Answer to Question #239597 in Mechanical Engineering for Kem

Question

An electrically driven road vehicle of mass 450kg has four wheels of effective diameter 400mm radius of gyration 125mm and mass 9kg. The armature of the electric motor has a mass of 65kg a radius of gyration of 100mm and rotates at 4 times the speed of the road wheels. The rolling resistance to motion of the vehicle is to be assumed constant at 225N. Determine the acceleration of the vehicle if it is allowed to run freely down a slope whose inclination to the horizontal is sin-¹ 0.2

Answer 1

m_v=450kg

d_w=0.4m

k_w=0.125m

m_w=9kg

m_a=65kg

k_a=0.1m

Rotates 4 times speed of road wheels

F_r=225N

a=?

From energy balance;

rate of P.E loss of car = rate of rotational K.E gain of armature + rate of rotational K.E gain of wheels + rate of K.E gain of car - rate of friction dissipation

Rate of P.E loss of car = m_v*g*vSin$\theta$

Rate of rotational K.E gain of armature = I_a*w_a*w_a'

Rate of rotational K.E gain of wheels = I_w*w*w'

Rate of K.E gain of car = m_v*v*v''

Rate of friction dissipation = F*v

0.2*m_v*g*v=64*I_a*(v)$\frac{(a)}{d^{2}}$ +4*I_w*(v)$\frac{(a)}{d^{2}}$ +m_v*v*a-F*v

0.2*m_v*g=64*I_a*$\frac{(a)}{d^{2}}$ +4*I_a*$\frac{(a)}{d^{2}}$ +m_v*a-F

0.2*450*9.81=a(64*65*$\frac{0.1^{2}}{0.4^{2}}$ +450)-225

a=$\frac{1107.9}{710}$

=$1.5604m/s^{2}$