Physics 2210 Fall smartphysics 16 Rotational Dynamics 11/13/2015

Size: px

Start display at page:

Download "Physics 2210 Fall smartphysics 16 Rotational Dynamics 11/13/2015"

Ethelbert Merritt
6 years ago
Views:

1 Physics 10 Fall 015 smartphysics 16 Rotational Dynamics 11/13/015

2 A rotor consists of a thin rod of length l=60 cm, mass m=10.0 kg, with two spheres attached to the ends. Each sphere has radius R=10 cm, and mass M=5.0 kg. A force of magnitude F=48 N is applied at a distance of l/4 to the right of the axle in the direction shown (a) Calculate the moment-of-inertia of the rotor. (b) Find the angular acceleration of the rotor. Example 15. (1/) Solution (a) The moment of inertia of the rotor (about the axle shown) is the sum of the 3 parts I = I rrr + I ssssss where I rrr = ml 1 since the axel goes through its CM perpendicularly, and each sphere has (by parallel axis theorem) I ssssss = I CC + MD = MR 5 + M R + l And to total is therefore I = 1 1 ml MR + M R + l 10 kg 0.60 m kg 0.10 m = kg 0.10 m m 1 5 = 0.30 kg m kg m kg m = 1.94 kg m l 10 l 4 F R

3 A rotor consists of a thin rod of length l=60 cm, mass m=10.0 kg, with two spheres attached to the ends. Each sphere has radius R=10 cm, and mass 5.0 kg. A force of magnitude F=48 N is applied at a distance of l/4 to the right of the axle in the direction shown (a) Calculate the moment-of-inertia of the rotor. (b) Find the angular acceleration of the rotor. Solution (b): The torque on the rotor is given by τ = rr sin φ From the zoom of the diagram we have r = l 4 and φ = 10 (not immediately obvious) And so the torque is Example 15. (/) 10 φ = 10 r = l 4 τ = m 48 N sin 10 lf sin φ = = 6.4 N m 4 4 The angular acceleration is given by α = τ I = 6.4 N m = 3.1 rad/s 1.94 kg m *** Units: N m kg m = kg m s m kg m = kg m m s kg m = 1 s = rad s l l 4 10 F R F

Rotational Dynamics Unfortunately as is usually the case with smartphysics, they give you a nice convenient formula and almost invites you to memorize it BUT IT IS MISLEADING and not entirely correct

4 Rotational Dynamics Unfortunately as is usually the case with smartphysics, they give you a nice convenient formula and almost invites you to memorize it BUT IT IS MISLEADING and not entirely correct Work done by a torque (often not associated with a conservative force): Remembering, however, that θ W = τττ, K = W θ 1 K = K CC + K for rigid bodies K = 1 MV CC + 1 I CCω Often more convenient to use (W NN =work done by none-conservative forces) E = K + U = W NN

5 Rotational Dynamics Problem: It is not easy to go from the nd box to the 3 rd NOT clear which equation they are talking about

6 In general, the linear motion of the center-of-mass and the rotational motion about the center-of-mass are independent of one another 3:53 4:03

7 Rolling (without slipping) See f S In rolling, rotational and translational are locked together (no slipping!!!): Displacement of CM in ONE rotation (π rad) = X CC = Circumference = ππ CM velocity and angular velocity about the CM are related by: V CC = Rω Static friction force f S is perpendicular to the direction of motion at all times: f S does NO work

8 Gravitational Potential Energy of an Extended Body For a collection of particles with mass m 1, m, m N at positions x 1, y 1, x, y, x N, y N, where the +y direction is assumed to be UP, and the reference ZERO for potential energy is taken at y = 0, The gravitational potential energy of the body is the SUM of those of the individual particles: U g = N U gg i=1 = g 1 N M M m iy i i=1 N = m i gy i i=1 N = g m i y i i=1 = MM 1 N M m iy i i=1 U g = MMY CC i.e. The gravitational potential energy of a body = gravitational potential energy of the center-of-mass (treated like a point particle)

Poll 11-13-01 M M A block and a ball have the same mass M and move with the same initial velocity across a floor and then encounter identical ramps.

9 Poll M M A block and a ball have the same mass M and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp? A. Block B. Ball C. Both reach the same height.

Poll 11-13-0 A cylinder and a hoop have the same mass and radius.

10 Poll A cylinder and a hoop have the same mass and radius. They are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first? A. Cylinder B. Hoop C. Both reach the bottom at the same time.

Poll 11-13-03 A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without

11 Poll A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first? A. Small cylinder B. Large cylinder C. Both reach the bottom at the same time.

12 Example 16.1 (1/4) A round wheel of mass M, radius R and I CC = γγr (0 < γ < 1) rolls without slipping down an incline at angle φ from the horizontal. Find the linear acceleration of the wheel down the incline. Solution: Solution: the kinetic energy of the wheel is given by I CC = γγr R MM φ K = 1 MV CC + 1 I CCω For rolling without slipping: V CC = Rω, and we have I CC = γγr : K = 1 MV CC + 1 γγr ω = 1 MV CC + 1 γγ(r ω ) = 1 MV CC + 1 γmv CC = γ MV CC If we start from rest, then by conservation of energy: E = K + U = 0 K = γ MV CC = U = MM H CC Taking down the ramp to be positive x direction then H CC = X CC sin φ. But: X CC = x and V CC v x γ Mv x = MM x sin φ H x φ

Solution (continued): Differentiating both sides with respect to time: 1 1 + γ M v x dv x dd = MM sin φ dd dd 1 a x = MMv x sin φ 1 + γ Mv x a x = + γ Mv xa x = MMv x

13 Example 16.1 (/4) A round wheel of mass M, radius R and I CC = γγr (0 < γ < 1) rolls without slipping down an incline at angle φ from the horizontal. Find the linear acceleration of the wheel down the incline. Solution (continued): Differentiating both sides with respect to time: γ M v x dv x dd = MM sin φ dd dd 1 a x = MMv x sin φ 1 + γ Mv x a x = + γ Mv xa x = MMv x sin φ g sin φ 1 + γ Example given in Main Point: Solid Ball/Sphere I = 5 MR γ = γ = γ = 7 5 a x = 7 g sin φ for a solid ball 5 Note that we used φ here for the angle of the inclined ramp instead of θ to avoid confusion between the rotation angle of the ball and the angle of the incline

14 Example 16.1 (3/4) A round wheel of mass M, radius R and I CC = γγr (0 < γ < 1) rolls without slipping down an incline at angle φ from the horizontal. Find the linear acceleration of the wheel down the incline. I CC = γγr φ Alternate Solution: Let +x be down slope and +y point perpendicularly out of the incline. We will chose CCW As the positive rotation sense. For rolling without slipping then we have: v x = Rω, a x = RR Wherev x and a x are the velocity and acceleration components down slope. Applying Newton s Second Law on the CM, first in the y-direction Ma y = N MM cos φ = 0 N = MM cos φ Which is not really used in this problem except possibly to verify that f s < μ s N In the x-direction : Ma x = MM sin φ f s 1 Next we look at rotation about the CM: I CC α = Rf s sin( 90 ) = Rf s *** We have used the fact that, for the purpose of calculating torque, the force of gravity acts at the center-of-mass (CM). This follows from I CC α = Rf s sin( 90 ) = Rf s f s R N MM y x

15 Example 16.1 (4/4) A round wheel of mass M, radius R and I CC = γγr (0 < γ < 1) rolls without slipping down an incline at angle φ from the horizontal. Find the linear acceleration of the wheel down the incline. Solution (continued): From last page Ma x = MM sin φ f s 1 I CC α = Rf s sin( 90 ) = Rf s But a x = RR, α = a x R, and so we have γγr a x R = Rf s, γγγ x = f s () substituting () back into (1) for f s : Ma x = MM sin φ γγγ x, 1 + γ Ma x = MM sin φ And so we have: a x = g sin φ 1 + γ

16 Example 16. (1/6) A 115 kg car is being unloaded by a winch. At the moment shown below, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 344 kg m and that of the pulley is 3 kg m. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water. Solution: Looking for the speed of something as it drops through a certain distance. Does it remind you of a problem to solve using conservation of energy? Except in this case, as the car acquires speed, both the pulley and the drum acquire angular speed (no slipping!!!): Drum: R D ω D = v C... (1), Pulley: R P ω P = v C (). The total kinetic energy of the system is the sum of those of the three objects: K = K D + K P + K C = 1 I Dω D + 1 I Pω P + 1 M Cv C (3) Substituting (1) and () into (3) gives us K = 1 I D v C R D + 1 I P v C R P + 1 M Cv C = 1 I D R D + I P R P + M C v C

17 Example 16. (/6) A 115 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg m, of pulley: 3 kg m. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. I D K = 1 R + I P D R + M C v C P No friction: total energy is conserved. The only potential energy is the gravitational potential energy of the car: so we have E = K + U = 0. We started from rest: K i = 0 So the final kinetic energy (after 5.0m drop) is then K f = K i + K = U = M C g h, h = 5.0m And so v CC = 1 M C g h I D R + I P D R + M C P = m s I D R + I P D R + M C P = v Cf = M C g h 115kg 9.8 m s ( 5.0m) 344 kg m 0.80 m + 3 = kg m 0.30 m + 115kg v CC = 8.10 m s kg m s kg

The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.

18 Example 16. (1/6) A 115 kg car is being unloaded by a winch. At the moment shown below, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 344 kg m and that of the pulley is 3 kg m. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water. Solution: *** The figure has been annotated *** (%i1) /* see the free body diagram for the car We choose down to be the positive direction for y Newton's nd Law in the y direction is then */ eqn1: m*a = m*g - T1; (%o1) a m = g m - T1 (%i) /* solve for T1 */ soln1: solve(eqn1, T1); (%o) [T1 = (g - a) m] (%i3) T1a: rhs(soln1[1]); (%o3) (g - a) m... continued

(%i4) /* next observe the free body diagram for the disk pulley. We choose Clock-wise (C) to be the positive sense for rotation. Let Rp = radius of pulley.

19 Example 16. (/6) A 115 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg m, of pulley: 3 kg m. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i4) /* next observe the free body diagram for the disk pulley. We choose Clock-wise (C) to be the positive sense for rotation. Let Rp = radius of pulley. angular acceleration of the pulley is then linked to the acceleration of the car by */ alpha_p: a/rp; a (%o4) -- Rp (%i5) /* Rotational Newton's nd Law for the pulley */ eqn: Ip*alpha_p = Rp*T1 - Rp*T; a Ip (%o5) ---- = Rp T1 - Rp T Rp (%i6) /* substitute T1 from eqn1 into eqn */ eqna: eqn, T1=T1a; a Ip (%o6) ---- = (g - a) m Rp - Rp T Rp... continued

20 Example 16. (3/6) A 115 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg m, of pulley: 3 kg m. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i7) /* solve fopr T */ solna: solve(eqna, T); (g - a) m Rp - a Ip (%o7) [T = ] Rp (%i8) Ta: rhs(solna[1]); (g - a) m Rp - a Ip (%o8) Rp... continued

21 Example 16. (4/6) A 115 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg m, of pulley: 3 kg m. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i9) /* Now look at the free body diagram for the drum. Again let the positive direction of rotation be clock-wise. Let Rd be the radius of the drum. Again the angular acceleration of the drum is linked to the linear acceleration of the car by the cord */ alpha_d: a/rd; a (%o9) -- Rd (%i10) /* Newton's nd law for the drum */ eqn3: Id*alpha_d = Rd*T; a Id (%o10) ---- = Rd T Rd (%i11) /* substittute in T from eqna */ eqn3a: eqn3, T=Ta; a Id Rd ((g - a) m Rp - a Ip) (%o11) ---- = Rd Rp... continued

22 Example 16. (5/6) A 115 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg m, of pulley: 3 kg m. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i1) /* now we can solve for "a" */ soln3a: solve(eqn3a, a); g m Rd Rp (%o1) [a = ] (m Rd + Id) Rp + Ip Rd (%i13) aa: rhs(soln3a[1]); g m Rd Rp (%o13) (m Rd + Id) Rp + Ip Rd (%i14) /* solve for a value */ aa_val: aa, g=9.81, m=115, Rd=0.80, Rp=0.30, Id=344, Ip=3; (%o14) continued

23 Example 16. (6/6) A 115 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg m, of pulley: 3 kg m. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i15) /* For this last part we can use vf^ - vi^ = *a*dy */ kill(a); (%o15) done (%i16) /* and since vi=0 then we have */ vf: sqrt(*a*dy); (%o16) (%i17) vf, a=aa_val, Dy=5.0, numer; sqrt() sqrt(a Dy) (%o17) Answer: vf = 8.10 m/s

Webreview Torque and Rotation Practice Test

Webreview Torque and Rotation Practice Test Please do not write on test. ID A Webreview - 8.2 Torque and Rotation Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A 0.30-m-radius automobile