Dynamics of Rotational Motion

Transcription

1 Chapter 10 Dynamics of Rotational Motion To understand the concept of torque. To relate angular acceleration and torque. To work and power in rotational motion. To understand angular momentum. To understand the conservation of angular momentum. To study how torques add a new variable to equilibrium. To see the vector nature of angular quantities.

2 10.1 Torque What is torque? When you apply a force to rotate an object about a pivot or an axis, the effectiveness of your action depends not only on the magnitude of the force you apply, but also on a quantity known as the moment arm. What is the moment arm? It is the perpendicular distance from the pivot or the axis to the line of action of the applied force. Torque = (Magnitude of Force) (Moment Arm) τ = Fl Units: N m ı റF

3 Torque is a Vector The Magnitude and the Direction of a Torque Using the examples in the figure on the right: (a) Torque by റF 1 τ 1 = F 1 l 1 (b) Torque by റF 2 τ 2 = F 2 l 2 counter-clockwise, positive clockwise, negative (c) Torque by റF 3 τ 3 = F 3 l 3 = 0

4 More on the Magnitude of Torque Three ways to calculate torque: = Fl = Frsin = F(rsin ) = Fr = (Fsin )r = F r

5 Conceptual Analysis 10.1 on page 284 A rod is pivoted at its center. Three forces of equal magnitude are apply as shown. Which of the following statements correctly describe the order in the magnitudes of the torques by these three forces with respect to the pivot? (a) 1 = 2 = 3 (b) 1 > 3 > 2 (c) 2 > 3 > 1

6 10.2 Torque and Angular Acceleration Again, cut the rigid body into many small pieces, A, B, C,. The force acting on piece A is റF A. Consider the motion of piece (particle) A. According to Newton s Second Law, F A,tan = m A a A,tan = m A r A α τ A = r A F A,tan = m A r A 2 α Sum over the torques for all the pieces: τ A + τ B + τ C = m A r A 2 α + m B r B 2 α + m C r C 2 α + = (m A r A 2 + m B r B 2 + m C r C 2 + )α or σ τ = Iα This is also known as Newton s Second Law for rotational motion.

7 How about torques by internal forces? σ τ int = 0 Therefore, we have σ τ ext = Iα

8 Re-visit an Earlier Problem Example 10.2 on page 288 A cable unwinding from a winch. റF y R Find: (a) Magnitude of the angular acceleration (b) Final angular velocity (c) Final speed of cable x Solution: The torque by the tension force = Fl = (9.0 N) (0.06 m) = 0.54 N m. The moment of inertia about the given axis I = 1 2 MR2 = 1 (50 kg)( m)2 = kg m 2 Therefore, the angular acceleration a = /I = (0.54 N m)/(0.090 kg m 2 ) = 6.0 rad/s 2 Angular Displacement q q 0 = s/r = (2.0 m)/(0.06 m) = 33 rad Final angular velocity ω = ω α(q q 0 ) = 2α(q q 0 ) = 20 rad/s Final speed of cable v = rω = (0.06 m)(20 rad/s) = 1.2 m/s

9 Re-visit Another Problem Example 10.3 on page 288 Given: M, R, m, and h Find: (a) a for winch, a for bucket, and tension force. (b) v for the bucket just before hitting the water (c) w for the winch just before hitting the water Solution: Linear motion of the bucket mg T = ma (1) Rotational motion of the winch = Ia or TR = ( 1 2 MR2 )a (2) From (2): T = 1 MRa = 1 Ma. (3) 2 2 Substitute (3) into (1): mg 1 Ma = ma 2 Linear acceleration: a = mg/(m + M/2) = g/(1 + M/2m) Tension force T = Ma/2 = Mg/(2 + M/m) Angular acceleration: a = a/r = g/[r(1 + M/2m) Final v and w calculated using the kinematic equations: v = 2gh 1+M/2m and w = 1 R 2gh 1+M/2m

10 Re-visit One more Problem y o R T x mg y Example 10.4 on page 290 Given: Given M, R, and h Find: (a) Center of mass acceleration and angular acceleration (b) Tension force (c) Velocity of the center of mass v cm Solution: Apply Newton s Second Law for linear motion Mg T = Ma (1) Apply Newton s Second Law for rotational motion = Ia or TR = ( 1 2 MR2 )a (2) From (2): T = 1 MRa = 1 Ma. (3) 2 2 Substitute (3) into (1): Mg 1 Ma = Ma 2 Linear acceleration: a = 2g/3 Angular acceleration: a = a/r = 2g/3R Tension force T = Mg/3 Final v cm using the kinematic equations v cm = 2ah = 4gh/3

11 Example 10.5 on page 291: Rolling without slipping What is the acceleration of a rolling bowling ball? Given: Given M, R, and b Find: (a) a cm and a (also, f s, required minimum m s ) Solution: Apply Newton s Second Law for linear motion Mgsinb f s = Ma cm (1) Apply Newton s Second Law for rotational motion = Ia or f s R = ( 2 5 MR2 )a.. (2) From (2): f s = 2 5 MRa = 2 5 Ma cm.. (3) Substitute (3) into (1): Linear acceleration: Angular acceleration: Static friction force Mgsinb 2 5 Ma cm = Ma cm a cm = 5 7 gsinb a = a cm = 5 gsinb R 7R f s = 2 MRa = 2 Mgsinb 5 7 Minimum static coefficient of friction m s = f s /n = ( 2Mgsinb )/Mgconb 7 = 2 tanb 7

12 10.3 Work and Power in Rotational Motion Again, cut the rigid body into many small pieces, A, B, C,. The force acting on piece A is റF A. If the angular position of A changes by θ, the work done by a constant force റF A acting on A is: W A = F A,tan R θ = τ A θ Sum over contributions from all the pieces, the total work W = τ θ = τ θ 2 θ 1 A few notes: (a) Applicable for constant torque τ. (b) Work has units N m. Power is the rate at which work is done: P = W θ = τ = τω t t (Compare with linear motion in which case P = Fv)

13 10.4 Angular Momentum In Chapter 8 we defined the momentum of a particle as റp = m റv, റp we could state Newton s Second Law as റF = lim. t 0 t Here we define the angular momentum of a rigid body: L = Iω Notes: It is also a vector, same as ω. Units: kg m 2 /s Angular momentum of a point particle: L = mvl (kg m 2 /s) Notes: (a) l is effective the moment arm (b) a particle moving along a straight line can still have an angular momentum about a pivot or rotational axis

14 Example 10.7 on page 295: A kinetic sculpture Given: Two small metal sphere of masses m 1 = m 2 = 2.0 kg Uniform metal rod of mass M = 3.0 kg and length s = 4.0 m Angular velocity 3.0 rev/minutes about a vertical axis through the middle Find: Angular momentum and Kinetic energy Solution: (a) Angular momentum Total moment of inertia I total = I sphere 1 + I sphere 2 + I rod = m 1 (s/2) 2 + m 2 (s/2) 2 + (1/12)Ms 2 = 20 kg m 2 Angular momentum L = Iω = (20 kg m 2 )(3.0 2p/60 s -1 ) = 6.2 kg m 2 /s (b) Kinetic energy K = (1/2)Iω 2 = 0.96 J

15 10.5 Conservation of Angular Momentum The Relationship Between Torque and Angular Momentum Since ω σ τ = Iα = I lim = lim I ω t 0 t t 0 t = lim (Iω) t 0 t We can state Newton s Second Law for rotational motion as L σ τ = lim t 0 t L = lim t 0 t Conservation of Angular Momentum If σ τ = 0, Then L = constant

16 Example 10.9 on page 299: Given: I body, i = 3.0 kg m 2 I body, f = 2.2 kg m 2 m dumbbell = 5.0 kg each R dumbbell, i = 1.0 m R dumbbell, f = 0.20 m ω i = 1 rev in 2 s = 2p/2 rad/s = p rad/s Find: (a) ω f (b) compare K f with K i Solution: (a) for ω f Total initial moment of inertia I total,i = I body, i + 2I dumbbell,i = (3.0 kg m 2 ) + 2[m dumbbell (R dumbbell, i ) 2 ] = 13 kg m 2 Total final moment of inertia I total,f = I body, f + 2I dumbbell,f = (2.2 kg m 2 ) + 2[m dumbbell (R dumbbell, f ) 2 ] = 2.6 kg m 2 Apply the conservation of angular momentum so that ω f = I total,i ω i /I total,f = 5.0p rad/s = 2.5 rev/s I total,i ω i = I total,f ω f (b) compare K f with K i K i = (1/2)I total,i ω i 2 = 64 J K f = (1/2)I total,f ω f 2 = 320 J

17 10.6 Equilibrium of a Rigid Body The equilibrium of a point particle is determined by the conditions of σ F x = 0 and σ F y = 0. The Equilibrium of a Rigid Body For the equilibrium of a rigid body, both its linear motion and rotational motion must be considered. Therefore, in addition to σ F x = 0 and σ F y = 0, We must add another condition regarding its rotational motion σ τ = 0. This third condition can be set up about any chosen axis.

18 General principle: Strategy for Solving Rigid Body Equilibrium Problems The net force must be zero. The net torque about any axis must be zero. Draw a diagram according to the physical situation. Analyze all the forces acting on each part of a rigid body. Sketch all the relevant forces acting on the rigid body. Based on the force analysis, set up a most convenient x-y coordinate system. Break each force into components using this coordinates. Sum up all the x-components of the forces to an equation: σ F x = 0. Sum up all the y-components of the forces to an equation: σ F y = 0. Based on the force analysis, set up a most convenient rotational axis. Calculate the torque by each force about this rotational axis. Sum up all the torques by the forces to an equation: σ τ = 0. Use the above three equations to solve for unknown quantities.

19 Example on page 304 Climbing a ladder (length is 5 m) Find: (a) Normal and friction forces at the base of the ladder (b) Minimum m s at the base (c) Magnitude and direction of the contact force at the base Solution: (a) σ F x = 0 σ F y = 0 f s n 1 = 0 n 2 (800 N) (180 N) = 0 n 2 = 980 N σ τ = 0 n 1 (4.0 m) (180 N)(1.5 m) (800 N)(1.0 m) = 0 n 1 = 268 N f s = 268 N (b) Minimum m s = (268 N)/(980 N) = 0.27 (c) F B = (268 N) 2 +(980 N) N = 1020 N θ = tan 268 N = 75