Answer:
1.10 m/s
Explanation:
Linear speed is given by
[tex]v=r\omega[/tex]
Kinetic energy is given by
[tex]KE=0.5I\omega^{2}[/tex]
Potential energy
PE= mgh
From the law of conservation of energy, KE=PE hence
[tex]0.5I\omega^{2}=mgh[/tex] where m is mass, I is moment of inertia, [tex]\omega[/tex] is angular velocity, g is acceleration due to gravity and h is height
Substituting m2-m1 for m and 0.5l for h, [tex]\frac {2v}{L}[/tex] for [tex]\omega[/tex] we obtain
[tex]0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)[/tex]
[tex] (\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I}[/tex] and making v the subject
[tex]v^{2}=\frac {gl^{3}(m2-m1)}{4I}[/tex]
[tex]v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}[/tex]
For the rod, moment of inertia [tex]I=\frac {ML^{2}}{12}[/tex] and for sphere [tex]I=MR^{2}[/tex] hence substituting 0.5L for R then [tex]I=M(0.5L)^{2}[/tex]
For the sphere on the left hand side, moment of inertia I
[tex]I=m1(0.5L)^{2}[/tex] while for the sphere on right hand side, [tex]I=m2(0.5L)^{2}[/tex]
The total moment of inertia is therefore given by adding
[tex]I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}[/tex]
Substituting [tex]\frac {L^{2}(M+3m1+3m2)}{12}[/tex] for I in the equation [tex]v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}[/tex]
Then we obtain
[tex]v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}} [/tex]
This is the expression of linear speed. Substituting values given we get
[tex]v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s[/tex]
