Answer:
Explanation:
Given
radius of circular path [tex]r=4\ in.[/tex]
Position is given by
[tex]\theta =\cos 2t---1[/tex]
Differentiate 1 to angular velocity we get
[tex]\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2[/tex]
Differentiate 2 to get angular acceleration
[tex]\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3[/tex]
Net acceleration is the vector summation of tangential and centripetal force
[tex]a_t=\alpha \times r[/tex]
[tex]a_t=-4\cos 2t\times 4=-16\cos 2t[/tex]
[tex]a_r=\omega ^2\cdot r[/tex]
[tex]a_r=(-2\sin 2t)^2\cdot 4[/tex]
[tex]a_r=16\sin^2(2t)[/tex]
[tex]a_{net}=\sqrt{a_r^2+a_t^2}[/tex]
[tex]a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}[/tex]
[tex]a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}[/tex]
