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Home , Kinematics, Translation (geometry)

Rigid Bodies

A is an extended object in which the Two-Dimensional Rotational between any two points in the object is constant in .

Springs or human bodies are non-rigid bodies. 8.01 W10D1

Rotation and Recall: Translational of of Rigid Body the Center of

Demonstration: Motion of a thrown baton • Total of system of particles sys total pV= m cm • External and of Translational motion: external force of acts on center of mass sys totaldp totaldVcm total FAext==mm = cm Rotational Motion: object rotates about center of dt dt mass

1 Main Idea: of Rigid Two-Dimensional Rotation Body produces about center of • Fixed axis rotation: mass Disc is rotating about axis τ total = I α passing through the cm cm cm center of the disc and is perpendicular to the I plane of the disc. cm is the of inertial about the center of mass • Plane of motion is fixed: α is the angular acceleration about center of mass cm For straight line motion, bicycle wheel rotates about fixed direction and center of mass is translating

Rotational Kinematics Fixed Axis Rotation: Angular for Fixed Axis Rotation Angle variable θ A point like particle undergoing at a non-constant has SI unit: [rad] dθ ω ≡≡ω kkˆˆ (1)An vector Angular velocity dt SI unit: −1 ⎣⎡rad⋅ s ⎦⎤ (2) an angular acceleration vector dθ Vector: ω ≡ Component dt dθ ω ≡ magnitude dt ω >+0, direction kˆ direction ω < 0, direction − kˆ

2 Fixed Axis Rotation: Angular Concept Question: Angular Acceleration Speed 2 ˆˆd θ Object A sits at the outer edge (rim) of a merry-go-round, and Angular acceleration: α ≡≡α kk2 object B sits halfway between the rim and the axis of rotation. The dt merry-go-round makes a complete revolution once every thirty SI unit −2 ⎣⎦⎡⎤rad⋅ s seconds. The magnitude of the angular velocity of Object B is Vector: d 2θ dω Component: α ≡ 2 ≡ dt dt dω α ≡ Magnitude: dt

1. half the magnitude of the angular velocity of Object A . dω >+0, direction kˆ 2. the same as the magnitude of the angular velocity of Object A . Direction: dt 3. twice the the magnitude of the angular velocity of Object A . dω 4. impossible to determine. <−0, direction kˆ dt

Rotational Kinematics: Constant Table Problem: Rotational Angular Acceleration Kinematics

A turntable is a uniform disc of mass m and a radius R. The The angular quantities θ, ω,andα turntable is spinning initially at a constant frequency f . The motor is turned off and the turntable slows to a stop in t seconds with constant angular deceleration. are exactly analogous to the quantities x, v ,anda x x a) What is the direction and magnitude of the initial angular velocity of for one-dimensional motion, and obey the same type of relations the turntable? t t ωωα()ttdt−= (′′ ) , θθω()ttdt−= (′′ ) . 0 ∫ 0 ∫ b) What is the direction and magnitude of the angular acceleration of 0 0 the turntable? Constant angular acceleration: c) What is the total angle in radians that the turntable spins while ω(t) = ω + α t slowing down? 0 2 2 ()ω()tt=+ωαθθ00 2 ( () − ) . 1 2 ⇒ θ(t) = θ0 + ω0 t + α t 2

3 of a Point : Angular Particle Momentum of a Point Particle • Point particle of mass m moving with a Direction velocity v • Momentum pv= m Right Hand Rule • Fix a point S • Vector r S from the point S to the location of the object

• Angular momentum about the point S LrpSS=× 2 -1 •SI Unit [kg ⋅ m ⋅s ]

Cross Product: Angular Momentum of a Point Particle Example Problem: Angular LrpSS=× Momentum and Cross Product Magnitude: A particle of mass m = 2 kg moves with a uniform velocity LrpSS= sinθ -1 ˆ -1 ˆ v = 3.0 m ⋅s i + 3.0 m ⋅s j a) moment arm At time t, the vector of the particle with respect ot the point S is rSS,⊥ = r sinθ r = 2.0 m ˆi + 3.0 m ˆj LpSS= r ,⊥ S b) Perpendicular Find the direction and the magnitude momentum of the angular momentum about the Lr= p pS,⊥ = p sinθ SS⊥ origin, (the point S) at time t.

4 Solution: Angular Momentum Table Problem: Angular and Cross Product Momentum and Circular Motion The angular momentum vector of the particle about the point S is given by: Consider a point particle of mass m moving in a circle of radius R with Lrprv=×=×m SS S velocity v = R ω θ ˆ . Find the direction =+×()()2.0m ˆˆi 3.0m j (2kg) 3.0m ⋅+⋅ s−−11 ˆi 3.0m s ˆj and magnitude of the angular momentum about the center of a circle 21−−ˆˆ 21 =+0 12kgms ⋅ ⋅kk0 − 18kgms ⋅ ⋅ ( − ) + in terms of the radius R, mass m, and =−6.0kg ⋅ m21 ⋅ s− kˆ . angular speed ω. ö The direction is in the negative k direction, and the magnitude is L =⋅⋅6.0 kg m21 s− . ijk×=, S ji×=− k, ii×=×=jj0

Angular Momentum and Circular Motion of a Point Particle Rigid Body Kinematics Fixed axis of rotation: z-axis for Fixed Axis Rotation Angular velocity ω ≡ ω kˆ Body rotates with angular velocity ω and angular acceleration α Velocity vrkr=×=ω ω ˆ ×RRˆ =ω θˆ Angular momentum about the point S ˆˆˆ2 LrprvSS=×=× SmRmvRmRmR = k =ω k = ω k

5 Rotational Kinetic and Divide Body into Small Elements Moment of Body rotates with angular velocity, ω Rotational about axis passing through S angular acceleration α 2 1 1 2 1 2 KI= ω 2 K = Δm v = Δm r ω cm2 cm cm rot,i 2 i tan,i 2 i ()⊥,i Individual elements of mass Δm i i= N 2 about S : IS =Δ∑ mi (r⊥,i ) Radius of orbit i=1 r⊥,i 2 SI Unit: ⎣⎡kg⋅ m ⎦⎤ Tangential velocity v = r ω i= N tan,i ⊥,i Δm → dm r → r ∑→ Continuous body: i ⊥,i ⊥,dm ∫ i=1 body Tangential acceleration atan,i = r⊥,iα 2 IS = dm (r⊥,dm ) 2 ∫ v body Radial Acceleration a = tan,i = r ω 2 Rotational Kinetic Energy: rad,i ⊥,i 2 ⎛ ⎞ r ⎛ 1 ⎞ 2 1 2 2 1 2 ⊥,i K = K = Δm r ω = dm (r ) ω = I ω rot ∑ rot,i ⎜ ∑ i ()⊥,i ⎟ ⎜ ∫ ⊥,dm ⎟ S i ⎝ i 2 ⎠ ⎝ 2 body ⎠ 2

Discussion: Moment of Inertia Concept Question: How does moment of inertia compare to the total mass and the center An object of mass m with velocity v of mass? moving in a straight line collides elastically with an identical object. The Different measures of the distribution of the mass. second object is initially at rest and is attached at one end to a string of length l and negligible mass. The other total Total mass: scalar m = ∫ dm end of the string is fixed (at the point S). body After the collision the second object undergoes circular motion with angular 1 Rr= dm speed ω . The kinetic energy of the Center of Mass: vector (three components) cm total ∫ m body second object after the collision is 1. (1/4) m l2 ω2 4. (1/4) ml ω2 Moment of Inertia about axis passing through S: (nine possible moments) 2. (1/2) ml2 ω2 5. (1/2) ml ω2 I = dm (r )2 S ∫ ⊥,dm body 3. (3/4) ml2 ω2 6. (1/4) ml ω

6 Strategy: Calculating Moment of Example: Moment of Inertia of a Inertia Disc Step 1: Identify the axis of rotation Consider a thin uniform disc of radius R and mass m. Step 2: Choose a What is the moment of inertia about an axis that pass perpendicular through the center of the disc? Step 3: Identify the infinitesimal mass element dm. da = rdrdθ

dm mtotal M r⊥,dm = r σ = = = r 2 Step 4: Identify the radius, ⊥, dm , of the circular orbit of da Area π R the infinitesimal mass element dm. M dm = σ rdrdθ = rdrdθ M r = R θ =2π 2 I = (r )2 dm = r 3 dθ dr π R cm ∫ ⊥,dm 2 ∫r =0 ∫θ =0 Step 5: Set up the limits for the integral over the body body π R in terms of the physical dimensions of the rigid body. M r = R θ =2π M r = R 2M r = R I = ⎛ dθ⎞ r 3dr = 2πr 3dr = r 3dr cm π R2 ∫r =0 ⎝ ∫θ =0 ⎠ π R2 ∫r =0 R2 ∫r =0 r = R Step 6: Explicitly calculate the . 4 4 2M r = R 2M r 2M R 1 I = r 3dr = = = MR2 cm R2 ∫r =0 R2 4 R2 4 2 r =0

Table Problem: Moment of Inertia Parallel Axis Theorem of a Rod Consider a thin uniform rod of length L and • Rigid body of mass m.

mass m. Calculate the moment of inertia • Moment of inertia I cm about axis through center of mass of about an axis that passes perpendicular the body. through the center of mass of the rod. • Moment of inertia I S about parallel axis through point S in body.

• dS,cm perpendicular distance between two parallel axes.

2 IS = Icm + mdS,cm

7 Summary: Moment of Inertia Concept Question: Kinetic

Moment of Inertia about S: Energy i= N I =Δm (r )2 = r 2 dm S ∑ i ⊥,S ,i ∫ ⊥,S A disk with mass M and radius R is spinning with i=1 body Examples: Let S be the center of mass angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. 2 1 2 Moment of inertia about cm is (1/2)MR. Its total • rod of length l and mass m Icm = ml 12 kinetic energy is: 1 • disc of radius R and mass m 2 2 2 Icm = mR 1. (1/4)MR ω 2 2 4. (1/4)MRω Parallel Axis theorem: I = I + md 2 2. (1/2)MR2 ω2 5. (1/2)MRω2 S cm S,cm 3. (3/4)MR2 ω2 6. (1/4)MRω

Summary: Fixed Axis Rotation Angular Momentum Kinematics for Fixed Axis Rotation θ Angular Momentum about the point S Angle variable ˆ ˆ LrpSi,,=×= Si i()rzp⊥ , i rkˆ + i × tan, iθ Angular velocity ω ≡ dθ / dt ˆ LkrSi,=−rp⊥ , i tan, i zp i tan, iˆ Angular acceleration 2 2 α ≡ d θ / dt Tangential component of momentum Mass element Δm p =Δm v =Δm r ω i tan,i i tan,i i ⊥,i Radius of orbit r⊥,i z-component of angular momentum i= N 2 2 about S: Moment of inertia I =Δm (r ) → dm(r ) S ∑ i ⊥,i ∫ ⊥ L = r p = r Δm r ω =Δm r 2ω i=1 body S ,z,i ⊥,i tan,i ⊥,i i ⊥,i i ⊥,i 2 i= N i= N Parallel Axis Theorem 2 IS = Md + Icm LS,z = ∑ LS ,z,i =Δ∑ mir⊥,i ω = ISω i=1 i=1

8 Concept Question: Angular Concept Question: Angular Momentum momentum A dumbbell is rotating at a constant angular speed about its center. Compared to the A disk with mass M and radius R is spinning with angular dumbbell's angular momentum about its center velocity ω about an axis that passes through the rim of the A, its angular momentum about point B (as disk perpendicular to its plane. The magnitude of its angular shown in the figure) is momentum is: 1 1 1. MR2ω 2 4. MR2ω 1. bigger. 4 4 1 1 2. the same. 2. MR2ω 2 5. MR2ω 3. smaller. 2 2 3 3 3. MR2ω 2 6. MR2ω 2 2

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