Chapter 09 - 9.1: a) 1 . 50 m 2 .50 m 0 . 60 rad 34.4 . b)...

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. : 91 a) . . 34 4 rad 60 . 0 m 50 2 m 50 . 1 ° = = . b) cm. 27 . 6 ) 180 rad )( ( 128 cm) 0 . 14 ( = ° ° π c) m. . 1 05 rad) m)( . 0 70 50 . 1 ( = . : 92 a) s. rad 199 s 60 min 1 rev rad 2 min rev 1900 = × π b) s. 10 . 3 07 s) rad 199 ( ) 180 rad ( 35 3 - × = ° × ° π . : 93 a) . s rad 42 s, . 3 5 at so , ) s rad ( . 12 0 2 3 = = = = α t t dt α z z The angular acceleration is proportional to the time, so the average angular acceleration between any two times is the arithmetic average of the angular accelerations. b) , ) s rad ( . 6 0 2 3 t ω z = so at s. rad . 73 5 s, . 3 5 = = z ω t The angular velocity is not linear function of time, so the average angular velocity is not the arithmetic average or the angular velocity at the midpoint of the interval. . : 94 a) . ) s rad 60 . 1 ( 2 t) ( 3 t βt α dt z z - = - = = b) . s rad . 4 80 s) 0 . 3 )( s rad 60 . 1 ( s) 0 . 3 ( 2 3 - = - = z α , s rad 40 . 2 s 0 . 3 s rad 00 . 5 s rad 20 . 2 s 0 . 3 ) 0 ( s) 0 . 3 ( . 2 av - = - - = - = - ω ω α z which is half as large (in magnitude) as the acceleration at s. . 3 0 = t . : 95 a) 2 3 2 ) s rad ( . 0 036 s) rad 400 . 0 ( 3 t βt γ ω z + = + = b) = = = γ ω t z , 0 At s. rad 70 . 0 so rad, . 3 50 s, rad 3 . 1 s, . 5 00 At c) s. rad 400 . 0 s . 5 00 rad 50 . 3 av = = = = = - z z ω θ ω t The acceleration is not constant, but increasing, so the angular velocity is larger than the average angular velocity.
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- - = - - = ) s rad 0 . 40 ( , ) s rad 50 . 4 ( ) s rad 0 . 40 ( s) rad 250 ( 2 z 2 3 2 α t t ω z at which time positive only the ; in quadratic a in results 0 Setting a) . ) s rad 00 . 9 ( 3 t ω t z = rev. 93.3 rad 586 s, 4.23 At ) c . s rad 78.1 s, 4.23 At b) s. 4.23 is 0 2 = = = - = = = = θ t α t t ω z z s. rad 138 e) s. rad 250 0, At d) s 4.23 rad 586 av = = = = - z z ω ω t 9.7: a) . , 0 Setting b) . 6 2 and 3 2 3 2 c b z dt dw z dt z t α ct b α ct bt ω z = = - = = - = = 9.8: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value. (b) The angular acceleration is 2 0 s rad 00 . 2 s 00 . 7 s) rad 00 . 6 ( s rad 00 . 8 = - - = - = t ω ω α Thus it takes 3.00 seconds for the wheel to stop ) 0 ( = z ω . During this time its speed is decreasing. For the next 4.00 s its speed is increasing from s rad 8.00 to s rad 0 + . (c) We have rad. 7.00 rad 49.0 rad 0 . 42 s) (7.00 ) s rad 00 . 2 ( s) (7.00 s) rad 00 . 6 ( 0 2 2 2 1 2 2 1 0 0 + = + - = + - + = α + ϖ + θ = t t θ Alternatively, the average angular velocity is s rad 00 . 1 2 s rad 00 . 8 s rad 00 . 6 = + - Which leads to displacement of 7.00 rad after 7.00 s. 9.9: a) ? s, 30.0 s, rev 8.333 min rev 500 rev, 200 0 0 = = = = = - ω t ω θ ω rpm 300 s rev 5.00 gives 2 0 0 = = + = - ω t ω ω θ θ b) Use the information in part (a) to find : α rev 312 gives 2 and 75.0 gives ? s, rev 333 . 8 , s rev 1111 . 0 0, Then s rev 1111 . 0 gives 0 0 0 0 0 2 2 0 = θ - θ + = -
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 09 - 9.1: a) 1 . 50 m 2 .50 m 0 . 60 rad 34.4 . b)...

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