Chapter13 - 13.1 Model The crankshaft is a rotating rigid...

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13.1. Model: The crankshaft is a rotating rigid body. Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s. The angular acceleration ( α ) graph is based on the fact that α is the slope of the ω -versus- t graph.
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13.2. Model: The turntable is a rotating rigid body. Solve: The angular velocity is the area under the α -versus- t graph: α ω ω α ω α = = = + d dt x dt ( ) 0 area under the graph The values of ω at selected values of time ( t ) are: t (s) ω (rad/s) 0 0 0.5 (5 + 3.75)(0.5)/2 = 2.18 1.0 (5 + 2.5)(1)/2 = 3.75 1.5 (5 + 1.25)(1.5)/2 = 4.68 2.0 (5 + 0)(2)/2 = 5.0 2.5 5.0 3.0 5.0
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13.3. Model: The wheel is a rotating rigid body. Solve: (a) The angular acceleration ( α ) is the slope of the ω -versus- t graph. (b) The car is at rest at t = 0 s. It gradually speeds up for 4 s and then slows down for 4 s. The car is at rest from t = 8 s to t = 12 s, and then speeds up in the opposite direction for 4 s.
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13.4. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body. Solve: (a) The α -versus- t graph has a positive slope of 5 rad/s 2 from t = 0 s to t = 2 s and a negative slope of 5 rad/s 2 from t = 2 s to t = 4 s. (b) The angular velocity is the area under the α -versus- t graph: α ω ω α ω α = = = + d dt x dt ( ) 0 area under graph.
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13.5. Model: Spinning skater, whose arms are outstretched, is a rigid rotating body. Visualize: Solve: The speed v = r ω , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2 π /60 rad/s = 6 π rad/s. Thus, v = (0.70 m)(6 π rad/s) = 13.2 m/s. Assess: A speed of 13.2 m/s 26 mph for the hands is a little high, but reasonable.
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13.6. Model: The magnetic computer disk is a rigid rotating body. Visualize: Solve: Using the rotational kinematic equation ω ω α f i = + t , we get ω 1 = 0 rad + (600 rad/s 2 )(0.5 s 0 s) = 300 rad/s ω 2 = (300 rad/s) + (0 rad/s 2 )(1.0 s – 0.5 s) = 300 rad/s The speed of the painted dot v 2 = r ω 2 = (0.04 m)(300 rad/s) = 12 m/s. The number of revolutions during the time interval t 0 to t 2 is θ θ ω α θ θ ω α 1 0 0 1 0 0 1 0 2 2 2 1 1 2 1 1 2 1 2 1 2 0 0 1 2 600 0 5 0 75 1 2 75 300 1 0 0 5 0 225 1 = + + = + + = = + + = + + = ( ) ( ) ( )( . ) ( ) ( ) ( )( . . ) ( ) t t t t t t t t rad rad rad/s s s rad rad rad/s s s rad = 225 rad rad 2 rev rad rev 2 35 8 π = .
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13.7. Model: The drill is a rigid rotating body. Visualize: The figure shows the drill’s motion from the top. Solve: (a) The kinematic equation ω f = ω i + α ( t f – t i ) becomes, after using ω i = 2400 rpm = (2400)(2 π )/60 = 251.3 rad/s, t f t i = 2.5 s – 0 s = 2.5 s, and ω f = 0 rad/s, 0 rad = 251.3 rad/s + α (2.5 s) = − α 100 5 . rad/s 2 (b) Applying the kinematic equation for angular position yields: θ θ ω α f i i f i f i 2 rad rad/s s s rad/s s s 314.2 rad = 50.0 rev = + + = + + = ( ) ( ) ( . )( . ) ( . )( . ) t t t t 1 2 0 251 3 2 5 0 1 2 100 5 2 5 0 2 2
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13.8. Model: The turbine is a rigid rotating body. Solve: The known values are ω i = 3600 rpm = (3600)(2 π )/60 = 120 π rad/s, t i = 0 s, t f = 10 min = 600 s, ω f = 0 rad/s, and θ i = 0 rad. Using the rotational kinematic equation ω f = ω i + α ( t f – t i ), we get 0 rad = (120 π rad/s) + α (600 s – 0 s).
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