316_PartUniversity Physics Solution

# 316_PartUniversity Physics Solution - Dynamics of...

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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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