Physics 07 Lectue 17 Goals: Lectue 17 Chapte 1 Define cente of mass Analyze olling motion Intoduce and analyze toque Undestand the equilibium dynamics of an extended object in esponse to foces Employ consevation of angula momentum concept Assignment: HW7 due tomoow Wednesday, Exam Review Physics 07: Lectue 17, Pg 1 Rotational otion: Statics and Dynamics oces ae still necessay but the outcome depends on the location fom the axis of otation This is in contast to the tanslational motion and acceleation of the cente of mass. Hee the position of these foces doesn t matte (doesn t alte the physics we see) Howeve: o otational statics & dynamics: we must efeence the specific position of the foce elative to an axis of otation. It may be necessay to conside moe than one otation axis! Vectos emain the key tool fo visualizing Newton s Laws Physics 07: Lectue 17, Pg Page 1
Physics 07 Lectue 17 A special point fo otation System of Paticles: Cente of ass (C) A suppoted object will otate about its cente of mass. Cente of mass: Whee the system is balanced! Building a mobile is an execise in finding centes of mass. + m 1 m + m m 1 mobile Physics 07: Lectue 17, Pg 3 System of Paticles: Cente of ass How do we descibe the position of a system made up of many pats? Define the Cente of ass (aveage position): o a collection of N individual point like paticles whose masses and positions we know: R C N i= 1 m i i 1 y R C m 1 m (In this case, N = ) x Physics 07: Lectue 17, Pg 4 Page
Physics 07 Lectue 17 Sample calculation: Conside the following mass distibution: m at ( 0, 0) N m at (1,1) R C m = i i i= 1 = X C î + Y X C = (m x 0 + m x 1 + m x 4 )/4m metes R C = (1,6) Y C = (m x 0 + m x 1 + m x 0 )/4m metes C ĵ + Z C (1,1) m X C = 1 metes Y C = 6 metes m m kˆ m at (4, 0) (0,0) (4,0) Physics 07: Lectue 17, Pg 5 System of Paticles: Cente of ass o a continuous solid, one can convet sums to an integal. N y dm x R R C C= i= 1 dm = dm whee dm is an infinitesimal mass element but thee is no new physics. m i i dm Physics 07: Lectue 17, Pg 6 Page 3
Physics 07 Lectue 17 Connection with motion... So a igid object that has otation and tanslation K K otates about its cente of mass! And Newton s Laws apply to the cente of mass N m TOTAL TOTAL = K Rotation + K = K + Rotation K o a point p otating: R p 1 = i Tanslation i= R 1 C VC 1 1 mpvp = mp( ωp ) i p p p V C p ω p p p Physics 07: Lectue 17, Pg 7 Wok & Kinetic Enegy: Recall the Wok Kinetic-Enegy Theoem: K = W NET This applies to both otational as well as linea motion. K = ( ω ) 1 f ωi + m( VC f VC i = WNET 1 ) I What if thee is olling? Physics 07: Lectue 17, Pg 8 Page 4
Physics 07 Lectue 17 Demo Example : A ace olling down an incline Two cylindes with identical adii and total masses oll down an inclined plane. The 1 st has moe of the mass concentated at the cente while the nd has moe mass concentated at the im. Which gets down fist? Two cylindes with adius R and mass m h θ who is 1 st? A) ass 1 B) ass C) They both aive at same time Physics 07: Lectue 17, Pg 9 Same Example : Rolling, without slipping, otion A solid disk is about to oll down an inclined plane. What is its speed at the bottom of the plane? h θ v? Physics 07: Lectue 17, Pg 10 Page 5
Physics 07 Lectue 17 Rolling without slipping motion Again conside a cylinde olling at a constant speed. V C V C C Physics 07: Lectue 17, Pg 11 otion Again conside a cylinde olling at a constant speed. Rotation only V Tang = ωr C V C Both with V Tang = V C V C C Sliding only C V C If acceleation a cente of mass = - αr Physics 07: Lectue 17, Pg 1 Page 6
Physics 07 Lectue 17 Example : Rolling otion A solid cylinde is about to oll down an inclined plane. What is its speed at the bottom of the plane? Use Wok-Enegy theoem Disk has adius R h θ v? gh = ½ v + ½ I C ω and v =ωr gh = ½ v + ½ (½ R )(v/r) = ¾ v v = (gh/3) ½ Physics 07: Lectue 17, Pg 13 Example: The ictionless Loop-the-Loop last time U b =mgh U=mgR h? To complete the loop the loop, how high do we have to elease a ball with adius ( <<R)? Condition fo completing the loop the loop: Cicula motion at the top of the loop (a c = v / R) Use fact that E = U + K = constant! ball has mass m & <<R R Recall that g is the souce of the centipetal acceleation and N just goes to zeo is the limiting case. Also ecall the minimum speed at the top is v Tangential = gr Physics 07: Lectue 17, Pg 14 Page 7
Physics 07 Lectue 17 Example: The Loop-the-Loop last time If olling then ball has both otational and C motion! E= U + K C + K Rot = constant = mgh (at top) E= mgr + ½ mv + ½ /5 m ω = mgh & v=ω E= mgr + ½ mgr + 1/5 m v = mgh h = 5/R+1/5R U b =mgh U=mgR h? ball has mass m & <<R v Tangential = gr R Just a little bit moe. Physics 07: Lectue 17, Pg 15 How do we econcile foce, angula velocity and angula acceleation? Physics 07: Lectue 17, Pg 16 Page 8
Physics 07 Lectue 17 Angula motion can be descibed by vectos With otation the distibution of mass mattes. Actual esult depends on the distance fom the axis of otation. Hence, only the axis of otation emains fixed in efeence to otation. We find that angula motions may be quantified by defining a vecto along the axis of otation. We can employ the ight hand ule to find the vecto diection Physics 07: Lectue 17, Pg 17 The Angula Velocity Vecto The magnitude of the angula velocity vecto is. The angula velocity vecto points along the axis of otation in the diection given by the ight-hand ule as illustated above. As ω inceased the vecto lengthens Physics 07: Lectue 17, Pg 18 Page 9
Physics 07 Lectue 17 om foce to spin (i.e., ω)? A foce applied at a distance fom the otation axis gives a toque θ Tangential = Tang sin θ adial If a foce points at the axis of otation the wheel won t tun Thus, only the tangential component of the foce mattes With toque the position & angle of the foce mattes a Tangential adial τ NET = Tang sin θ Physics 07: Lectue 17, Pg 19 Rotational Dynamics: What makes it spin? A foce applied at a distance fom the otation axis τ NET = Tang sin θ Toque is the otational equivalent of foce Toque has units of kg m /s = (kg m/s ) m = N m τ NET = Tang = m a Tang = m α = (m ) α o evey little pat of the wheel a Tangential adial Physics 07: Lectue 17, Pg 0 Page 10
Physics 07 Lectue 17 o a point mass τ NET = m α The futhe a mass is away fom this axis the geate the inetia (esistance) to otation (as we saw on Wednesday) τ NET = I α This is the otational vesion of NET = ma a Tangential andial oment of inetia, I equivalent of mass. Σ i m i i, is the otational If I is big, moe toque is equied to achieve a given angula acceleation. Physics 07: Lectue 17, Pg 1 Rotational Dynamics: What makes it spin? A foce applied at a distance fom the otation axis gives a toque τ NET = Tang sin θ a Tangential adial A constant toque gives constant angula acceleation if and only if the mass distibution and the axis of otation emain constant. Physics 07: Lectue 17, Pg Page 11
Physics 07 Lectue 17 Toque, like ω, is a vecto quantity agnitude is given by (1) sin θ () tangential (3) pependicula to line of action Diection is paallel to the axis of otation with espect to the ight hand ule cos(90 θ) = Tang. line of action sin θ a 90 θ θ adial And fo a igid object τ = I α Physics 07: Lectue 17, Pg 3 Example : Rolling otion Newton s Laws: N f g Σ x Σ Στ y C x di θ = a x = gsinθ f = a y = 0 = N gcosθ = Iα C = 1 R α C αc R = a x = fr Notice otation CW (i.e. negative) when a x is positive! Combining 3 d and 4 th expessions gives f = a x / Top expession gives a x + f = 3/ a x = g sin θ So a x =/3 g sin θ Physics 07: Lectue 17, Pg 4 Page 1
Physics 07 Lectue 17 Execise Toque agnitude In which of the cases shown below is the toque povided by the applied foce about the otation axis biggest? In both cases the magnitude and diection of the applied foce is the same. Remembe toque equies, and sin θ o the tangential foce component times pependicula distance L A. Case 1 B. Case C. Same L axis case 1 case Physics 07: Lectue 17, Pg 5 Execise Toque agnitude In which of the cases shown below is the toque povided by the applied foce about the otation axis biggest? In both cases the magnitude and diection of the applied foce is the same. Remembe toque equies, and sin θ o the tangential foce component times pependicula distance (A) case 1 L (B) case (C) same L axis case 1 case Physics 07: Lectue 17, Pg 6 Page 13
Physics 07 Lectue 17 Example: Rotating Rod Again A unifom od of length L=0.5 m and mass m=1 kg is fee to otate on a fictionless pin passing though one end as in the igue. The od is eleased fom est in the hoizontal position. What is the initial angula acceleation α? L m Physics 07: Lectue 17, Pg 7 Example: Rotating Rod A unifom od of length L=0.5 m and mass m=1 kg is fee What is its initial angula acceleation? 1. o foces you need to locate the Cente of ass C is at L/ ( halfway ) and put in the oce on a BD. The hinge changes eveything! L m Σ = 0 occus only at the hinge mg but τ z = I α z = - sin 90 at the cente of mass and I End α z = - (L/) mg and solve fo α z Physics 07: Lectue 17, Pg 8 Page 14
Physics 07 Lectue 17 Statics Equilibium is established when Tanslational motion Σ Net = 0 0 Rotational motion Στ Net = In 3D this implies SIX expessions (x, y & z) Physics 07: Lectue 17, Pg 9 Example Two childen (60 kg and 30 kg) sit on a hoizontal teete-totte. The lage child is 1.0 m fom the pivot point while the smalle child is tying to figue out whee to sit so that the teete-totte emains motionless. The teete-totte is a unifom ba of 30 kg its moment of inetia about the suppot point is 30 kg m. Assuming you can teat both childen as point like paticles, what is the initial angula acceleation of the teete-totte when the lage child lifts up thei legs off the gound (the smalle child can t each)? o the static case: Rotational motion Στ Net = 0 Physics 07: Lectue 17, Pg 30 Page 15
Physics 07 Lectue 17 Example: Soln. Use Στ Net = 0 30 kg N 30 kg 60 kg 0.5 m 1 m 300 N 300 N 600 N Daw a ee Body diagam (assume g = 10 m/s ) 0 = 300 d + 300 x 0.5 + N x 0 600 x 1.0 0= d + 1 4 d = 1.5 m fom pivot point Physics 07: Lectue 17, Pg 31 Recap Assignment: HW7 due tomoow Wednesday: eview session Physics 07: Lectue 17, Pg 3 Page 16