Respuesta :
Answer:
1) [tex]\sum T_{externalhoop}=0.4418Nm[/tex]
2) [tex]\sum T_{externaldisc}=0.2209Nm[/tex]
Explanation:
Using the second equation of angular motion we have
[tex]\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}[/tex]
Since the wheels start from rest we ahve [tex]omega _{o}=0[/tex]
Applying the given values in the equation we have
[tex]14=\frac{1}{2}\alpha \times 8^{2}\\\\\therefore \alpha =\frac{28}{64}=0.4375rad/s^{2}[/tex]
Now by newton's second law of motion in angular motion we have
[tex]\sum T_{external}=I\alpha[/tex]
1) For Hoop We have
[tex]I_{hoop}=Mr^{2}\\\\\therefore I_{hoop}=4.4\times(0.48)^{2}=1.01376kgm^{2}[/tex]
Thus
[tex]\sum T_{external}=1.01376\times 0.4375[/tex]
[tex]\sum T_{external}=0.4418Nm[/tex]
2)For disc We have
[tex]I_{disc}=\frac{Mr^{2}}{2}\\\\\therefore I_{hoop}=2.2\times(0.48)^{2}=0.506kgm^{2}[/tex]
Thus
[tex]\sum T_{external}=0.506\times 0.4375[/tex]
[tex]\sum T_{external}=0.2209Nm[/tex]
