Answer to Question #162708 in Mechanical Engineering for loy

Question #162708

A race car enters the circular portion of a track that has a radius of 70 m. When the car enters the curve at point P, it is traveling with a speed of 120 km/h that is increasing at 4 m/s2 . Four seconds later, determine the x and y components of velocity and acceleration of the car.


1
Expert's answer
2021-02-15T05:40:23-0500

Radius of car= 70m,travelling with speed=120 km/h= 120"\\times \\frac{5}{18}=33.33\\frac{m}{s}" ,

acceleration=4m/s2, time=t=4s,

we take inital position of particles as

"x=rcos\\theta {i} +rsin\\theta {j},"

"v=-r\\omega sin\\theta{i}+r\\omega cos\\theta{j}"

and

acceleration

"a=-r(\\omega^2 cos\\theta+\\alpha sin\\theta){i}+r(\\omega^2sin\\theta+\\alpha cos\\theta){j}"

Now, we know that initially angle "\\theta = 0^o"

"v_i=" "r\\omega {i}" , "\\omega_i=\\frac{33.33}{70}=0.476 rad\/s"

Now, after time t= 4 ,

"\\omega_f= \\frac{16+33.33}{70}=0.7047 rad\/s"


"\\alpha=\\frac{\\omega_f-\\omega_i}{4}=0.057 \\frac{rad}{s^2}"

this is angular acceleration of car

"\\theta=\\omega_i t+0.5\\alpha t^2=0.476\\times 4+0.5\\times 0.057 \\times 16=135.286 ^o"

this is the angular dispalcement of car after 4 second

Now,

"V_f=-70\\times .7047 sin(135.286){i}+70\\times 0.7047 cos (135.286){j}"


V"_f" ="-34.706 {i} -35.054{j}"


So here x component of velocity will be "v_x=" -34.706 m/s and y component of velocity is "v_y=" 35.054 m/s


Now acceleration of particles will be

"a=-70(0.7047^2 cos(135.285)+ 0.057sin(135.286){i}+70(0.7047^2sin(135.286)+0.057cos \n (135.286)){j}"

a=-21.89 i- 21.89j

a=30.96 m/s2


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