EAS208_HW07_110110_Solutions

EAS208_HW07_110110_Solutions - EAS 208 (Fall 2010) Homework...

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EAS 208 (Fall 2010) Homework #7 – Solutions Problem 17–6 (a) If the bicycle’s 120 sprocket wheel rotates through one revolution, through how many revolutions does the 45 gear turn? mm mm (b) If the angular velocity of the sprocket wheel is 1 rad s , what is the angular velocity of the gear? Answer: Assume that there is no slippage between the chain – sprocket wheel and the chain – gear. (a) Any point on the gears travels the same distance (the chain is inextensible). For the sprocket wheel and the small gear use the abbreviation and . s g 1
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Thus, the distance that the chain travels on the sprocket wheel is equal to the distance that the chain travels on the small gear: sg Distance Distance = RR s g θ = s g R R g s = 120 2 1 45 g rad rev rev π ⎛⎞ = ⎜⎟ ⎝⎠ 16.76 g rad = or 1 16.76 2.67 2 g rev rad rev rad == If the bicycle’s 120 sprocket wheel rotates through one revolution, the gear turns 2.67 . mm 45 mm rev (b) Similarly, the tangential velocity of a point on the teeth, in contact with the chain, of the sprocket wheel is the same as the tangential velocity of a point on the teeth, in contact with the chain, of the small gear: ,, s tg vv t = s g R R ω = s g s g R R = 120 1 45 g rad s = 2.67 g rad s = If the angular velocity of the sprocket wheel is 1 rad s , the angular velocity of the gear is 2.67 g rad s = . 2
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Problem 17–12 Consider the bar shown in Problem 17.11. If 3 A m s = v and 2 28 A m s = a , what are B v and B a ? Answer: The velocity of point A is given by: AA R ω = v A A R = v 22 3 5.30 0.4 0.4 rad s == + The acceleration of point A is given by: ,, n nA t aa t = + aee A t 2 n R R ωα =+ ae e 3
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The normal component of the acceleration at point is: A 2 , An A aR ω = 22 2 2 , 5.30 0.4 0.4 15.89 m a s =+ = The magnitude of the acceleration for point A is given to be 2 28 A m s = a . The magnitude of the acceleration for point is: A 2 ,, 28 AA n A t m aa s = a 28 At =− 2 2 , 28 15.89 23.05 m a s = But, the tangential component of the acceleration at point is given by the relation: A , A α = , A a R = 2 23.05 40.75 0.4 0.4 rad s == + (angular acceleration) Therefore, the angular velocity and angular acceleration of the rod system are, respectively: 5.30 rad s = 2 40.75 rad s = Note that we can only calculate the magnitude of the angular velocity and angular acceleration of the system and not the direction. This is because we are given information for the magnitudes of velocity and acceleration of point A as initial conditions. 4
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The velocity of point B is given by: BB R ω = v 22 5.30 0.4 0.2 B =+ v 2.37 B m s = v The acceleration of point B is given by: ,, n nB t aa t = + aee B t 2 n R R ωα ae e 2 2 2 5.30 0.4 0.2 40.75 0.4 0.2 Bn ++ t e t e 12.56 18.22 ( ) 2 m s Therefore, the magnitude of velocity at point B is: n a B t 12.56 18.22 B a 2 22.13 B m s = a Finally, the magnitude of velocity and acceleration of point B are, respectively: 2.37 B m s = v 2 22.13 B m s = a 5
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Problem 17–30 Points A and B of the bar slide on the plane surfaces. Point 2 m B is moving to the right at 3 m s . What is the velocity of the midpoint of the bar? G Answer: The velocity of point A relative to point B is: ABA B = + vvv AB A B r ω = vv r r (1)
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EAS208_HW07_110110_Solutions - EAS 208 (Fall 2010) Homework...

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