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Naya [18.7K]
3 years ago
14

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

gular velocity when it strikes a small curb. Assume no rebound of the right-side tires and estimate the minimum speed v1 which will cause the vehicle to roll completely over to its right side. The mass of the SUV is 2140 kg and its mass moment of inertia about a longitudinal axis through the mass center G is 875 kg·m2. Assume d = 895 mm, h = 765 mm.
Calculate v1.
Physics
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

v_1  = 3.5 \ m/s

Explanation:

Given that :

mass of the SUV is  = 2140 kg

moment of inertia about G , i.e I_G = 875 kg.m²

We know from the conservation of angular momentum that:

H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

\omega_2 = \frac{1637.1 v_1}{3841.575}

\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

[\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

706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

v_1^2  =  \frac{8657.49}{706.93 }

v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

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