Chapter 10: Rotation

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

position zero angular position angular position: θ = s/r s: length of the arc, r:

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 Δθ = θ 2 θ 1 direction: clockwise is negative Angular velocity average: ω avg = Δθ/Δt instantaneous: ω = dθ/dt unit: rad/s, rev/s magnitude of angular velocity = angular speed Angular acceleration average: α avg = Δω/Δt 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. clockwise is negative Direction of ω is given by 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 α (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 = 2θr/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 i 2 = 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

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

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 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. units of τ: N. m Torque direction: clockwise (CW) is negative because the angle is decreasing

13 Sample Problem 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: Sample Problem 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 θ 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 20 cm). 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 θ 1 = (20)Fsin(90 o ) = 20F (CCW)

16 τ = r x F Magnitude: τ = r F sin θ 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 20 cm). 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 θ 2 = (0)Fsin(60 o ) = 0 (no direction)

17 τ = r x F Magnitude: τ = r F sin θ 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 θ 3 = (20)Fsin(90 o ) = 20F (CW)

18 τ = r x F Magnitude: τ = r F sin θ 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 20 cm). 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 θ 4 = (20)Fsin(60 o ) = 17.3F (CW)

19 τ = r x F Magnitude: τ = r F sin θ 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 θ 5 = (80)Fsin(0 o ) = 0 (no direction)

20 = r x F = r F sin θ 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

21 Newton s second law for rotation I: rotational inertia α: angular acceleration Compare to the linear equation:

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

Linear Motion vs. Rotational Motion

Linear Motion vs. Rotational Motion Linear motion involves an object moving from one point to another in a straight line. Rotational motion involves an object rotating about an axis. Examples include a