YF_ISM_10

# YF_ISM_10 - DYNAMICS OF ROTATIONAL MOTION 10 EXECUTE = Fl l...

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10-1 D YNAMICS OF R OTATIONAL M OTION 10.1. IDENTIFY: Use Eq.(10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in Fig.(10.4) to calculate the torque direction. (a) SET UP: Consider Figure 10.1a. EXECUTE: Fl τ = sin (4.00 m)sin90 lr φ == ° 4.00 m l = (10.0 N)(4.00 m) 40.0 N m Figure 10.1a This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector G is directed out of the plane of the figure. (b) SET UP: Consider Figure 10.1b. EXECUTE: Fl = sin (4.00 m)sin120 ° 3.464 m l = (10.0 N)(3.464 m) 34.6 N m Figure 10.1b This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector G is directed out of the plane of the figure. (c) SET UP: Consider Figure 10.1c. EXECUTE: Fl = sin (4.00 m)sin30 ° 2.00 m l = (10.0 N)(2.00 m) 20.0 N m Figure 10.1c This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector G is directed out of the plane of the figure. (d) SET UP: Consider Figure 10.1d. EXECUTE: Fl = sin (2.00 m)sin60 1.732 m ° = (10.0 N)(1.732 m) 17.3 N m Figure 10.1d This force tends to produce a clockwise rotation about the axis; by the right-hand rule the vector G is directed into the plane of the figure. (e) SET UP: Consider Figure 10.1e. EXECUTE: Fl = 0 r = so 0 l = and 0 = Figure 10.1e 10

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10-2 Chapter 10 (f) SET UP: Consider Figure 10.1f. EXECUTE: Fl τ = sin , lr φ = 180 , so 0 l = and 0 = Figure 10.1f EVALUATE: The torque is zero in parts (e) and (f) because the moment arm is zero; the line of action of the force passes through the axis. 10.2. IDENTIFY: Fl = with sin = . Add the two torques to calculate the net torque. SET UP: Let counterclockwise torques be positive. EXECUTE: 11 1 (8.00 N)(5.00 m) 40.0 N m Fl =− . 22 2 (12.0 N)(2.00 m)sin30.0 12.0 N m =+ = ° . 12 28.0 N m ττ τ =+ = . The net torque is 28.0 N m , clockwise. EVALUATE: Even though FF < , the magnitude of 1 is greater than the magnitude of 2 , because 1 F has a larger moment arm. 10.3. IDENTIFY and SET UP: Use Eq.(10.2) to calculate the magnitude of each torque and use the right-hand rule (Fig.10.4) to determine the direction. Consider Figure 10.3 Figure 10.3 Let counterclockwise be the positive sense of rotation. EXECUTE: 123 (0.090 m) (0.090 m) 0.1273 m rrr === + = 1 1 sin (0.1273 m)sin135 0.0900 m == ° = 1 (18.0 N)(0.0900 m) 1.62 N m 1 G is directed into paper 2 2 sin (0.1273 m)sin135 0.0900 m ° = 2 (26.0 N)(0.0900 m) 2.34 N m 2 G is directed out of paper 33 3 3 sin (0.1273 m)sin90 0.1273 m ° = 3 (14.0 N)(0.1273 m) 1.78 N m 3 G is directed out of paper 1.62 N m 2.34 N m 1.78 N m 2.50 N m ττ τ τ =++ = ⋅+ ⋅= EVALUATE: The net torque is positive, which means it tends to produce a counterclockwise rotation; the vector torque is directed out of the plane of the paper. In summing the torques it is important to include + or signs to show direction. 10.4. IDENTIFY: Use sin Fl rF to calculate the magnitude of each torque and use the right-hand rule to determine the direction of each torque. Add the torques to find the net torque.
Dynamics of Rotational Motion 10-3 SET UP: Let counterclockwise torques be positive. For the 11.9 N force ( 1 F ), 0 r = . For the 14.6 N force ( 2 F ), 0.350 m r = and 40.0 φ = ° . For the 8.50 N force ( 3 F ), 0.350 m r = and 90.0 = ° EXECUTE: 1 0 τ = .

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YF_ISM_10 - DYNAMICS OF ROTATIONAL MOTION 10 EXECUTE = Fl l...

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