Answer:
[tex]v_1 = 3.5 \ m/s[/tex]
Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e [tex]I_G[/tex] = 875 kg.m²
We know from the conservation of angular momentum that:
[tex]H_1= H_2[/tex]
[tex]mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2[/tex]
[tex]2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2[/tex]
[tex]1637.1 v_1[/tex] [tex]= 3841.575 \omega_2[/tex]
[tex]\omega_2 = \frac{1637.1 v_1}{3841.575}[/tex]
[tex]\omega _2 = 0.4626 \ v_1[/tex]
From the conservation of energy as well;we have :
[tex]T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0[/tex]
[tex][\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0[/tex]
[tex]706.93 \ v_1^2 - 8657.49 =0[/tex]
[tex]706.93 \ v_1^2 = 8657.49[/tex]
[tex]v_1^2 = \frac{8657.49}{706.93 }[/tex]
[tex]v_1 ^2 = 12.25[/tex]
[tex]v_1 = \sqrt{ 12.25[/tex]
[tex]v_1 = 3.5 \ m/s[/tex]
