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Chapter 12 Rotation of a Rigid Body Vector (or “cross”) Product Cross Product is a vector perpendicular to the plane of vectors A and B A B A B sin Cross Product of Vectors A B A sin B sin B A B A B sin Position Force A B sin A Right hand rule: Curl your right hand around the center of rotation with the fingers going from the first vector to the second vector and the thumb will be pointing in the torque direction A ×B ≠ B× A AP Physics C 3 Cross Product Problem F in d : E = D × C if C = 2 N a n d D = 1 m Csin(110) C 110° D AP Physics C E = C D s in θ E = 2 1 s in 1 1 0 = 1 .8 8 N m 4 Test your Understanding Which way will it rotate once the support is removed? 1. 2. 3. 4. Clockwise. Counter-clockwise. Not at all. Not sure what will happen. Torque If the forces are equal, which will open the heavy door more easily? AP Physics C 6 Interpretation of torque Measures tendency of any force to cause rotation Torque is defined with respect to some origin – must talk about “torque of force about point X”, etc. Torques can cause clockwise (+) or anticlockwise rotation (-) about pivot point Torque con’t AP Physics C 8 Torque con’t AP Physics C 9 Definition of Torque: r F where r is the vector from the reference point (generally either the pivot point or the center of mass) to the point of application of the force F . If r and F are not perpendicular then: | | r F sin where is the angle between the vectors r and . Definition of Torque: | | r F sin Torque Problem Adrienne (50 kg) and Bo (90kg) are playing on a 100 kg rigid plank resting on the supports seen below. If Adrienne stands on the left end, can Bo walk all the way to the right end with out the plank tipping over? If not, How far can he get past the support on the right? AP Physics C 12 Torque Problem con’t N1 2m N2 3m 4m x 50kg 100kg 90kg τ N et = 90x - 100( .5) - 50( 5)= 0 x = 3 .3 AP Physics C 13 Moments M1 M2 d1 d2 Suppose we have masses m1 and m2 on the seesaw at distances d1 and d2, respectively, from the fulcrum, when does the seesaw balance? By Archimedes’ Law of the lever, this occurs when m1d1 + m2d2 = 0 AP Physics C 14 Moments con’t M1 M2 x1 x2 If we place a coordinate system so that 0 is at the fulcrum and if we let xi be the coordinate at which is placed then: m1x1 + m2x2 = m1d1 + m2d2 = 0 AP Physics C 15 Moments con’t M1 M2 M3 M4 M5 x1 x2 x3 x4 x5 More generally, if we place masses m1, m2, …, mr at points x1, x2, …. , xr, respectively, then the see saw balances with the fulcrum at the origin, if and only if m1x1 + m2x2 + …+ mrxr = 0 AP Physics C 16 Moments con’t M1 M2 M3 x1 x2 x3 M4 M5 x5 x4 Now, suppose that we place masses m1, m2, … , mr at points x1, x2, … xr, respectively, then where should we place the fulcrum so that the seesaw balances? The answer is that we place the fulcrum at x-bar where: m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )= 0 AP Physics C 17 Moments con’t M1 M2 M3 x1 x2 x3 M4 M5 x5 x4 m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) ) is called the moment about x-bar. x-bar Moment is from the Greek word for movement, not time. If positive, movement is counter-clockwise, negative it is clockwise. AP Physics C 18 Moments con’t M1 M2 M3 x1 x2 x3 x-bar M4 M5 x4 x5 0 Suppose that m 1 x 1 x m 2 x 2 x ... m r x r x We want to solve for x m 1 x 1 m 2 x 2 ... m r x r m 1 x m 2 x ... m r x 0 m 1 x 1 m 2 x 2 ... m r x r m 1 m 2 ... m r x 0 r x AP Physics C m i xi i 1 r m i 1 i 19 Center of Mass M1 M2 M3 x1 x2 x3 x-bar M4 M5 x4 x5 r x m i xi i 1 r m M0 m to t i i 1 AP Physics C 20 Center of Mass con’t Suppose m1, m2, … , mr are masses located at points (x1, y1), (x2, y2), … , (xr, yr). M1 M4 M2 r The moment about the y-axis is: y My m i xi m i yi i 1 r The moment about the x-axis is: Mx i 1 r M3 x x m r i yi i 1 r i 1 mi My m to t y m i xi i 1 r mi Mx m to t i 1 My Mx , Center of Mass is m m to t to t AP Physics C 21 Center of Mass con’t Now lets find the center of mass of a thin plate with uniform density, ρ. b First we need the mass of the plate: M A rea f ( x ) g ( x )d x a AP Physics C 22 Center of Mass con’t Next we need the moments of the region: M x 1 2 y m Mx 1 2 1 2 f ( x ) f ( x ) x 1 2 g ( x ) g ( x ) x b 2 2 f ( x ) g ( x ) d x f(x) a M y x m x f ( x ) g ( x ) x g(x) ∆x b M y x f ( x ) g ( x ) d x a To find the center of mass we divide by mass: y Mx m to t 1 2 b 2 2 f ( x ) g ( x ) d x a b b a b f ( x ) g ( x )d x 2 f ( x ) g ( x )d x a x AP Physics C My M f ( x )2 g ( x )2 d x 1 2A b a f ( x )2 g ( x )2 d x a 1 A b a x f ( x ) g ( x ) d x 23 Center of Mass Problem Determine the center of mass of the region bounded by y = 2 sin (2x) and y = 0 on the interval, [0, π/2] Given the symmetry of the curve it is obvious that x-bar is at π/4. First find the area. A 2 0 2 sin (2 x ) d x co s (2 x ) |0 2 2 Using the table of integrals: y 1 2A b a AP Physics C 1x 1 2 2 f ( x )2 g ( x )2 d x sin (2 x ) d x sin 4 x 4 0 4 2 4 0 1 2 24 Moment of Inertia a r F T T T F ma F r m ra 2 mr r mr I ma AP Physics C M r CR F 25 Moment of Inertia con’t M w ( y )d y I AP Physics C b a y w y dy 2 26 Calculating Moment of Inertia AP Physics C 27 Calculating Moment of Inertia dm dA M dm A M dA A d A 2 r d r an d A R dm I AP Physics C M R 2 2 rd r r dm 2 2M R 2 R 0 2M R 2 rd r r dr 3 2 MR 2 2 28 Moment of Inertia con’t AP Physics C 29 Parallel Axis Theorem I || I C M M d AP Physics C 2 30 ||-axis Theorem Proof I x dm 2 x ' d d m x ' d m 2d x ' d m d 2 I IC M M d AP Physics C 2 dm 2 31 Rotational Kinetic Energy K ro t 1 2 m 1v K ro t 1 2 m r 1 2 2 1 2 1 1 2 1 2 2 2 m 2v 1 2 m r ... 2 2 m i ri i 2 2 2 1 2 K ro t E m ech AP Physics C ... 1 2 I 1 2 2 2 I 2 I M g y cm 2 32 Rotational Dynamics Ft = m a t = m rα rFt = m r α 2 = mr α 2 N et = mr 2 i i α i N e t Iα AP Physics C 33 Rotation About a Fixed Axis AP Physics C 34 Bucket Problem A 2.0 kg bucket is attached to a mass-less string that is wrapped around a 1.0 kg, 4.0 cm diameter cylinder, as shown. The cylinder rotates on an axel through the center. The bucket is released from rest 1.0 m above the floor, How long does it take to reach the floor AP Physics C 35 Bucket Problem con’t FB = m a y = T - m g N et = T R net = α I = T R α= TR 2 = 2T .5 M R MR 2T 2T a y = -α R = R =MR M AP Physics C I = TR 36 Bucket Problem Con’t 2T ay = - M may = - ay = Δy = AP Physics C 1 2 T = May 2 2m g -a y M 2 - mg a y = -7 .8 4 m s 2 2m + M ay Δt 2 Δ t = 0 .5 0 s 37 Static Equilibrium FN e t = 0 AP Physics C = 0 N et 38 Statics Problem A 3.0 m ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of μs, that prevents the ladder from slipping? F x F y = n 2 - fs = 0 = n1 - m g = 0 τ N e t = d 1FG - d 2 n 2 = 1 L c o s 6 0 M g - L s in 6 0 n o 2 = .5 ( .7 5 ) M g AP Physics C o 3 32n 2 2 =0 0 39 Statics Problem n 2 = fs = μ s n1 fs = n2 = n1 = M g Mg 2 ta n 6 0 o 2 ta n 6 0 o = μ sn1 μ s M g μsM g = μs = Mg Mg 2 ta n 6 0 o 1 2 ta n 6 0 o μ s ≥ 0 .2 9 AP Physics C 40 Balance and Stability θc h t/2 θc h t/2 AP Physics C 41 Rolling Motion Δ x cm = 2π R = v cm T AP Physics C v cm = 2πR T = Rv 42 Rolling Motion con’t ri = rc m + ri,re l v i = v c m + v i,re l AP Physics C 43 Rolling Motion con’t AP Physics C 44 Rolling Motion con’t AP Physics C 45 Rolling Kinetic Energy K ro t,P = I ω 1 2 P IP = I c m + M R K = AP Physics C 2 I ω + 1 2 cm 2 1 2 M R ω 46 2 Great Downhill Race AP Physics C 47 Downhill Race con’t 2 I ω + 1 2 cm 1 2 Ic m = c M R P article c = 0 S phere c = 2 / 5 C ylin d e r c = 1 / 2 H oop c = 1 AP Physics C 1 2 2 Mv v cm = = M gh 2 cM R ω + 2 2 cm 2 1 2 v cm M = M gh R 2gh 1+ c 48 Angular Momentum p = mv L = Iω = m rv L = r × p = r m v sin AP Physics C 49 Angular Momentum AP Physics C 50 Angular Momentum L z = m rv t = r × p dL dt = d dt r × p = dr dt p+r dp dt = v × p + r × Fn e t dL dt AP Physics C = net 51 Conservation of Angular Momentum 1 2 2 Lf = Li m lf ω f = 2 2 m li ω i 1 2 2 i lf ω f = l ω i Two equal masses are at the ends of a massless 500 cm long rod. The rod spins at 2.0 Rev/s about an axis through its midpoint. If the rod lengthens to 160 cm, what is the angular velocity 2 2 li 50 ωf ωi = 2 = .2 0 160 lf AP Physics C re v s 52 Testing Understanding There is no torque on the buckets so angular momentum is conserved. Increased mass in buckets increases inertia so angular velocity must decrease. AP Physics C 53 Problem 1 An 18 cm long bicycle crank arm with a pedal at one end is attached to a 20 cm diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 60 rpm to 90 rpm in 10 s. a. What is the tangential acceleration of the pedal? b. Want length of chain passes over the top of the sprocket during this interval? AP Physics C 54 Problem 1 con’t a. Since at = rα, find α first. With 60 rpm = 6.28 rad/s and 90 rpm = 9.43 rad/s: α= Δω Δt = 9 .4 3 ra d - 6 .2 8 ra d / s 10 s = 0 .3 1 4 ra d / s 2 The angular acceleration of the sprocket and chain are the same. a t = rα = 0 .1 8 m 0 .3 1 4 ra d / s 2 = 0 .0 5 7 m / s 2 b. Since L = r∆θ, find r∆θ. θ f = θ i + ω iΔ t + 1 2 α Δt 2 θ f - θ i = Δ θ = 6 .2 8 ra d / s 1 0 s + 1 0 .3 1 4 ra d / s 1 0 s 2 2 2 = 7 8 .5 ra d The length of chain which has passed over the top of the sprocket is L e n g th = 0 .1 0 m 7 8 .5 ra d = 7 .9 m AP Physics C 55 Problem 2 A 100 g ball and a 200 g ball are connected by a 30 cm mass-less rigid rod. The balls rotate about their center of mass at 120 rpm. What is the speed of the 100g ball? x cm = 1 0 0 g 0 c m + 2 0 0 g 3 0 c m 100 g + 200 g = 20 cm 2π rad m in v 1 = rω = x cm ω = 0.20 m 120 rev / m in rev 60 s AP Physics C = 2.5 m / s 56 Problem 3 A 300 g ball and a 600 g ball are connected by a 40 cm mass-less rigid rod, The structure rotates about its center of mass at 100 rpm. What is its rotational kinetic energy? ( 3 0 0 g ) ( 0 c m+)( 6 0 0 g ) ( 4 0 c m ) x cm = 300 g + 600 g I =( 3 0 0 g ) ( x 2 )+(c 6 00 g) ( 40 cm - x m = 2 6 .6 7 c m ) 2 cm 2 2 =( 0 .3 0 0 k g ) ( 0 .2 6 6 7 m )+( 0 .6 0 0 k g ) ( 0 .1 3 3 3 K rot AP Physics C 1 1 2 = Iω = ( 0 .0 3 2 kg m 2 2 2 m )= 0 .0 3 2 k g m 2 2 100 × 2π ) ra d / s = 1 .7 5 J 60 57 Problem 4 A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door’s moment of inertia for (a) rotation on its hinges (b) rotation about a vertical axis inside the door, 15 cm from one edge. I= 1 3 2 5 k g 0 .9 1 m 2 = 6 .9 k g m 2 0 .9 1 m d= - 0 .1 5 m = 0 .3 0 5 m . 2 2 I = Ic m + M d = AP Physics C 1 12 2 5 k g 0 .9 1 m 2 2 + 2 5 k g 0 .3 0 5 c m = 4 .1 k g m 2 58 Problem 5 An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.o kg. What is the magnitude of the torque about his shoulder if he holds the ball (a) straight out to his side, parallel to the floor and (b) straight, but 45 degrees below horizontal. τ = τ b a ll + τ a rm = m b g rb s in 9 0 ° + m a g r a s in 9 0 ° = 3 .0 k g 9 .8 m / s 2 0 .7 0 m + 4 .0 k g 9 .8 m / s 0 .3 5 m 2 = 34 N m AP Physics C 59 Problem 5 con’t τ = τ b a ll + τ a rm = m b g rb s in 4 5 ° + m a g r a s in 4 5 ° = 3 .0 k g 9 .8 m / s 2 0 .7 0 m 0 .7 0 7 + 4 .0 k g 9 .8 m / s 0 .3 5 m 0 .7 0 7 2 = 24 N m AP Physics C 60 Problem 6 Starting from rest, a 12 cm diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular acceleration is constant. The disk’s moment of inertia is 2.5x10-5 kgm2. (a) How much torque is applied? (b) How many revolutions does it make before reaching full speed? AP Physics C 61 Problem 6 con’t a. Using the rotational kinematic equation ω 1 = ω 0 + α ( t1 - t 0 ) 2π ( 2 0 0 0 rp m ) rad / s = 0 ra d + 60 α= 200π 9 ra d / s 2 τ = Iα =( 2 .5 × 1 0 = 1 .7 5 × 1 0 -3 -5 Nm 2 200π k g m ) ra d / s 9 b. θ 1 = θ 0 + ω 0( t1 - t 0 )+ = 1 0 0 π ra d = AP Physics C α 3 .0 s - 0 s 1 2 100π 2π 2 2 α ( t1 - t0 )= 0 ra d + 0 ra d + 1 200π 2 2 ra d / s 3 .0 s - 0 s 2 9 re v o lu tio n s = 5 0 re v 62 Problem 7 The two objects in the figure are balanced on the pivot. What is the distance d? - FG 1 - FG 2 + P = 0 N P = FG 1 + FG 2 = 1 .0 k g 9 .8 m / s 2 + 4 .0 k g 9 .8 m / s 2 = 49 N net = 0 N m P d - w (1 1 .0 m-) w ( 1 .5 m=) 0 N m 2 4 9 N d - 1 .0 k g 9 .8 m / s 2 1 .0 m - 4 .0 k g 9 .8 m / s 1 .5 m = 0 N 2 d = 1 .4 0 m AP Physics C 63