Rotational Kinematics

Transcription

1 Rotational Kinematics 1

2 Linear Motion Rotational Motion all variables considered positive if motion in counterclockwise direction displacement velocity acceleration angular displacement (Δθ) angular velocity (ω) angular acceleration (α) Δθ swept out by radial line as rigid body turns about rotation axis ω = Δθ/Δt α = Δω/Δt ω in rad/s α in rad/s 2 θ = s/r θ in radians 2

3 Angular Variables Example The Sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit is 2.2 x m, and the angular speed of the sun is 1.1 x rad/s. How long (in years) does it take for the Sun to make one revolution around r ω the center? looking for: Δt (in years) solve for Δt: to find Δθ: plug in: ω = Δθ/Δt Δt = Δθ/ω Δθ = 2π rad Δt = (2π rad)/ω given: r = 2.2 x m ω = 1.1 x rad/s Δt = (2π rad)/(1.1 x rad/s) Δt = 5.7 x s (min/60 s)(hr/60min)(d/24 hr)(yr/365 d) Δt = 1.8 x 10 8 yr 3

4 Equations of Rotational Kinematics for use when angular acceleration is constant 4

5 Using the Equations of Rotational Kinematics read & interpret problem list givens & unknown make sure units of givens are consistent choose kinematic equation rearrange equation if needed substitute givens into equation solve for unknown remember: when motion divided into segments, final angular velocity of one segment is initial angular velocity of next segment From Physics Concepts and Connections Book 2 remember: there may be two possible answers-- choose the answer that makes sense physically 5

6 Rotational Kinematics Example A figure skater is spinning with an angular velocity of +15 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is +5.1 rad. Determine (a) her average angular acceleration and (b) the time top view: during which she comes to rest. ω1 given: ω1 = +15 rad/s ω2 = 0 rad/s Δθ = +5.1 rad looking for: (a) α (b) Δt part (a): ω2 2 = ω αΔθ solve for α: 2αΔθ = ω2 2 - ω1 2 α = (ω2 2 - ω1 2 )/(2Δθ) α = [(0 rad/s) 2 - (15 rad/s) 2 ]/[2(5.1 rad)] α = -22 rad/s 2 6

7 Rotational Kinematics Example A figure skater is spinning with an angular velocity of +15 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is +5.1 rad. Determine (a) her average angular acceleration and (b) the time top view: during which she comes to rest. ω given: ω1 = +15 rad/s ω2 = 0 rad/s Δθ = +5.1 rad looking for: (a) α (b) Δt solve for Δt: to find Δω: part (b): α = Δω/Δt Δt = Δω/α Δt = (ω2 - ω1)/α Δt = [(0 rad/s) - (15 rad/s)]/(-22 rad/s 2 ) Δt = 0.68 s 7

8 Tangential Velocity and Acceleration vt = rω if vt increases, we have acceleration: at (tangential acceleration) angular quantities describe motion of entire rigid object, tangential quantities describe motion of single point on object at = (vt - vt0)/t at = [(rω) - (rω0)]/t at = r[(ω - ω0)/t] T = (ω - ω0)/t at = r ( in rad/s 2 ) 8

9 Centripetal Acceleration & Angular Speed remember: centripetal acceleration applies to uniform circular motion (constant tangential speed) ac = vt 2 /r vt = rω ac = (rω) 2 /r = rω 2 (ω in rad/s) 9

10 Nonuniform Circular Motion when tangential speed is changing direction of vt still changing--ac ac = rω 2 magnitude of vt changing also now--at at = r also FT = mat total acceleration a = ac + at 10

11 Rolling Motion no slipping where rolling object meets the ground example: tires on a normally moving car not rolling motion: slipping tires of race car when it starts 11

12 Rolling Motion when no slipping, there is a relationship between angular speed of tire rotation and linear speed at which car moves forward linear speed: v = d/t vt = s/t v = vt = rω (ω in rad/s) a = at = r ( in rad/s 2 ) 12

13 Directions of Angular Vectors ω & are vectors direction of ω points along axis of rotation right-hand rule: grab the rotation axis with right hand with fingers pointing in direction of rotation--extended thumb points along axis in direction of ω no part of the rotating object moves in the direction of ω comes about when ω changes-- is also along rotation axis in direction of Δω (for increasing ω, points in same direction as ω; for decreasing ω, points in opposite direction as ω) 13

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17 Rotational Dynamics Torque, Equilibrium, & Center of Gravity 1

18 Translational vs. Rotational Motion translational motion: all points on the object travel on parallel paths (not necessarily straight lines) happens when a force on an object is applied through the center of mass rotational motion: object turning about an axis an additional motion that occurs when a force on an object is applied not through the center of mass 2

19 Torque rotational effect on a body caused by an applied force magnitude of torque: τ = (magnitude of force) (lever arm) lever arm: = distance between line of action and axis of rotation line of action: line extended out from force vector τ = F convention: positive when force produces counterclockwise rotation, negative when force produces clockwise rotation SI unit: [N] [m] = [N m] 3

20 Rigid Body Equilibrium A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero: Newton s Second Law: ΣF = ma For rigid body equilibrium: ΣFx = 0 ΣFy = 0 Στ = 0 4

21 Applying Conditions of Equilibrium to a Rigid Body 1. Draw free-body diagram for rigid body that shows all external forces acting on the body. 2. Resolve all forces into x & y components. 3. If a = 0, set ΣFx = 0 & ΣFy = If there is ω, choose an axis of rotation for the body. Find where each external force acts on the body & calculate τ produced by each force about the axis. 5. If = 0, set Στ = Solve all equations for unknowns. 5

22 Center of Gravity The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. for a symmetrically shaped object with uniform weight distribution: center of gravity not necessarily within object itself center of gravity located at geometrical center 6

23 Finding Center of Gravity for a group of objects, you can find a collective center of gravity: similar to center of mass xcg = (W1x1 + W2x2)/(W1 + W2) add more terms for more objects also not necessarily within any object 7

24 Balancing Center of Gravity To an object (or system of objects) at rest on a horizontal surface, what force(s) is/are being applied? FG, FN For a = 0, F = 0 FN FG center of gravity support must be under center of gravity 8

25 Equilibrium Example The drawing shows a person ( weight W = 584 N) doing pushups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. Nfeet = 0 axis FNfeet +y W W FNhands given: W = 584 N looking for: FN for each hand & foot (1/2)FNfeet (1/2)FNhands Nhands 2) 5) 4) 6) 3) 1) if calculate draw resolve α = for free-body 0, forces set unknowns torque ΣFx Στ into = diagram for 0 and x- each and ΣFy force y-components = 0 to find FNhands: ΣFy = FNfeet + FNhands - W τ = F τnfeet = 0 τw = -W W τnhands = FNhands Nhands = 0 Στ = FNhands Nhands - W W = 0 FNhands = (W W)/( Nhands) FNhands = [(584 N)(0.840 m)]/(0.840 m m) FNhands = 392 N (1/2)FNhands = 196 N 9

26 Equilibrium Example The drawing shows a person ( weight W = 584 N) doing pushups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. Nfeet = 0 axis FNfeet +y W W FNhands given: W = 584 N looking for: FN for each hand & foot (1/2)FNfeet (1/2)FNhands Nhands 6) solve for unknowns to find FNfeet: ΣFy = FNfeet + FNhands - W = 0 τ = F τnfeet = 0 τw = -W W τnhands = FNhands Nhands Στ = FNhands Nhands - W W = 0 FNfeet = 192 N FNfeet = W - FNhands FNfeet = 584 N N (1/2)FNfeet = 96 N 10