CAn anyone help me with my physics formulas and help with the questions, thanks?

Question

The specifications are, Assuming a designers requires a motor veichle to accelerate its 670mm diamater drive wheels from a standing start to 40rads/sec in 6 seconds. If the mass is 1500kg the friction coefficient between road and tyres is 0.35 and taking its acceleration due t gravity as 10m/s/s.

Help with these question:
1) The angular acceleration of the wheels to attain cruising speed.
2)The forward velocity of the veichle when up to cruising speed.
3) The distance travelled to attain this speed.
4) The total load required to be produced by the engine.
5) The power developed by the engine in getting up to speed.

Thanks, full points given to the best.

Answer 1

Note: The friction is only relevant if the tires of the vehicle slip. See below where this is tested.
Let's test to see if nonslipping condition is satisfied:

The angular speed is related to angular acceleration as
w(t)=w0+a*t
where a is the angular acceleration and w is angular velocity
in this case
w(6)=40=0+a*6
a=40/6 rad/sec^2

Now that we know a, we can see if this a is great enough to cause the wheels to slip.
The force gets computed as F=m*a
F=1500*.670*40/6 N
F=6700 N

The normal force of the vehicle is
1500*10*
or 15000
and the maximum friction without slipping is
15000*.35
or 5250 N

The wheels do in fact slip under this acceleration
so this isn't even possible to accelerate the car this fast since the greates force that can be applied at any given time is 5250 N, which is ot enough to achieve the above.

However, if friction were great enough to hold then:

2. translational velocity is related to angular speed as
v=w*r
in this case
v=40*.670

3. The total angle traversed by the wheels is related by
th=th0+w0*t+.5*a*t^2
where w and a are as above and th is the angle w/r/t starting vertical
since th0 and w0 are 0, and t=6, and a=40/6
then
th=.5*(40/6)*6^2
or
th=40*6/2

(Note that this th units is in radians)

since th*r=s
then
.670*40*6/2 is the distance the motor vehicle has traveled.

4. Force is the load developed by the engine
since F=m*a
and alpha*r=a
then
F=1500*.670*40/6 N

5. Average power is F*v, since the vectors line up and the average v is
.670*40/2
then multiply by the Force to get power

j

Answer 2

hold please

1) a=dw/dt
dw= 40rad/s dt= 6 seconds

so a=40/6 rad/s

2) v=w(cross)r =v r sin (theta) = v r since theta is 90 degrees.

w= 40 rad/s r= 670 /2 mm

so v = 13400 mm/s (linear velocity)

3) from position equation:
y=at^2 + vot + yo with vo=0 and yo=0\
(a) so y= at^2

and from velocity equation:
v=at + vo with vo =0
(b) so t = a/v

solve for a from angular velocity
you get a=40/6 rad/s * 670/2 mm = 6700/3 mm/ s^2
plug this value for a into eqtn (b) then plug this value into (a)

this gives you y

4) sum of Forces = ma = Fe - Ff
with Fe=force(load) of engine and
Ff = force of friction = umg where u- coef. friction
and a = acceleration of vehicle.

you get ma + umg = Fe

so Fe=m(a+ ug) then plug in value of a, u, m, and g

the guy above is wrong, he didnt account for friction.

5) Work = F dx
and Power = Work / time

So Power= F dx/dt
plug in F from 4) and velocity (dx/dt) from 2)

-