T1a7s - Assignment 7 solutions (Fall term Physics 102) 7.9...

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7.9 Main Rotor: () 2 r a d rev 1 min 3.80 m 450 min 1 rev 60 s vr π ω ⎛⎞ ⎛⎞ ⎛ = == ⎜⎟ ⎜ ⎜⎟ ⎝⎠ ⎝ ⎝⎠ 179 m s m = 179 = s3 4 3 m s sound v v 0.522 sound v Tail Rotor: r a d 0.510 m 4138 ⎛⎞⎛ = ⎜⎟⎜ ⎝⎠⎝ 221 m s m = 221 = 4 3 m s sound v v 0.644 sound v 7.21 Friction must supply the needed centripetal force. Hence, it is necessary that ( ) cs max Ff , or 2 t s v mm r µ g and the maximum safe speed is ( ) ( ) 2 0.70 20 m 9.80 m s ts max g = 12 m s . 7.23 (a) ( ) 2 2 2.00 m 3.00 rad s c ar = 2 18.0 m s (b) ( ) ( ) 2 50.0 kg 18.0 m s cc Fm a = 900 N (c) We know the centripetal acceleration is produced by the force of friction. Therefore, the needed static friction force is 900 N s f = . Also, the normal force is . Thus, the minimum coefficient of friction required is 490 N nm g ( ) 900 N = 490 N s max s f n 1.84 . So large a coefficient of friction is unreasonable, and he will not be able to stay on the merry-go-round. 7.25 (a) Since the 1.0-kg mass is in equilibrium, the tension in the string is ( ) ( ) 2 1.0 kg 9.8 m s Tm g = 9.8 N .
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T1a7s - Assignment 7 solutions (Fall term Physics 102) 7.9...

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