316_PartUniversity Physics Solution

316_PartUniversity Physics Solution - Dynamics of...

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Dynamics of Rotational Motion 10-11 (c) From either of the above relations between f and cm , a cm s s 22 sin cos 57 fm a m g nm g βμ μ β == = . () s 27tan . μ EVALUATE: If s 0 = , cm sin am g = . cm a is less when friction is present. The ball rolls farther uphill when friction is present, because the friction removes the rotational kinetic energy and converts it to gravitational potential energy. In the absence of friction the ball retains the rotational kinetic energy that is has initially. Figure 10.26 10.27. (a) IDENTIFY: Use Eq.(10.7) to find z α and then use a constant angular acceleration equation to find . z ω SET UP: The free-body diagram is given in Figure 10.27. EXECUTE: Apply z z I τ = to find the angular acceleration: z FR I = 2 2 (18.0 N)(2.40 m) 0.02057 rad/s 2100 kg m z FR I = Figure 10.27 SET UP: Use the constant z kinematic equations to find . z ?; z = 0 z (initially at rest); 2 0.02057 rad/s ; z = 15.0 s t = EXECUTE: 2 0 0 (0.02057 rad/s )(15.0 s) 0.309 rad/s zz z t ωω α =+= + = (b) IDENTIFY and SET UP: Calculate the work from Eq.(10.21), using a constant angular acceleration equation to calculate 0 , θ or use the work-energy theorem. We will do it both ways. EXECUTE: (1) z W (Eq.(10.21)) 2 11 00 0 (0.02057 rad/s )(15.0 s) 2.314 rad tt θθθ ω Δ=− = + =+ = (18.0 N)(2.40 m) 43.2 N m z tF R = Then (43.2 N m)(2.314 rad) 100 J. z W τθ =Δ= = or (2) tot 2 1 WK K =− (the work-energy relation from chapter 6) tot , WW = the work done by the child 1 0; K = 2 2 (2100 kg m )(0.309 rad/s) 100 J KI = Thus 100 J, W = the same as before. EVALUATE: Either method yields the same result for W . (c) IDENTIFY and SET UP: Use Eq.(6.15) to calculate av P EXECUTE: av 100 J 6.67 W 15.0 s W P t Δ = Δ EVALUATE: Work is in joules, power is in watts. 10.28. IDENTIFY: Apply P τω = and W . SET UP: P must be in watts, Δ must be in radians, and must be in rad/s. 1 rev 2 rad π = . 1 hp 746 W = . rad/s 30 rev/min = . EXECUTE: (a) ( ) 175 hp 746 W / hp 519 N m. rad/s 2400 rev/min 30 rev/min P = ⎛⎞ ⎜⎟ ⎝⎠ (b) ( ) 519 N m 3260 J W =Δ = =
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10-12 Chapter 10 EVALUATE: 40 rev/s ω = , so the time for one revolution is 0.025 s. 5 1.306 10 W P , so in one revolution, 3260 J WP t == , which agrees with our previous result. 10.29. IDENTIFY: Apply z z I τ α = and constant angular acceleration equations to the motion of the wheel. SET UP: 1 rev 2 rad π = . rad/s 30 rev/min = . EXECUTE: (a) 0 z z zz II t τα . () ( ) 2 rad s 1 2 1.50 kg 0.100 m 1200 rev min 30 rev min 0.377 N m 2.5 s z ⎛⎞ ⎜⎟ ⎝⎠ (b) ( ) av 600 rev/min 2.5 s 25.0 rev 157 rad. 60 s/min t Δ= = = (c) (0.377 N m)(157 rad) 59.2 J W τθ =Δ = = . (d) 2 22 11 r a d / s (1/ 2)(1.5 kg)(0.100 m) (1200 rev/min) 59.2 J 3 0 r e v / m i n KI = . the same as in part (c). EVALUATE: The agreement between the results of parts (c) and (d) illustrates the work-energy theorem 10.30. IDENTIFY: The power output of the motor is related to the torque it produces and to its angular velocity by z z P τω = , where z must be in rad/s. SET UP: The work output of the motor in 60.0 s is 2 (9.00 kJ) 6.00 kJ 3 = , so 6.00 kJ 100 W 60.0 s P . 2500 rev/min 262 rad/s z . EXECUTE: 100 W 0.382 N m 262 rad/s z z P = EVALUATE: For a constant power output, the torque developed decreases and the rotation speed of the motor increases.
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316_PartUniversity Physics Solution - Dynamics of...

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