Question #e6799

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Answer 1 · 2018-02-22T15:56:50

See the explanation below

Apply the Equations of motion (rotational)

$omega=omega_0+alphat$ .......... $(1)$

$Deltatheta=omega_0+1/2alphat^2$ .............. $(2)$

The initial angular velocity is $omega_0=?$

The time is $t=4s$

The angle is $Deltatheta=162rad$

The final angular velocity is $omega=108rads^-1$

Substituting those values in equations $(1)$ and $(2)$ and solving for $omega_0$

$Deltatheta=omega_0+1/2*(omega-omega_0)/2*t^2$

$162=omega_0+1/4*(108-omega_0)*16$

$162=omega_0+432-16omega_0$

$15omega_0=432-162=270$

$omega_0=270/15=18rads^-1$

The angular acceleration is

$alpha=(omega-omega_0)/t=(108-18)/4=22.5rads^-2$

The tangential acceleration is

$a_("tangential")=r*alpha=0.12*22.5=2.7ms^-2$

The velocity is tangential $v=omegar$ and the angle is $=0^@$

The total acceleration is

$a=sqrt(a_T^2+a_C^2)$

The centripetal acceleration is $=a_C$

And

$tanphi=a_T/a_C$