- #1

Heyxyz

- 8

- 3

- Homework Statement
- A toy car and a solid metal sphere are rolled down an incline, why does the car reach the bottom first?

- Relevant Equations
- KE (total) = KE + KE (rotational) = 1/2mv^2 + 1/2Iw^2

Sphere: I = 2/5mr^2

v = wr

w = omega

Hello,

I tried to put it in an equation, but it didn't really work out. In this situation, the car was about the size of a model, and, while not exact, the radius of each wheel couldn't have been more than like a centimeter. Conversely, the ball was like twice the size of the car and had a diameter of 10 - 15 cm.

Ball:

=1/2mv^2 + 1/2I(V/R)^2, I = 2/5mr^2

= 1/2mv^2 + (1/2)(2/5)(m)(r^2)(v^2/r^2)

=7/10mv^2

Car:

Since wheels are practically cylinders, I figured I = 1/2mr^2. There are four wheels, so 4 * 1/2mr^2 = 2mr^2

=1/2mv^2 + 1/2I(V/R)^2, I = 2mr^2

= 1/2mv^2 + 2m(r^2)(v^2/r^2)

=1/mv^2 + mv^2

= 3/2mv^2I don't understand where I'm going. If I follow my above equations, it would appear that the ball reaches the bottom sooner due to the smaller coefficient, but I know that isn't true. Maybe the smaller radius of each wheel has something to do with it?

Thank you.

I tried to put it in an equation, but it didn't really work out. In this situation, the car was about the size of a model, and, while not exact, the radius of each wheel couldn't have been more than like a centimeter. Conversely, the ball was like twice the size of the car and had a diameter of 10 - 15 cm.

Ball:

=1/2mv^2 + 1/2I(V/R)^2, I = 2/5mr^2

= 1/2mv^2 + (1/2)(2/5)(m)(r^2)(v^2/r^2)

=7/10mv^2

Car:

Since wheels are practically cylinders, I figured I = 1/2mr^2. There are four wheels, so 4 * 1/2mr^2 = 2mr^2

=1/2mv^2 + 1/2I(V/R)^2, I = 2mr^2

= 1/2mv^2 + 2m(r^2)(v^2/r^2)

=1/mv^2 + mv^2

= 3/2mv^2I don't understand where I'm going. If I follow my above equations, it would appear that the ball reaches the bottom sooner due to the smaller coefficient, but I know that isn't true. Maybe the smaller radius of each wheel has something to do with it?

Thank you.