Angular Displacement. θ i. 1rev = 360 = 2π rads. = "angular displacement" Δθ = θ f. π = circumference. diameter
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1 Rotational Motion
2 Angular Displacement π = circumference diameter π = circumference 2 radius circumference = 2πr Arc length s = rθ, (where θ in radians) θ 1rev = 360 = 2π rads Δθ = θ f θ i = "angular displacement"
3 Angular Kinematics ω avg = θ 2 θ 1 t 2 t 1 ω = dθ dt = Δθ Δt θ α avg = Δω α = dω dt Δt = ω ω f i t
4 Angular Direction? Based on right-handed coordinate system y θ x z
5 Angular Kinematics v = Δx avg Δt a = Δv avg Δt s = rθ v = rω a = rα ω = Δθ avg Δt α = Δω avg Δt v f = v i + at x f = x i + v i t at 2 v f 2 = v i 2 + 2ax ω f = ω i + αt θ f = θ i + ω i t αt 2 ω f 2 = ω i 2 + 2αθ
6 Example 1 The turntable shown starts from rest, and begins rotating in the direction shown. 10 seconds later, it is rotating at 33.3 revolutions per minute a) What is the turntable s final angular velocity (mag and dir)? b) What is the turntable s average acceleration (mag and dir)? c) What distance in meters does a bug located 10cm from the axis of rotation, travel during the time the turntable is accelerating?
7 Example 2 A wheel rotates with a constant angular acceleration of 3.50 rad/s 2. At time t=0, its angular velocity is 2.00 rad/s. a) What angle does the wheel rotate through in 2.00 s? b) What is the angular speed at that moment?
8 Rotational K How can we determine the kinetic energy associated with the distributed mass of a rotating object? K rotational = K i 1 K rotational = 2 m v 2 i i 1 K rotational = m (r i iω) 2 2 K rotational = 1 2 K rotational = 1 2 Iω 2 ( 2 m i r i )ω 2 I = moment of inertia
9 Moment of Inertia I = 2 m r i i The rotational quantity I is analogous to the linear quantity m, and represents an object s resistance to change in its rotation.
10 Example 3 2 children (m=20 kg) and 2 adults (m=70 kg) sit in seats on the teacup ride at Disneyland. The center of mass for the adults is located 70 cm from the axis of rotation, while the cm for the children is 80 cm from the axis of rotation. a) Calculate the cup s moment of inertia. b) Calculate the cup s rotational energy if it spins at 1.3 revolutions/second. I=94.2 kg m 2, K=3140 J
11 I for Continuous Dist. I = 2 m r i i dm 2 I = lim Δm r i i Δm i 0 I = r 2 dm r dm = λ dr dm = σ da dm = ρ dv
12 Example 4 Calculate the moment of inertia for a uniform hoop of mass M and radius R. I = I = r 2 dm R 2 dm I = R 2 dm I = MR 2
13 Example 5 Calculate the moment of inertia for a uniform rod, of mass M, with length L, rotating about its center of mass. I = r 2 dm I = r 2 λdr +L / 2 I = λ 1 3 r3 L / 2 dm = λdr I = M L ( 1 ( L 3 2 )3 1 ( L 3 2 )3 ) I = 1 12 ML2
14 Example 6 Calculate the moment of inertia for a uniform rod, of mass M, with length L, rotating about an axis at one end of the rod. I = r 2 dm I = r 2 λdr L I = λ 1 3 r3 0 I = M $ L 3 ' & ) L % 3 ( I = 1 3 ML2 dm = λdr
15 Example 7 Calculate the moment of inertia around the central axis of a uniform, solid cylinder, with mass M, radius R, and length L. I = r 2 dm I = r 2 ρ dv I = r 2 ρ2πrl dr I = 2Lπρ r 3 dr dm = ρdv dv = 2πrL dr I = 2Lπρ 1 R LπρR 4 4 r4 = 0 2 ρ = M V = M πr 2 L I = LπR4 % M ( ' * = 1 2 & πr 2 L) 2 MR2
16 Parallel-Axis Theorem If one knows the moment of inertia about the center of mass of an object, one can determine the moment of inertia about any other axis parallel to the original one. I = I cm + MD 2
17 Example 8 Use the parallel axis theorem to calculate the moment of inertia for a uniform rod, of mass M, with length L, rotating about an axis at one end of the rod. I = r 2 dm I cm = 1 12 ML2 I end = I cm + MD 2 I end = 1 # 12 ML2 + M L & % ( $ 2 ' I end = 1 12 ML ML2 2 I = 1 3 ML2
18 Torque F applied θ F sinθ moment arm (r) τ = rf sinθ τ = Fd
19 Example 9 A one-piece cylinder, shown here, has a coresection that protrudes from a larger drum. A rope wrapped around the large drum of radius R 1 exerts a force F 1 to the right, while a rope wrapped around the core, radius R 2, exerts a force F 2 downward. a) Calculate the net torque acting on the cylinder in terms of the variables given. b) If F 1 =5.0 N, R 1 =1.0m, F 2 =6.0N, and R 2 =0.50m, which way does the cylinder rotate?
20 Torque Cross Product W = F s W = Fscosθ τ = r F τ = rf sinθ
21 Determinants A B = i j k A x A y A z B B B x y z A B = (A B A B ) i + y z z y (A B A B ) j + z x x z (A B A B ) k x y y x
22 Cross Products in 3-d ixi = jxj = kxk = 0 ixj = k -jxi =k jxi = -k -ixj = -k jxk = i -kxj = i kxj = -i -jxk = -i C = A B C = ABsinθ kxi = j -kxi = -i ixk = -j -ixk = i A B θ
23 Example 10 a) Find the cross product of 5i+6j and -3i + -2j b) What is the angle between these two vectors? 8k 164 =( )
24 Torque & α F tangential m τ = F tangential r F radial τ = (ma )r tangential τ = (m(αr))r τ = (mr 2 )α τ = Iα
25 Example 11 Calculate the linear acceleration of the object m, the angular acceleration of the wheel, and the tension in the cord (using m, r, and I). m r M I Mass : F net = ma F g F T = ma Wheel : τ = Iα r F T = Iα F T = mg ma r(mg ma) = I a r a = r2 mg r 2 m + I r(mg ma) = r(mg mrα) = Iα α = rmg I + r 2 m
26 Example 12 A uniform rod has a mass M and length L, and can pivot as shown. WALL Rod on Pivot a) If the rod is released from rest in the position shown what is the initial angular acceleration? τ = r F = Iα L 2 (mg) = 1 3 ML2 α α = 3g 2L b) What is the initial linear acceleration of the right end of the rod? α = 3g 2L, so a r = 3g 2L, or a L = 3g 2L a = 3 2 g
27 Rotational Work & Power How much Work is done when a Force is applied to a point, causing the object to rotate a distance ds? dw = F d s dw = F sinθ ds dw = F sinθ r dφ dw = τ dφ W = τ dφ τ = rf sinθ r F sin θ θ i How much Power is required to do that work? dw = τ dθ dt dt P = τ ω F φ f
28 Work K? W = W = W = m F d s ma ds s f dv ds = m dv dt dx s i s f s i dx dt dx W = m v f v i v dv = 1 2 mv 2 f 1 2 mv i2
29 Example 13 Show that the Work done by Torque produces a change in an object s rotational K. τ = Iα τ = I dω dt = I dω dθ dθ dt τ = I dω dθ ω (by chain rule) dw = τ dθ, so τ = dw dθ, and dw dθ = I dω dθ ω dw = Iω dω W = I ω f ω dω = 1 2 Iω f Iω i 2 ω i
30 Example 14 A rod of length L is free to rotate on a frictionless pin as shown, the rod is released from rest in the horizontal position. WALL Rod on Pivot a) What is the angular speed of the rod in its lowest position (right before it hits the wall)? omega = (3g/L) b) What is the linear speed of the center of mass at this lowest position? v(cm)=1/2 (3gL) c) What is the linear speed of the end of the rod at this lowest position? v(end)= (3gL)
31 Example 15 Two masses are connected to a pulley as shown and the system is released from rest. Find the linear speeds of the masses after m 2 has descended a distance h, and the angular speed of the pulley at that moment. m 1 m 2 U 1i +U 2i + K 1i + K 2i + K rotational = U 1 f +U 2 f + K 1 f + K 2 f + K rotational = m 1 gh + m 2 g( h) m v m v Iω 2 v = rω, and speed v 1 = v 2, so gh(m 2 m 1 ) = 1 2 m 1(rω) m 2(rω) Iω 2 2gh(m 2 m 1 ) = ω 2 (m 1 r 2 + m 2 r 2 + I) ω = 2gh(m 2 m 1 ) m 1 r 2 + m 2 r 2 + I
32 Translational Rotational v = dx dt a = dv dt F = ma v f = v i + at x f x i = v i t at 2 v f 2 = v i 2 + 2a(x f x i ) W = K = 1 2 mv 2 F dx P = dw dt = Fv p = mv F = dp dt
33 Fixed Axle Rolling! x cm = rθ v cm = ds dt = d(rθ) dt a cm = dv cm dt = d(rω) dt = r dθ dt = rω = r dω dt = rα
34 K rolling body =? The total K energy of an object undergoing rolling motion is the sum of the rotational K energy about the center of mass, and the translational K energy of the center of mass. K total = K rotational + K translational
35 Example 16 Find the acceleration, and speed at the bottom, of a sphere rolling down an inclined plane as shown, using energy concepts. I cm = 2 5 MR2 U g = 1 2 I cmω mv 2 cm Solve to get v = 10 7 gh, then use v 2 = v i 2 + 2ax, to get a = 5 7 gsinθ h x θ
36 Example 17 Find the acceleration, and speed at the bottom, of a sphere rolling down an inclined plane as shown, using a Force analysis. h x θ F net = ma cm F // F friction = ma cm mgsinθ F friction = ma cm τ = I cm α & rf friction = 2 ) ( ' 5 mr2 * + & ( a) ' r + * & F friction = ( 2 ' 5 ma ) + * & mgsinθ ( 2 ' 5 ma ) + = ma * cm a cm = 5 7 gsinθ
37 Example 18 Based on the previous analysis, what is the minimum µ necessary for the object to roll? What happens if the µ is less than that minimum value? h x θ " F friction = $ 2 # 5 ma % ' & a & cm = 5 gsinθ, so 7 F friction = 2 7 mgsinθ µ = F friction F Normal = 2 7 mgsinθ mgcosθ = 2 7 tanθ A smaller µ implies that the sphere will be slipping as it rolls, losing some energy to frictional heat losses. Thus, it will have less total K by the time it reaches the bottom of the ramp. Its rotational speed will be less (why? Do a Force analysis) but its translational speed will be greater (why?)
38 Example 19 A hoop, a ball, and a solid cylinder are all released from the top of a ramp at the same time so that they roll down. Which one reaches the bottom first? Does it depend on the mass of the objects? Does it depend on their relative radii?
39 Tests Comments Be careful what you re differentiating with respect to. (Work= F dx, Impulse= F dt) F Normal mg every time! Simplify when possible. When giving answers in variable form: don t indicate units don t write 9.8 for g Don t leave two paths to two answers on paper Don t cross out work if you re not replacing it with something else.
40 Photogates
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