Muscle Head and Pip Squeak?

Question

Suppose you had a car that was mostly wheels and another car that had tiny wheels. if the cars have the same total mass and their center-of-masses are equal distances from the ground and each goes from zero to 40 mph in ten seconds, which one will nose up the most?

a) Muscle Head
b) Pip Squeak
c) Both the same

I'm glad the posts are stimulating discussion among everyone. That is my main purpose as well as to help students who have an interestand whoe enjoy challenging themselves. I also enjoy the different approaches or ways of thinking about each questionone and always learn something myself.

I'm glad the posts are stimulating discussion among everyone. That is my main purpose as well as to help students who have an interestand whoe enjoy challenging themselves. I also enjoy the different approaches or ways of thinking about each questionone and always learn something myself.

The answer is a. The wheels did not come into the NOSING CAR story at all -- except that they moved the car -- so you might expect that the size of the wheels makes no difference here. Even if the car had no wheels and was dragged on a sled it would still nose up when the sled accelerated!

Nevertheless you might have a gut feeling that the big-wheel buggy should nose up the most -- and you are right. So far, though, only half of the "nosing" story has been told. To visualize the other half, suppose the car was floating out in intersteller space and you hit the gas and made the wheels spin. What would the car do? Well, it certainly would not go anyplace. Why? Because there is no road to push on -- no friction! But it would not just do nothing. If the wheels were made to spin clockwise the rest of the car body would spin anti-clockwise. Why? Because this is the rotational counterpart to action and reaction and momentum conservation.

Now back on earth that effect still operates.

There are two effects which make the accelerating car nose up. One depends only on the car's acceleration and has nothing to do with the wheels. The second depends only on the spinning wheels and has nothing to do with the accleration. if the mass of the wheels is very mcuh less than the mass of the car only the first effect counts for much. But as the mass of the wheels becomes closer to the mass of the vehicle the second effect becomes more important.

Answer 1

A, assuming Muscle Head is the car with the huge wheels. Both cars experience identical nose-up torques due to the horizontal reaction forces from the road to the drive wheels applied at the road surface. These torques are equal to the road force * the vertical distance from the road to the CM.
In addition, because of Muscle Head's larger wheels, it experiences a larger reaction torque at the gearbox (differential) serving the drive wheels. This torque is in the same direction as the road-force torque, and is equal to the the angular acceleration of the wheel (= acceleration of car A / wheel radius r) * wheel moment of inertia I. For similar wheels with mass m concentrated near the radius, I/r increases as the 4th power of the wheel size (mr^2/r = mr ~ L^4) . So Muscle Head noses up more than Pip Squeak. And incidentally, Muscle Head must produce more power for the same acceleration because of the additional rotational energy of the wheels (~ m ~ L^3).
EDIT: I have to respnd to linlyons's conception of the front springs "absorbing" the nose-up tendency. If the downforce on the front springs is reduced, they will extend further and the nose will rise, whether the wheels remain on the ground or not. Also, a point about torque. Consider a pure torque, i.e., one generated by a force "couple", two equal and opposed but offset forces. As long as the offset (lever arm) and the forces remain the same, this torque has the same effect on the body it's applied to, no matter where the force couple is applied. In other words, if you could apply a torque through, let's say, a perfectly flexible, massless cable (sort of like a roto-rooter or a Dremel tool), you can apply it anywhere on the body and the effect is the same. So, considering torque only, front-wheel and rear-wheel effects are identical.
P.S. I don't intend this as argumentative or promoting my answer, it's just that Dr. H's posts seem to stimulate what I hope is helpful discussion.
Edit again: I rewrote the first part since it was wrong about the source of the additional large-wheel torque.

Answer 2

There are many other factors that affect the ability to nose up. If the body is the same and we can ignore the weight of the wheels, then correct answer is c.
The main factor affecting the ability to nose up is ratio of the distance to the center of mass from rear and front wheels. Let's show this. As a pivot point consider point at the road level just below the center of mass. For this point there are just 2 forces N1 and N2 - reaction forces on front and rear wheels - that have non-zero torque. Relative to this pivot point the car has tangential acceleration a, that is translated into angular acceleration A=a/H, where H is the height of the center of mass. This angular acceleration is created by torque T and is equal A=T/I, where I is moment of inertia. So T = aI/H. If distances to front and rear wheels are D1 and D2, then torque is equal to D2*N2 - D1*N1 = aI/H. The sum of reaction forces is equal to weight of the car: N1 + N2 = m*g. Substituting N1 = m*g - N2 into first equation we get D2*m*g - (D1+D2)*N1 = aI//H. N1 = (D2*mg - aI/H)/(D1 + D2).
When N1 becomes <=0 the car will nose up. This condition can be written as D2*m*g - aI/H <= 0, or a>=D2*m*g*H/I.
As you can see this expression does not include the wheel size but includes distance to rear wheels, height of the center of gravity and moment of inertia. We can express I = I0 + m*H^2, where I0 is moment of inertia relative to the center of mass, then a>=D2*m*g*H/(I0 + m*H^2) to see clearly that higher center of mass means less stability.