chap09 - 9.1: a) . 34.4 rad 60 . m 50 2 m 50 . 1 ° = = ....

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Unformatted text preview: 9.1: a) . 34.4 rad 60 . m 50 2 m 50 . 1 ° = = . b) cm. 27 . 6 ) 180 rad )( (128 cm) . 14 ( = ° ° π c) m. 1.05 rad) m)(0.70 50 . 1 ( = 9.2: 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 9.3: a) . s rad 42 s, 3.5 at so , ) s rad (12.0 2 3 = = = = α t t dt dω 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. 9.4: a) . ) s rad 60 . 1 ( 2 t) ( 3 t βt α dt dω z z- =- = = b) . s rad 4.80 s) . 3 )( s rad 60 . 1 ( s) . 3 ( 2 3- =- = z α , s rad 40 . 2 s . 3 s rad 00 . 5 s rad 20 . 2 s . 3 ) ( s) . 3 ( 2. av- =-- =- =- ω ω α z which is half as large (in magnitude) as the acceleration at s. 3.0 = t 9.5: a) 2 3 2 ) s rad (0.036 s) rad 400 . ( 3 t βt γ ω z + = + = b) = = = γ ω t z , At s. rad 70 . so rad, 3.50 s, rad 3 . 1 s, 5.00 At c) s. rad 400 . 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. 9.6: -- =-- = ) s rad . 40 ( , ) s rad 50 . 4 ( ) s rad . 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 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 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) . , Setting b) . 6 2 and 3 2 3 2 c b z dt dw z dt dθ 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 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 ) ( = 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 + . (c) We have rad. 7.00 rad 49.0 rad . 42 s) (7.00 ) s rad 00 . 2 ( s) (7.00 s) rad 00 . 6 ( 2 2 2 1 2 2 1 + = +- = +- + = 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 = = = = =- ω t ω θ ω rpm 300 s rev 5.00 gives 2 = = + =- 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, Then s rev 1111 . gives 2 2 = θ- θ + =- = + = = =- = α =- = α + = t ω ω θ...
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap09 - 9.1: a) . 34.4 rad 60 . m 50 2 m 50 . 1 ° = = ....

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