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= 120×518=33.33ms\times \frac{5}{18}=33.33\frac{m}{s} ,

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

we take inital position of particles as

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

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

and

acceleration

a=r(ω2cosθ+αsinθ)i+r(ω2sinθ+αcosθ)ja=-r(\omega^2 cos\theta+\alpha sin\theta){i}+r(\omega^2sin\theta+\alpha cos\theta){j}

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

vi=v_i= rωir\omega {i} , ωi=33.3370=0.476rad/s\omega_i=\frac{33.33}{70}=0.476 rad/s

Now, after time t= 4 ,

ωf=16+33.3370=0.7047rad/s\omega_f= \frac{16+33.33}{70}=0.7047 rad/s


α=ωfωi4=0.057rads2\alpha=\frac{\omega_f-\omega_i}{4}=0.057 \frac{rad}{s^2}

this is angular acceleration of car

θ=ωit+0.5αt2=0.476×4+0.5×0.057×16=135.286o\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,

Vf=70×.7047sin(135.286)i+70×0.7047cos(135.286)jV_f=-70\times .7047 sin(135.286){i}+70\times 0.7047 cos (135.286){j}


Vf_f =34.706i35.054j-34.706 {i} -35.054{j}


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


Now acceleration of particles will be

a=70(0.70472cos(135.285)+0.057sin(135.286)i+70(0.70472sin(135.286)+0.057cos(135.286))ja=-70(0.7047^2 cos(135.285)+ 0.057sin(135.286){i}+70(0.7047^2sin(135.286)+0.057cos (135.286)){j}

a=-21.89 i- 21.89j

a=30.96 m/s2


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