Answer:v=41.23 m/s
Explanation:
Given
mass of heavy object [tex]m_1=52 kg[/tex]
distance of [tex]m_1[/tex] from the axle [tex]r_1=14 cm[/tex]
mass of rock [tex]m_2=123 gm[/tex]
Length of rod [tex]=4.1 m [/tex]
distance of [tex]m_2[/tex] from axle [tex]r_2=4.1-0.14=3.96 m[/tex]
Net torque acting is
[tex]T_{net}=m_1gr_1-m_2gr_2[/tex]
[tex]T_{net}=52\times 0.14\times g-0.123\times 3.96\times g[/tex]
[tex]T_{net}=6.793\times 9.8[/tex]
[tex]T_{net}=66.57 N-m[/tex]
Work done by [tex]T_{net}[/tex] is converted to rock kinetic Energy
thus
[tex]T_{net}\times \theta =\frac{mv^2}{2}[/tex]
Where [tex]\theta =angle\ turned =\frac{\pi }{2}[/tex]
[tex]v= velocity\ at\ launch[/tex]
[tex]66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}[/tex]
[tex]v^2=66.57\times \pi [/tex]
[tex]v=\sqrt{1700.511}[/tex]
[tex]v=41.23 m/s[/tex]