Two-Dimensional Rotational Kinematics

Two-Dimensional Rotational Kinematics

Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid bodies.

Rotation and Translation of Rigid Body Demonstration: Motion of a thrown baton Translational motion: external force of gravity acts on center of mass Rotational Motion: object rotates about center of mass

Recall: Translational Motion of the Center of Mass Total momentum of system of particles p sys = m External force and acceleration of center of mass F dp dt total V dv cm sys total = = total cm total ext m = m dt A cm

Main Idea: Rotation of Rigid Body Torque produces angular acceleration about center of mass total τ cm = I cm α cm I cm α cm is the moment of inertial about the center of mass is the angular acceleration about center of mass

Two-Dimensional Rotation Fixed axis rotation: Disc is rotating about axis passing through the center of the disc and is perpendicular to the plane of the disc. Plane of motion is fixed: For straight line motion, bicycle wheel rotates about fixed direction and center of mass is translating

Cylindrical Coordinate System Coordinates Unit vectors (r,θ, z) ( rˆ, θˆ, zˆ )

Circular Motion for point-like particle vector description Use plane polar coordinates Position r( t) = Rrˆ ( t) Velocity dθ v ( t) = R θ ˆ( t) = R ω θ ˆ( t) dt Acceleration ˆ t a = a rˆ + a θ a = rα a = r = v r r 2 2, r ω ( / ) t

Rotational Kinematics for Fixed Axis Rotation A point like particle undergoing circular motion at a non-constant speed has (1) An angular velocity vector (2) an angular acceleration vector

Fixed Axis Rotation: Angular Velocity Angle variable SI unit: Angular velocity SI unit: Vector: Component magnitude direction ω > ω < θ [rad] ˆ dθ ω ω k kˆ dt rad s 1 ω dθ dt ω dθ dt 0, direction 0, direction + kˆ kˆ

Fixed Axis Rotation: Angular Angular acceleration: SI unit Vector: Component: Magnitude: Direction: Acceleration α α kˆ α d 2 θ dt 2 α rad s 2 dω dt dω dt dω > 0, direction dt dω < 0, direction dt 2 d θ kˆ 2 dt + k ˆ k ˆ

Checkpoint Problem: Angular Velocity Consider the uniformly rotating object shown in the figure below. What is the direction of the angular velocity of the object?

Rotational Kinematics: Constant Angular Acceleration The angular quantities θ, ω, and α are exactly analogous to the quantities x, v x, and a x for one-dimensional motion, and obey the same type of integral relations t 0 = t dt t 0 0 ω( t) ω α( ʹ ) ʹ, Constant angular acceleration: θ ( ) θ = ω( tʹ ) dtʹ. t 0 ω(t) = ω 0 + α t θ(t) = θ 0 + ω 0 t + 1 2 α t 2 2 2 ( ω t ) = ω + α ( θ t θ ) ( ) 2 ( ). 0 0

Checkpoint Problem: Rotational Kinematics A turntable is a uniform disc of mass m and a radius R. The turntable is initially spinning clockwise when looked down on from above at a constant frequency f. The motor is turned off and the turntable slows to a stop in t seconds with constant angular deceleration. a) What is the direction and magnitude of the initial angular velocity of the turntable? b) What is the direction and magnitude of the angular acceleration of the turntable? c) What is the total angle in radians that the turntable spins while slowing down?

Summary: Kinematics of Circular Motion Arc length s = Rθ Tangential velocity v tan = ds dt = R dθ dt = Rω Tangential acceleration a tan = dv dt = R d 2 θ dt = Rα 2 Centripetal Rotational kinetic energy a rad = vω = v2 R = Rω 2 K rot = 1 2 mv 2 = 1 tan 2 mr2 ω 2

Worked Example: Simple Pendulum Simple Pendulum: bob of mass m hanging from end of massless string of length I pivoted at S. Angular velocity Angular acceleration Kinetic energy of rotation ω = dθ dt ˆk α 2 d θ = kˆ 2 dt K = 1 2 ml 2 ω 2

Worked Example: Simple Pendulum: Mechanical Energy A simple pendulum is released from rest at an angle θ 0. Find angular speed at angle θ

Velocity Worked Example Simple Pendulum: Mechanical Energy Kinetic energy Initial energy Final energy Conservation of energy K f v tan = l dθ dt = 1 2 mv 2 = 1 tan 2 m l dθ dt E 0 = K 0 + U 0 = mgl(1 cosθ 0 ) E f = K f + U f = 1 2 m l dθ dt 1 2 m l dθ dt 2 2 2 + mgl(1 cosθ) + mgl(1 cosθ) = mgl(1 cosθ 0 ) dθ dt = 2 l ( mg(1 cosθ) + g(1 cosθ 0 ))

Rigid Body Kinematics for Fixed Axis Rotation Body rotates with angular velocity acceleration α ω and angular

Divide Body into Small Elements Body rotates with angular velocity, angular acceleration Individual elements of mass Radius of orbit ω α Δm i r,i Tangential velocity Tangential acceleration Radial Acceleration v tan,i = r,i ω a tan,i = r,i α a rad,i = v 2 tan,i = r r,i ω 2,i

Rotational Kinetic Energy and Moment of Inertia Rotational kinetic energy about axis passing through S K rot,i = 1 2 Δm v 2 = 1 i tan,i 2 Δm ( r ) 2 i,i ω 2 K = 1 2 I ω 2 cm cm cm Moment of Inertia about S : SI Unit: kg m Continuous body: Rotational Kinetic Energy: 2 i= N I S = Δm i (r,i ) 2 i=1 I S = dm (r,dm ) 2 body i= N Δm i dm r,i r,dm i=1 body K rot = K rot,i = i i 1 ( ) 2 2 Δm i r,i ω 2 = 1 2 body dm (r,dm ) 2 ω 2 = 1 2 I ω 2 S

Discussion: Moment of Inertia How does moment of inertia compare to the total mass and the center of mass? Different measures of the distribution of the mass. Total mass: scalar m total = Center of Mass: vector (three components) body dm R cm 1 = total m body r dm Moment of Inertia about axis passing through S: (nine possible moments) I S = dm (r,dm ) 2 body

Checkpoint Problem All of the objects below have the same mass. Which of the objects" has the largest moment of inertia about the axis shown?" 1) Hollow Cylinder 2) Solid Cylinder 3) Thin-walled Hollow Cylinder"

Strategy: Calculating Moment of Inertia Step 1: Identify the axis of rotation Step 2: Choose a coordinate system Step 3: Identify the infinitesimal mass element dm. Step 4: Identify the radius, r, of the circular orbit of the infinitesimal mass element,dm dm. Step 5: Set up the limits for the integral over the body in terms of the physical dimensions of the rigid body. Step 6: Explicitly calculate the integrals.

Worked Example: Moment of Inertia for Uniform Disc Consider a thin uniform disc of radius R and mass m. What is the moment of inertia about an axis that pass perpendicular through the center of the disc?

Worked Example: Moment of Inertia of a Disc Consider a thin uniform disc of radius R and mass m. What is the moment of inertia about an axis that pass perpendicular through the center of the disc? r,dm = r I cm = M π R 2 I cm = (r,dm ) 2 dm = M π R 2 body r = R θ =2π dθ θ =0 r 3 dr = M r =0 π R 2 I cm = 2M R 2 r = R r =0 r 3 dr r = R r =0 = 2M r 4 r = R R 2 4 r =0 r = R r =0 θ =2π θ =0 r 3 dθ dr 2πr 3 dr = 2M R 2 da = r dr dθ σ = dm da = m total Area = M π R 2 dm = σ r dr dθ = M r dr dθ 2 π R = 2M R 2 R 4 4 = 1 2 MR2 r = R r =0 r 3 dr

Checkpoint Problem: Moment of Inertia of a Rod Consider a thin uniform rod of length L and mass M. a) Calculate the moment of inertia about an axis that passes perpendicular through the center of mass of the rod. b) Calculate I about an axis that passes perpendicular through the end of the rod.

Parallel Axis Theorem Rigid body of mass m. I cm Moment of inertia about axis through center of mass of the body. I S Moment of inertia about parallel axis through point S in body. d S,cm perpendicular distance between two parallel axes. 2 I S = I cm + md S,cm

Summary: Moment of Inertia Moment of Inertia about S: i= N I S = Δm i (r,s,i ) 2 2 = r,s i=1 body dm Examples: Let S be the center of mass rod of length l and mass m disc of radius R and mass m Parallel Axis theorem: I cm = 1 12 ml 2 I cm = 1 2 mr2 2 I S = I cm + md S,cm

Checkpoint Problem: Kinetic Energy A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The moment of inertia about the cm is (1/2)M R 2. What is the kinetic energy of the disk?

Summary: Fixed Axis Rotation Kinematics Angle variable Angular velocity Angular acceleration Mass element Radius of orbit θ ω dθ / dt α d 2 θ / dt 2 Δm i Moment of inertia Parallel Axis Theorem i= N r,i I S = Δm i (r,i ) 2 dm(r ) 2 i=1 body I S = Md 2 + I cm

Checkpoint Problem: Moment of Inertia Wheel Using energy techniques, Calculate the speed of block 2 as a function of distance that it moves down the inclined plane using energy techniques. Let I P denote the moment of inertia of the pulley about its center of mass. Assume there are no energy losses due to friction and that the rope does slip around the pulley.

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