Chapter 10: Rotation

Transcription

1 Chapter 10: Rotation Review of translational motion (motion along a straight line) Position x Displacement x Velocity v = dx/dt Acceleration a = dv/dt Mass m Newton s second law F = ma Work W = Fdcosφ Kinetic energy K = ½ mv 2 What about rotational motion?

2 Rotational variables We will focus on the rotation of a rigid body about a fixed axis Reference line: pick a point, draw a line perpendicular to the rotation axis Angular position zero angular position angular position: θ = s/r s: length of the arc, r: radius Unit of θ: radians (rad) 1 rev = 360 o = 2πr/r= 2π rad 1 rad = 57.3 o = rev

3 Angular displacement θ = θ 2 θ 1 direction: clockwise is negative Angular velocity average: ω avg = θ/ t instantaneous: ω = dθ/dt unit: rad/s, rev/s magnitude of angular velocity = angular speed Angular acceleration average: α avg = ω/ t instantaneous: α = dω/dt unit: rad/s 2

4 Angular velocity and angular acceleration are vectors. For rotation along a fixed axis, we need not consider vectors. We can just use + and sign to represent the direction of ω and α. clockwise is negative Direction of ω : right hand rule

5 Rotation with constant angular acceleration The equations for constant angular acceleration are similar to those for constant linear acceleration replace θ for x, ω for v, and α for a, missing v = v 0 + at ω = ω 0 + αt θ θ 0 x x 0 = v 0 t + ½ at 2 θ θ 0 = ω 0 t + ½ αt 2 ω v 2 = v a(x x 0 ) ω 2 = ω 02 +2α (θ θ 0 ) t and two more equations x x 0 = ½ (v 0 + v)t θ θ 0 = ½(ω 0 ω)t α x x 0 = vt ½ at 2 θ θ 0 = ωt ½ αt 2 ω 0

6 Relating the linear and angular variables The linear and angular quantities are related by radius r The position θ *in* radians! distance s = θ r The speed ds/dt = d(θ r)/dt = (dθ/dt)r v = ω r Time for one revolution T = 2πr/v = 2π/ω Note: θ and ω must be in radian measure

7 Acceleration dv/dt = d(ωr)/dt = (dω/dt)r tangential component a t = α r ( α = dω/dt ) α must be in radian measure radial component a r = v 2 /r = (ωr) 2 /r = ω 2 r Note: a r is present whenever angular velocity is not zero (i.e. when there is rotation), a t is present whenever angular acceleration is not zero (i.e. the angular velocity is not constant)

8 Cp11-3: A cockroach rides the rim of a rotating merry-goaround. If the angular speed of this system (merry-go-around+ cockroach) is constant, does the cockroach have (a ) radial acceleration? (b) tangential acceleration? If the angular speed is decreasing, does the cockroach have (c) radial acceleration? (d) tangential acceleration?

9 Kinetic Energy of Rotation Consider a rigid body rotating around a fixed axis as a collection of particles with different linear speed, the total kinetic energy is K = Σ ½ m i v i2 = Σ ½ m i (ω r i ) 2 = ½ ( Σ m i r i 2 ) ω 2 Define rotational inertia ( moment of inertia) to be I = Σ m i r i 2 r i : the perpendicular distance between m i and the given rotation axis Then K = ½ I ω 2 Compare to the linear case: K = ½ m v 2

10 Rotational inertia involves not only the mass but also the distribution of mass for continuous masses Calculating the rotational inertia I = r 2 dm

11 Parallel-Axis theorem If we know the rotational inertia of a body about any axis that passes through its center-of-mass, we can find its rotational inertia about any other axis parallel to that axis with the parallel axis theorem I = I c.m. + M h 2 h: the perpendicular distance between the two axes

12 Torque The ability of a force F to rotate a body depends not only on its magnitude, but also on its direction and where it is applied. Torque is a quantity to measure this ability Torque is a VECTOR τ = r F sin φ F is applied at point P. τ = r x F r : distance from P to the rotation axis. unit: N. m direction: clockwise (CW) is negative because the angle is decreasing

13 Torque The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F

14 Force producing the greatest (most positive) torque: Torque 1. F 1 2. F 2 F 1 = F 2 = F 3 = F 4 = F 5 =F 3. F 3 4. F 4 5. F 5 6. F 1 and F 3 7. need more information

15 τ = r x F Magnitude: τ = r F sin φ Checkpoint 10-6 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F F 1 : τ 1 = r 1 F 1 sin φ 1 = (20)Fsin(90 o ) = 20F (CCW) v v v τ1 = r1 F1 = + 20F

16 τ = r x F Magnitude: τ = r F sin φ Checkpoint 10-6 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F F 2 : τ 2 = r 2 F 2 sin φ 2 = (0)Fsin(60 o ) = 0 (no direction) v v v τ2 = r2 F2 = 0F = 0

17 τ = r x F Magnitude: τ = r F sin φ Checkpoint 10-6 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F F 3 : τ 3 = r 3 F 3 sin φ 3 = (20)Fsin(90 o ) = 20F (CW) v v v τ3 = r3 F3 = 20F

18 τ = r x F Magnitude: τ = r F sin φ Checkpoint 10-6 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F F 4 : τ 4 = r 4 F 4 sin φ 4 = (20)Fsin(60 o ) = 17.3F (CW) v v v τ4 = r4 F4 = 17.3F

19 τ = r x F Magnitude: τ = r F sin φ Checkpoint 10-6 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F F 5 : τ 5 = r 5 F 5 sin φ 5 = (80)Fsin(0 o ) = 0 (no direction) v v v τ5 = r5 F5 = 0F = 0

20 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F 1 = F 2 = F 3 = F 4 = F 5 =F τ = r x F τ = r F sin φ Checkpoint 10-6 v v v τ1 = r1 F1 = + 20F v v v τ2 = r2 F2 = 0F = 0 v v v τ5 = r5 F5 = 0F = 0 v v v τ4 = r4 F4 = 17.3F v v v τ = F = 20F 3 r3 3

21 Newton s second law for rotation ma F F n 1 i i net v v v = = = Compare to the linear equation: α = = τ = v v v v I F r n 1 i i i net I: rotational inertia α: angular acceleration

22 Balanced Torques Forces F 1 and F 2 are applied on a meter stick which is free to rotate around the pivot point. Only F 1 is shown. F 2 is perpendicular to the stick (in the same plane as F 1 and the stick) and is applied at the right end (point labeled 2 ). The stick is not to rotate. r 1 r 2 2 F 2 As viewed from above, what should be the direction of F 2?

23 Balanced Torques-1 As viewed from above, what should be the direction of F 2 so the rod doesn t rotate? r 1 r 2 2 F 2 1. up 2. down 3. not enough information

24 Forces F 1 and F 2 are applied on a meter stick which is free to rotate around the pivot point. Only F 1 is shown. F 2 is perpendicular to the stick (in the same plane as F 1 and the stick) and is applied at the right end (point labeled 2 ). The stick is not to rotate. r 1 r 2 2 F2 As viewed from above, what should be the direction of F 2? DOWN or else both forces will cause the rod to rotate CW

25 Forces F 1 and F 2 are applied on a meter stick which is free to rotate around the pivot point. Only F 1 is shown. F 2 is perpendicular to the stick (in the same plane as F 1 and the stick) and is applied at the right end (point labeled 2 ). The stick is not to rotate. r 1 r 2 2 F2 Should F 2 be greater than, less than, or equal to F 1?

26 Balanced Torques-2 Should F 2 be greater than, less than, or equal to F 1? r 1 r 2 2 F2 1. F 2 > F 1 2. F 2 < F 1 3. F 2 = F 1 4. F 2 can t balance F 1 5. none of the above

27 Forces F 1 and F 2 are applied on a meter stick which is free to rotate around the pivot point. Only F 1 is shown. F 2 is perpendicular to the stick (in the same plane as F 1 and the stick) and is applied at the right end (point labeled 2 ). The stick is not to rotate. r 1 r 2 2 F2 Should F 2 be greater than, less than, or equal to F 1? r 2 > r 1 => F 2 must be less that F 1 so the torques cancel

28 Sample Problem 10-9 This figure shows a uniform disk, with mass M=2.5kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. Note: R = 20 cm = 0.2m y a

29 This figure shows a uniform disk, with mass M=2.5kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. The key points are following: For the block: T mg = ma (1) For the pulley: τ = I α RT = (1/2)MR 2 α (2) acceleration a of the block is equal to a t at the rim of the pulley a = a t = α R (3) Three equations and three unknowns: a, α, T, so a unique solution exists. a y

30 Sample Problem This figure shows a uniform disk, with mass M=2.5kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. Equations 2&3: T = (1/2)Ma Equations 2&3 and 1: a = αr = (Ma)/2m g Equations 1,3 and 2: a = αr = g MR α/2m 1 M 2m + a = g a = g = 2m ( M + 2m) a 2 4.8m/s y

31 Sample Problem This figure shows a uniform disk, with mass M=2.5kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. Then, y T = 1/2Ma = 6.0N and, a α = a/r = 24 rad/s 2

32 Work and Rotational Kinetic Energy Work-kinetic energy theorem : W = K = K f K i K = ½ I ω f2 ½ I ω i2 (if there is only rotation) θ Work done = f x v W τ dθ (compare to v W = f F dx ) θi if τ is constant, W = τ (θ f θ i ) x i Power P = dw/dt P = τ dθ/dt = τ (dθ/dt) = τ ω compare to P = v v F

33 translational motion Quantity Rotational motion x Position θ x Displacement θ v = dx/dt Velocity ω=dθ/dt a = dv/dt Acceleration m Mass Inertia I F = ma W x v v F dx = f x i Newton s second law Work τ = r x F K = ½ mv 2 Kinetic energy K = ½ Iω 2 v v P = F Power (constant F or τ) P = τω W θ = f θ i τ dθ

34 Chapter 10: Rotation Review of translational motion (motion along a straight line) Position x Displacement x Velocity v = dx/dt Acceleration a = dv/dt Mass m Newton s second law F = ma Work W = Fdcosφ Kinetic energy K = ½ mv 2 What about rotational motion?

35 Kinetic Energy of Rotation Consider a rigid body rotating around a fixed axis as a collection of particles with different linear speed, the total kinetic energy is K = Σ ½ m i v i2 = Σ ½ m i (ω r i ) 2 = ½ ( Σ m i r i 2 ) ω 2 Define rotational inertia (moment of inertia) to be I = Σ m i r i 2 r i : the perpendicular distance between m i and the given rotation axis Then K = ½ I ω 2 Compare to the linear case: K = ½ m v 2

36 Rotational inertia (moment of inertia) Define rotational inertia (moment of inertia) to be I = Σ m i r 2 i or 2 I = r i : the perpendicular distance between m i and the given rotation axis r dm m 1 m 2 x 1 x 2 Moment of inertia I = m x m2x 2

37 Rotational inertia (moment of inertia) Define rotational inertia (moment of inertia) to be I = r i : the perpendicular distance between m i and the given rotation axis dm = ρ rdθ where ρ = M/2πr r 2 dm r = a I = Moment of inertia = r r 3 2 dm ρ 2π 0 = 2π 0 dθ = r 2 ρrdθ = 2πr 3 2π 0 r 3 ρ = 2πr ρdθ 3 M 2πr = Mr 2

38 Rotational inertia involves not only the mass but also the distribution of mass for continuous masses Calculating the rotational inertia I = r 2 dm

39 Parallel-Axis theorem If we know the rotational inertia of a body about any axis that passes through its center-of-mass, we can find its rotational inertia about any other axis parallel to that axis with the parallel axis theorem I = I c.m. + M h 2 h: the perpendicular distance between the two axes

40 Work and Rotational Kinetic Energy Work-kinetic energy theorem: W = K = K f K i K = ½ I ω f2 ½ I ω i2 (if there is only rotation) = f θ θ Work done W τ dθ (compare to W = f F dx ) i if τ is constant, W = τ (θ f θ i ) x v x i v Power P = dw/dt P = τ dθ/dt = τ (dθ/dt) = τ ω. compare to P = F v

41 Summary -- Translation - Rotation translational motion Quantity Rotational motion x Position θ x Displacement θ v = dx/dt Velocity ω=dθ/dt a = dv/dt Acceleration α m Mass Inertia I F = ma Newton s second law τ = r x F x v θ v W = F dx Work W = τ dθ x θi i K = ½ mv 2 Kinetic energy K = ½ Iω 2 v v P = F Power (constant F or τ) P = τω

42 Rolling viewed as a combination of pure rotation and pure translation Rolling viewed as pure rotation v top = (ω)(2r) = 2 v com Different views, same conclusion

43 The Kinetic Energy of Rolling View the rolling as pure rotation around P, the kinetic energy K = ½ I P ω 2 parallel axis theorem: I p = I com +MR 2 so K = ½ I com ω 2 + ½ MR 2 ω 2 since v com = ωr K = ½ I com ω 2 + ½ M(v com ) 2 ½ I com ω 2 : due to the object s rotation about its center of mass ½ M(v com ) 2 : due to the translational motion of its center of mass

44 Torque revisited For a fixed axis rotation, torque τ = r F sinθ direction : right-hand rule v τ = v r v F magnitude : τ = rfsin Φ = rf = r F