q q Physics 01, Lecture 18 Rotational Dynamics Torque Exercises and Applications Rolling Motion Today s Topics Review Angular Velocity And Angular Acceleration q Angular Velocity (ω) describes how fast an object rotstes, it has two components: Angular speed: and ω lim Δθ ω ave Δθ Δt direction of ω: + counter clockwise - clockwise è Angular velocity ω is a vector! (define direction next page) Δt 0 Δt = dθ dt q Hope you have previewed Chapter 10. (really!) Ø All particles of the rigid object have the same angular velocity q Angular Acceleration (α): è Angular acceleration α α ave Δω α Δt and lim Δω = dω is also a vector! Δt 0 Δt dt ote: the similarity between (θ,ω,α) and (x, v, a) Review: Moment of Inertia q Moment of Inertia of an object about an Moment of Inertia: I m i r i another form: I whole object r dm Moments Of Inertial Of Various Objects I = m i r i (= r dm) (unit of I : kgm ) Ø I depends on rotation, total mass, and mass distribution. 1
Which Has Larger Moment of Inertia? q Which of the following configurations has larger I Ø Central or side? (Trivial) Quick Quiz q A force is acting on a rigid rod around a fixed. Ø Which of the follow case(s) will not turn. F F F F Turning counter clockwise Turning clockwise o Turn o Turn Ø Effective Turning: Force, direction, acting point(action length). q Torque: Effect of Force on Rotation Torque: Magnitude: τ = Fsinφ r, depends on F, r, and sinφ Direction: conventional: clockwise = -, counter-clockwise = + More strictly: Right Hand Rule for cross-product q The angular acceleration of an object is proportional to the torque acting on it τ = r F Στ = I α I: Moment if inertia (Tuesday) (pivot) q τ Fsinφ r = F d Lever Arm d = r sinφ Lever Arm acting point The lever arm, d, is the perpendicular distance from the of rotation to a line drawn from the direction of the force
q τ Fsinφ r = F r Alternative View of Torque Summary: Two Views of Torque q τ Fsinφ r = F r = F r (pivot) (pivot) acting point acting point F r F r q Only the perpendicular component of the force contributes to the torque Torque Has a Direction q Torque is a vector. It has a magnitude and a direction. q For fixed rotation, the direction of torque can be described by a sign (+/-) in one dimension; (and by right hand rule in general) Ø Convention: Counter-clockwise : +, clockwise -. τ > 0 τ > 0 The Acting point of Gravity: Center of Gravity q The force of gravity acting on an object must be considered in determining equilibrium q In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point called center of gravity (cg) q Effectively, assuming gravitation field is uniform, the CG of an object is the same as its CM (that is usually true at the Phy103 level) τ < 0 τ > 0 x cg = Σm ix i Σm i and y cg = Σm iy i Σm i See demo: finding CG 3
Quick Exercise: Calculating Torque q As shown, a pencil is falling down under gravity. What is the torque (about the pivot) by the gravity? Rotational Dynamics Στ = I α Rotational Dynamics compared to 1-D Dynamics q Answer: + F grav cosθ L/ (note: why cosθ?) θ L/ pivot Angular Linear τ F α = a = I m ω = ω 0 + αt v v + at θ = θ + 1 0 + ω0t αt ω0 + αδθ ω = = 0 x = x + v 1 0 + v0t at = v + aδx KE = 1 Iω KE = 1 mv 0 Exercise: Pulley with Mass q A crate of mass m crate is hanging on a pulley of mass m pulley and radius R pulley as shown. What is the acceleration of the crate? Solution: 1: per FBD for the crate T-m crate g = m crate a : for the pulley: τ = -TR = I pulley α 3: connection: a=αr Please make sure you understand this Solve: a = - m crate g/(m crate +I pulley /R pulley ) Combined Translational and Rotational Motion q Generally, the motion of an extended object is a combination of the translational motion of the CM and the rotation about the CM Get T and α yourself after class KE tot = KE trans + KE rot = ½ Mv CM + ½ Iω 4
Combined Translation and Rotation q Combined translational and rotational motion. v rot C R A B v CM Everything on the wheel rotation about CM CM: moving linearly (1-D) v = v CM + v rot Rolling/Sliding Conditions q Recall: v bottom = v CM - Rω q Depending relative size of V cm and Rω, there can be three classes of rolling /sliding conditions. ω v cm v bottom = v cm Rω<0 ω v cm v bottom = v cm - Rω =0 ω v cm v bottom = v cm Rω>0 Ø At point A (top). v rot = Rω to the right. v top = v CM + Rω Ø At point B (bottom). v rot = Rω to the left. v bottom = v CM - Rω v cm < Rω v cm = Rω v cm > Rω Ø At arbitrary point C v v CM v rot spinning pure rolling (rolling w/o slipping) sliding Pure Rolling Motion (Rolling without Slipping) q Rolling motion refers to a form of combined translational and rotational motion. Quick Quiz and Demo q Consider a wheel in pure rolling without slipping. After a full resolution (i.e. in period T), how far the CM moves? R wheel: rotation about CM q R, R, πr, πr, other o slipping on road i.e. v bottom =0 v CM CM: moving linearly (1-D) R v CM = Rω=Rπ/T Δx CM =V CM *T =πr CM: moving linearly (1-D) Ø Condition for rolling w/o slipping: v CM = ωr and a CM = αr See Demo 5
Trajectory of a Point on the Rim of a Pure-Rolling Wheel Quick Quiz: Rolling Without Slipping q Consider a wheel rolling down a (not smooth) hill without slipping. How many (external) forces are acting on the wheel?, 3, more than 3, other fs q In the process, the work down by friction is Positive, negative, zero q ow consider a wheel rolling on a (flat but non smooth) horizontal plane without slipping. How many external forces are acting on it?, 3, more than 3, other Why? See next slide Exercise: Rolling w/o Slipping Down a Slope q A uniform disc (or wheel, or sphere) of mass M, radius R, and moment of inertia I is rolling down a slope without slipping as shown. Calculate its CM acceleration. q Solution: Ø Step 1: FBD as shown Ø Step : Set up as shown Ø Step 3: Dynamics for CM (x direction): sinθ f s = ma CM Ø Step 4: Dynamics for rotation: -f s R =- Iα Ø Step 5: rolling w/o slipping: Rα=a CM Ø Solve for unknowns: a CM = gsinθ sinθ 1+ I, f s = mr +1 mr I fs θ x Results Discussion: Rolling Down a Slope (w/o slipping) Consider a wheel rolling down a flat (but not smooth) slope. fs θ x a CM = gsinθ sinθ 1+ I, f s = mr +1 mr I Ø On a slope, a friction is necessary to keep it from slipping Ø For same mr, the larger the I, the slower it moves. Ø Spheres (or wheels or discs) of the same shape and mass distribution roll at the same speed regardless of their size and mass. Ø On a horizontal flat surface, the friction reduces to zero and the rolling can go forever even when the surface is not smooth! 6
Moments Of Inertial Of Various Objects I = m i r i (= r dm) q General Work: Work and Power by Torque (Self Reading) W = F Δs = F in_direction_of_δs Δs Δs q For an object rotating about a fixed, all mass elements are moving in the tangential direction : W = F t Δs = rf t Δs/r = τ Δθ P = W/Δt = τ Δθ/Δt = τω Ø Rotational version of Work-Energy theorem Δθ Δs See demo for rolling of wheels with different I τδθ = W = ΔKE 1 Iω f rot 1 Iω i 7