N-m into the page. 11. (a) Moment of inertia I of a rod about one end is 1/3 mL2. K = 1/2 Iω2 = 1/2(1/3mL2)ω2. (b) Let the initial position i be at the lowest position and the final position f be at the highest position. Take the gravitational potential energy of the center of the rod to be zero at the lowest position, that is, Ui = 0. The rod comes momentarily at rest at the highest position, that is, Kf = 0. From conservation of energy, Uf - Ui = (mg)change in height of center of mass = Kf - Ki = 1/6 (mL2)ω2. Change in height of center of mass = (L2ω2/6g). 12. Assume the center of mass of the pencil is at its center L/2 (Fig for # 12). From conservation of energy, Ui + Ki = Uf + Kf mgL/2 + 0 = 0 + 1/2 Iω2 ω = (mgL/I)1/2 = (mgL/1/3 mL2)1/2 = (3g/L)1/2 vtop of pencil = ωL = (3g/L)1/2L = (3gL)1/2 13. (a) See Fig. for #13 to the right, τabout pivot = Iabout pivot α L/2(Mg)sin 900 = 1/3 ML2 α α = 3/2 (g/L) (b) a = α L = 3/2(g/L) (L) = 3g/2 14. (a) τ = rF sin r,F = r sin r,F (F) = r (F), as shown in the upper figure of Fig. for #14a (b) τ = rF sin r,F = F sin r,F (r) = F(r), as shown in the lower figure of Fig. for #14b 15. The forces on the wheel are shown in Fig. for #15a above. Fh and Fv are the horizontal and vertical forces, respectively, of the curb on the wheel. These forces are not shown in Fig. for #15b above because they do not produce a torque about axis O. Figure for #15b shows the perpendicular distance to F and mg from O. The perpendicular distance from the axis O to the line of action of force F is (R - h). The torque due to F = (R - h)F into the page. The perpendicular distance from O to the line of action of mg = {R2 - (R - h)2}1/2 ={R2 - R2 + 2Rh - h2}1/2 = {2Rh - h2}1/2. The torque due to mg = {2Rh - h2}1/2mg out of the page. To rotate the wheel clockwise over the curb, the torque into the page must be slightly larger than the torque out of the page: (R - h)F > {2Rh - h2}1/2mg, or F must be slightly greater than {2Rh - h2}1/2mg/(R - h). 16. (a) L = r x mv. Because r x v is out of the page (Fig. for #16), L is out of the page. L = rmv sin r, v= rmv sin 900 = rmv. (b) Since v = ωr, L = (mr2)ω = Iω. 17. (a) L = rp sin r, p = p(r sin r, p) = pr (b) L = r(p sin r, p) = r p 18. L = r x mv L is into the page L = rmv sin r v = mv(r sin r, v) = mvb 19. L = r x mv L is out of the page for both particles (See Fig. 7 to the right) L = rmv sin r v = rmv sin 900 Total L = 0.50 m(5.0 m/s)(3.0 + 4.0) kg = 17.5 kg-m2/s 20. τ = r x F τ = r F sin r, F For the satellite orbiting about the earth as its axis, r, F = 1800. sin 1800 = 0 and the torque equals zero. When no torque acts on a system, its angular momentum is conserved. The angular momentum of the satellite at P1 = the angular momentum of the satellite at P2. 21. (a) The frictional force f acts at a point, not through a distance so it does no work and energy is conserved. We take the gravitational potential energy of the sphere equal to 0 at the bottom of the hill, that is, Uf = 0. The center of mass of the sphere at the top of the hill is a distance h above where it is at the bottom of the hill: Ui = Mgh. As the sphere rolls down the hill, it has kinetic energy of translation of the center of mass and it has rotational kinetic energy. From conservation of energy, Ui + Ki = Uf + Kf Mgh + 0 = 0 + 1/2 Mv2 + 1/2 Iω2 = 0 + 1/2 Mv2 + 1/2(2/5MR2)(v/R)2 Mgh = 7/10 Mv2 or v = (10gh/7)1/2 (b) (Fnet)x = ma Mg sin Θ - f = Ma (Equation 1) Take the torques about the center of mass. Since the weight passes through this axis, it produces no torque. τ = Iα Rf sin 900 = (2/5 MR2)(a/R); f = 2/5 Ma (Equation 2) (i) Substituting Eq. 2 into Eq. 1: Mg sin Θ - 2/5 Ma = Ma or a = 5g sin Θ/7 (ii) From Eq. 2, f = 2/5 Ma = (2/5)M(5g sin Θ/7) = 2 Mg sin Θ/7 (iii) v2 = vo2 + 2as = 0 + 2(5g sin Θ/7)s = (10g/7)(s sin Θ) = (10g/7)(h) v = (10gh/7)1/2 as found in Part (a). 22. For equilibrium, Στ = 0. Taking the axis at the bottom of the ladder, with τ = r F sin r, F, Στ = L Fwall sin Θ - (L/2)W cos Θ = 0 or tan Θ = W/(2 Fwall) (Equation 1) Note that neither N nor f contributes to the torque when the axis is at the bottom of the ladder because both forces pass through the axis. Also for equilibrium, σ Fx = 0 and σ Fy = 0 f - Fwall = 0 f = Fwall N - W = 0 N = W but f = µN so f = µW and µW = Fwall 51.30 (Equation 2) Substituting Eq. 2 into Eq. 1: tan Θ = W/2µW = 1/2µ = 1/0.80 = 1.25 Θ = 51.30 23. (a) Forces are shown in Fig. 10 above. The forces in addition to those shown in the statement of the problem are the attraction of the earth for the board = mg = 40 N and the force of the fulcrum on the board. For equilibrium, the vector sum of the forces = 0. 290 N - 50 N - 100 N - 40 N - 100 N = 0. (b) Taking the axis at O, the net torque = (1 m) 50 N + (0.5 m) 100 N - (1 m)100 N = 0 Neither the weight of the board nor the force of the fulcrum on the board produce rotation about O because they pass through the axis. (c) About O", the sum of the torques = -0.5 m(100 N) –1 m(40 N) – 2 m (100N) + 1 m(290 N) - 50 N-m - 40 N-m - 200 N-m + 290 N-m = 0. The force of 50 N at O" produces no torque. It does not make any difference where you take the axis as long as you stick with it through a single calculation. 24. (a) σ Fy = 0 Fnormal - mg = 0 or Fnormal = mg = 30 N (b) Take torques about the point where the object touches the floor. Then neither the frictional force nor the Fnormal contribute to the torque because these forces pass through the axis. The perpendicular component of the distance rW from the axis to the point of application of FW is shown in Fig. 5 above to be 1.50m. The perpendicular component of the distance rmg from the axis to the point of application of mg is shown in Fig. 5 to be 0.50m. Στ = 0 = +(0.5 m)30 N - (1.50 m)(FW). FW = 10 N. (c) σFx = 0 = FW - f or FW = 10 N = f (d) µ = f/Fnormal = 10 N/30 N - 1/3