The angular position of a point on the rim of a rotating wheel isgiven by = 7.0

Question

The angular position of a point on the rim of a rotating wheel isgiven by = 7.0t -3.0t2 +t3, where is in radians andt is in seconds.
(a) What is the angular velocity at t = 4 s?
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(b)What is the angular velocity at t = 6.0 s?
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(c) What is the average angular acceleration for the time intervalthat begins at t = 4 s and endsat t = 6.0 s?
3
(d) What is the instantaneous angular acceleration at the beginningof this time interval?
4
(e)What is the instantaneous angular acceleration at the end ofthis time interval?

Explanation / Answer

Angular velocity is :            = d / dt                = d/dt (  7.0t -3.0t2 +t3 )                =   7.0 - 6.0 t + 3 t2 (a)       angular velocity att = 4 s   is:                = 7.0 - 6.0 (4) + 3* (4)2                    =  31 rev /s (b)       angular velocity att = 6.0 s is :                 =    7.0 - 6.0 (6.0) + 3*(6.0)2                    = 79 rev /s (c)       Average angular acceleration is:                   = d / d t                      = -6.0 + 6 t      at t = 4 s is :   -6.0 + 6*4 = 18 rev/s2      at t  = 6 s is :   -6.0 + 6*6 = 30 rev /s2    Average is :  ( 18 + 30 ) /2   = 24 rev /s2 (d)       instantaneous angularacceleration at the beginning of this time interval is :          at t = 4 s is :   -6.0 + 6*4 = 18 rev /s2 (e)          instantaneousangular acceleration at the end of this time interval is :            at t  = 6 s is :   -6.0 +6*6 = 30 rev /s2 Hope this helps u!                                           (a)       angular velocity att = 4 s   is:                = 7.0 - 6.0 (4) + 3* (4)2                    =  31 rev /s (b)       angular velocity att = 6.0 s is :                 =    7.0 - 6.0 (6.0) + 3*(6.0)2                    = 79 rev /s (c)       Average angular acceleration is:                   = d / d t                      = -6.0 + 6 t      at t = 4 s is :   -6.0 + 6*4 = 18 rev/s2      at t  = 6 s is :   -6.0 + 6*6 = 30 rev /s2    Average is :  ( 18 + 30 ) /2   = 24 rev /s2 (d)       instantaneous angularacceleration at the beginning of this time interval is :          at t = 4 s is :   -6.0 + 6*4 = 18 rev /s2 (e)          instantaneousangular acceleration at the end of this time interval is :            at t  = 6 s is :   -6.0 +6*6 = 30 rev /s2 Hope this helps u!                                                              =  31 rev /s (b)       angular velocity att = 6.0 s is :                 =    7.0 - 6.0 (6.0) + 3*(6.0)2                    = 79 rev /s (c)       Average angular acceleration is:                   = d / d t                      = -6.0 + 6 t      at t = 4 s is :   -6.0 + 6*4 = 18 rev/s2      at t  = 6 s is :   -6.0 + 6*6 = 30 rev /s2    Average is :  ( 18 + 30 ) /2   = 24 rev /s2 (d)       instantaneous angularacceleration at the beginning of this time interval is :          at t = 4 s is :   -6.0 + 6*4 = 18 rev /s2 (e)          instantaneousangular acceleration at the end of this time interval is :            at t  = 6 s is :   -6.0 +6*6 = 30 rev /s2 Hope this helps u!                      Average is :  ( 18 + 30 ) /2   = 24 rev /s2 (d)       instantaneous angularacceleration at the beginning of this time interval is :          at t = 4 s is :   -6.0 + 6*4 = 18 rev /s2 (e)          instantaneousangular acceleration at the end of this time interval is :            at t  = 6 s is :   -6.0 +6*6 = 30 rev /s2 Hope this helps u!                                          

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