Answer:
2 revolutions.
Explanation:
We need to obtain the angular velocity when he is in the air.
The angular momentum is given by:
[tex]L_1=I_1*\omega_1\\L_1=1.7kg.m^2*0.7rev/s*\frac{2\pi rad}{1rev}=7.5kg.m^2/s[/tex]
Because of angular momentum conservation L1=L2, so the final angular velocity is given by:
[tex]\omega_2=\frac{L1}{I_2}=\frac{7.5kg.m^2/s}{0.7kg.m^2}=10.7rad/s[/tex]
We need to calculate the time the skater is in the air, so we need to use the formula of parabolic motion:
[tex]y=y_o+v_y*t-\frac{1}{2}*g*t^2\\0=0+8.8m/s*sin(45^o)*t-4.9m/s^2*t^2\\0=6.22t-4.9t^2\\solving:\\t=1.26s\\t=0[/tex]
so the time taken is 1.26s
the angular displacement is given by:
[tex]\theta=\omega*t\\\theta=13.5rad[/tex]
the number of revolutions is given by:
[tex]rev=\frac{\theta}{2\pi}=\frac{13.5}{2\pi}=2.1rev[/tex]
the skater execute two complete revolutions in the air.
