08-203_hmw-Ch10-11

08-203_hmw-Ch10-11 - Chapter 10: Rotational Kinematics and...

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Chapter 10: Rotational Kinematics and Energy Solutions to Problems 6. Picture the Problem : The tire rotates about its axis through a certain angle. Strategy: Use equation 10-2 to find the angular displacement. Solution: Solve equation 10-2 for θ : 1.75 m 5.3 rad 0.33 m s r θ = = = Insight: This angular distance corresponds to 304° or 84% of a complete revolution. Picture the Problem : The propeller rotates about its axis with constant angular acceleration. Strategy: Use the kinematic equations for rotating objects and the given formula to find the average angular speed and angular acceleration during the specified time intervals. By comparison of the formula given in the problem, ( ) ( ) 2 2 125 rad/s 42.5 rad/s t t = + ,with equation 10-10, 2 1 0 0 2 t t ω α = + + , we can identify 0 125 rad/s = and 2 1 2 42.5 rad/s = . Solution: 1. (a) Use equations 10-3 and 10-10 to find av : ( )( ) ( )( ) 2 1 0 0 2 0 av 2 2 2 av t 125 rad/s 0.010 s 42.5 rad/s 0.010 s 0 0.010 s 125 rad/s 1.3 10 rad/s t t t t + - - Δ = = = Δ + - = = = × 2. (b) Use equations 10-3 and 10-10 to find av : ( )( ) ( )( ) ( )( ) ( )( ) 2 2 2 1 0 2 2 2 2 1 0 0 0 0 2 0 av 0 125 rad/s 1.010 s 42.5 rad/s 1.010 s 169.60 rad 125 rad/s 1.000 s 42.5 rad/s 1.000 s 167.50 rad 169.60 167.50 rad 210 rad/s 2.1 1.010 1.000 s t t t t t t t = + = + = = + = + = - Δ - = = = = = × Δ - - 2 10 rad/s 3. (c) Use equations 10-3 and 10-10 to find av : ( )( ) ( )( ) ( )( ) ( )( ) 2 2 2 1 0 2 2 2 2 1 0 0 0 0 2 0 av 0 125 rad/s 2.010 s 42.5 rad/s 2.010 s 422.95 rad 125 rad/s 2.000 s 42.5 rad/s 2.000 s 420.00 rad 422.95 420.0 rad 295 rad/s 3.0 1 2.010 2.000 s t t t t t t t = + = + = = + = + = - Δ - = = = = = × Δ - - 2 0 rad/s 4. (d) The angular acceleration is positive because the angular speed is positive and increasing with time. 5. (e) Apply equation 10-6 directly: 2 0 av 210 125 rad/s 85 rad/s 1.00 0.00 s t - - = = = Δ - 6. Apply equation 10-6 directly: 2 0 av 295 210 rad/s 85 rad/s 2.00 1.00 s t - - = = = Δ - Insight: We violated the rules of significant figures in order to report answers with two significant figures. Such problems arise whenever you try to subtract two large but similar numbers to get a small difference. The answers are only known to one significant figure, but we reported two in order to show clearly that the angular acceleration is constant. Of course we could also have figured from the equation given in the problem that since 2 1 2 42.5 rad/s = , it must be true that 2 85.0 rad/s . = 14. Picture the Problem : The propeller rotates about its axis, increasing its angular velocity at a constant rate. Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration. Solution: Solve equation 10-11 for : ( ) ( ) ( ) 2 2 2 2 0 26 rad/s 12 rad/s 17 rad/s 2 2 2.5 rev 2 rad rev π 2 - - = = = Δ × Insight: A speed of 26 rad/s is equivalent to 250 rev/min, indicating the motor is turning pretty slowly. A typical outboard motor is designed to operate at 5000 rev/min at full throttle.
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26. Picture the Problem : The two children sit in different places on the same merry-go-round, which is rotating about its axis at a constant rate.
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This homework help was uploaded on 04/14/2008 for the course PHYSICS 203 taught by Professor Croft during the Spring '08 term at Rutgers.

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08-203_hmw-Ch10-11 - Chapter 10: Rotational Kinematics and...

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