Answer:
a). Angular acceleration 32 [tex]\frac{rev}{min^{2} }[/tex]
b). Final angular speed in rpm = 38.4 rpm
Explanation:
To calculated angular speed the distance is give 23 revolutions and time of 1.2 minutes so using equations:
a).
[tex]S_{f}= S_{o} + W_{o}*t +\frac{1}{2} * a * t^{2} \\W_{o}= 0 \frac{rev}{min}\\ S_{o}= 0 rev\\S_{f}= \frac{1}{2} * a * t^{2}\\a= \frac{S_{f}*2}{t^{2} }\\ a= \frac{23rev*2}{1.2min^{2} }=\frac{46 rev}{1.44 min^{2} } \\a= 31.94 \frac{rev}{min^{2} } \\[/tex]
a ≅ 32 [tex]\frac{rev}{min^{2} }[/tex]
b).
[tex]w_{f} = w_{o}+ a*t\\w_{f} = 0+ 3.2 \frac{rev}{min^{2} } *1.2 min\\w_{f} = 38.4 \frac{rev}{min} = 38.4[/tex]rpm
Comprobation:
Using a different equation but replacing a= 32 [tex]\frac{rev}{min^{2} }[/tex], [tex]S_{f}= 23 rev[/tex], [tex]V_{o}= 0 rev[/tex], [tex]v_{f}= 38.4 rev[/tex]
[tex]w_{f} ^{2} =w_{o} ^{2}+2*a (S_{f} -S_{o})\\38.4 ^{2} =0^{2}+2*32 (23 -0)\\38.4=\sqrt{2*32*23} \\38.4=\sqrt{1472} \\38.4=38.4[/tex]
