Chapter 8. Rotational Equilibrium and Rotational Dynamics

Save this PDF as:

Size: px
Start display at page:

Download "Chapter 8. Rotational Equilibrium and Rotational Dynamics"


1 Chapter 8 Rotational Equilibrium and Rotational Dynamics

2 Wrench Demo

3 Torque Torque, τ, is the tendency of a force to rotate an object about some axis τ = Fd F is the force d is the lever arm (or moment arm) Units are Newton-m

4 Direction of Torque Torque is a vector quantity Direction determined by axis of twist Perpendicular to both r and F In -d, torque points into or out of paper Clockwise torques point into paper. Defined as negative Counter-clockwise torques point out of paper. Defined as positive

5 Non-perpendicular forces Only the y- component, Fsinφ, produces a torque

6 Non-perpendicular forces τ = Fr sinφ F is the force r is distance to point where F is applied Φ is the angle between F and r

7 Torque and Equilibrium Forces sum to zero (no linear motion) ΣF x = 0 and ΣF = y 0 Torques sum to zero (no rotation) = 0 Στ

8 Meter Stick Demo

9 Axis of Rotation Torques require point of reference Point of reference can be anywhere Use same point for all torques Pick the point to make problem least difficult

10 Example Given M = 10 kg. Neglect the mass of the beam. a) Find the tension in the cable b) What is the force between the beam and the wall

11 Solution a) Given: M = 10 kg, L = 10 m, x = 7 m Find: T Basic formula τ = 0 TL Mgx = 0 Axis Solve for T = 84 N

12 Solution b) Given: M = 10 kg, L = 10 m, x = 7 m, T = 84 N Find f Basic formula F = 0 y T Mg + f = 0 f Solve for f = 353 N

13 Alternative Solution b) Given: M = 10 kg, L = 10 m, x = 7 m Find f Basic formula τ = 0 Mg( L x) fl = 0 f Axis Solve for f = 353 N

14 Another Example Given: W=50 N, L=0.35 m, x=0.03 m Find the tension in the muscle W Basic formula τ = 0 Fx WL = 0 x L F = 583 N

15 Center of Gravity Gravitational force acts on all points of an extended object However, it can be considered as one net force acting on one point, the center-of-gravity, X. i (m g) x i i = X i m g i X = i i m x i m i i Weighted Average

16 Example Consider the 400-kg beam shown below. Find T R

17 Solution Find T R Basic formula τ = 0 to solve for T R, choose axis here MgX + T R L = 0 T R = 1 11 N X = m L = 7 m

18 One Last Example Given: x = 1.5 m, L = 5.0 m, w beam = 300 N, w man = 600 N Find: T Fig 8.1, p.8 Slide 17 x L

19 Solution Given: x, L, w beam,w man Find: T First, find T y Basic formula τ = 0 Fig 8.1, p.8 Slide 17 w man x w beam L / + T L = y 0 T y = 330 N Now, find T T y = T sin( 53 ) T = 413 N x L

20 Baton Demo Moment-of_Inertia of_inertia Demo

21 Torque and Angular Acceleration When τ 0, rigid body experiences angular acceleration Relation between τ and α is analagous to relation between F and a F = ma, = τ Iα Moment of Inertia

22 Moment of Inertia This mass analog is called the moment of inertia, I, of the object I i m i r i r is defined relative to rotation axis SI units are kg m

23 More About Moment of Inertia I depends on both the mass and its distribution. If an object s mass is distributed further from the axis of rotation, the moment of inertia will be larger.

24 Demo: Moment of Inertia Olympics

25 Moment of Inertia of a Uniform Ring Divide ring into segments The radius of each segment is R I = Σm r = i i MR

26 Example What is the moment of inertia of the following point masses arranged in a square? a) about the x-axis? b) about the y-axis? c) about the z-axis?

27 Solution a) Find I about the x-axis? Given: M =, M 3 =3 kg, L=0.6 m Basic formula I = m r i i r = 0.6 sin(45 ) First, find distance to -kg masses I = M r + M r =0.7 kg m

28 Solution b) Find I about the y-axis? Same as before, except you use the 3-kg masses r = 0.6 sin(45 ) I = M r + 3 M r 3 =1.08 kg m

29 Solution c) Find I about the z-axis? r = 0.6 sin(45 ) Use all the masses I = M + r M 3 + M r + r M 3 r =1.8 kg m

30 Other Moments of Inertia

31 Example Treat the spindle as a solid cylinder. a) What is the moment of Inertia of the spindle? b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel? c) What is the acceleration of the bucket? M d) What is the mass of the bucket?

32 Solution a) What is the moment of Inertia of the spindle? Given: M = 5 kg, R = 0.6 m Moments of Inertia MR, cylindrical shell 1 MR, solid cylinder MR, solid sphere 5 MR, spherical shell M 3 1 ML, rod, about end I = MR = 0.9 kgm ML, rod, about middle 1

33 Solution b) If the tension in the rope is 10 N, what is α? Given: I = 0.9 kg m, T = 10 N, r = 0.6 m Basic = formula τ τ = rf Iα rt = Iα Solve for α α=6.67 rad/s c) What is the acceleration of the bucket? Given: r=0.6 m, α = 6.67 rad/s M a Basic = formula αr a=4 m/s

34 Solution d) What is the mass of the bucket? Given: T = 10 N, a = 4 m/s Basic F = formula ma Ma = Mg T T M = g a M = 1.7 kg M

35 Example A cylindrical space station of radius R = 1 m and mass M = 3400 kg is designed to rotate around the axis of the cylinder. The space station s moment of inertia is 0.75 MR. Retrorockets are fired tangentially to the surface of space station and provide an impulse of.9x10 4 N s. a) What is the angular velocity of the space station after the rockets have finished firing? b) What is the centripetal acceleration at the edge of the space station?

36 τ = Solution Given: M = 3400, R = 1, I = 0.75 MR = 3.67x10 5 F t =.9x10 4. a) Find: ω Basic formula Iα = I Basic formula τ = rf ω t τ t RF = t Iω = Iω Solve for ω. ω= rad/s b) Find centripetal acceleration Basic formula a = ω r a = ω R = 10.8 m/s

37 Rotational Kinetic Energy Each point of a rigid body rotates with angular velocity ω. KE 1 1 = m v 1 i i = miri ω = Iω Including the linear motion 1 KE = mv + 1 Iω KE due to rotation KE of center-of-mass motion

38 Example What is the kinetic energy of the Earth due to the daily rotation? Given: M earth =5.98 x10 4 kg, R earth = 6.63 x10 6 m. Basic formula ω = π T Solid sphere First, find ω ω = π = 7.7 x10-5 rad/s I = MR 5 Basic formula 1 5 KE = MR ω =.78 x10-9 KE = 1 Iω

39 Example A solid sphere rolls down a hill of height 40 m. What is the velocity of the ball when it reaches the bottom? (Note: We don t know r or m!) Basic formula 1 mgh = mv + For solid sphere I = mr 5 1 Iω m cancels gh = 1 v + 5 r ω v Basic = ωr formula 1 gh gh v + v, v = 5 7 / 5 = v = 3.7 m/s

40 Suitcase Demo

41 Angular Momentum L = Iω Rigid body L = mvr Point particle Analogy between L and p Angular Momentum L = Iω τ = L/ t Conserved if no net outside torques Linear momentum p = mv F = p/ t Conserved if no net outside forces

42 Rotating Chair Demo

43 Angular Momentum and Kepler s nd Law For central forces, e.g. gravity, τ = 0 and L is conserved. Change in area in t is: A L A t = = = 1 r( v mrv 1 L m t)

44 Example A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

45 Solution Known: M, R, m, v 0 Find: ω F First, find L 0 Basic formula L = mvr L = mvr 0 Next, find I tot Solid cylinder I Particle I = = 1 MR MR ITot = + mr 1 MR Now, given I tot and L 0, find ω Basic formula L = Iω ω = L 0 I Tot =.71 rad/s

46 Example Two twin ice skaters separated by 10 meters skate without friction in a circle by holding onto opposite ends of a rope. They move around a circle once every five seconds. By reeling in the rope, they approach each other until they are separated by meters. a) What is the period of the new motion? b) If each skater had a mass of 75 kg, what is the work done by the skaters in pulling closer?

47 Solution a) Find: T f L ω = Given: R 0 =5 m, R f =1 m, T 0 =5 s Basic formula = Iω π T Point particle I = mr First, find expression for initial L ( skaters) L 0 = mr 0 π T 0 Next, apply conservation of L mr π π mr f T0 Tf R f T T 0 = f = = T 0 /5 = 0. s 0 R0

48 Solution b) Find: W Given: m=75 kg, R 0 =5 m, R f =1 m, T 0 =5 s, T f =0. s First, find moments of inertia I = 0 = mr0, I f mr f Basic formula 1 KE = Iω Next, find values for ω ω π, ω 0 = f = T0 π T Finally, find change in KE 1 1 W = I fω f I0ω 0 = 7.11x10 5 J f