Rotational energy on an incline

[ZachAW]

Homework Statement

A cart whose body has mass M = 1.5 kg is set on four tires each of which has mass m = 0.3 kg and radius r = 0.1 meters. Each tire can be treated as a solid disk with rotational inertia mr2/2. The cart is set on an incline a height h = 1.2 meters high and released. At the bottom, the cart runs into a spring whose force constant is 600 N/m. What is the speed of the cart when it reaches the bottom of the incline? How much is the spring compressed when the cart comes to rest?

Homework Equations

Mgh = 1/2*MV^2 + 1/2*Iω^2
U(s) = 1/2*Kx^2
I = 4(m*r^2)

The Attempt at a Solution

a)
I = 4(0.3*0.1^2)
I = 0.006 Kg m^2

2*Mgh = V^2 (M + I/r^2)
V = √(2*Mgh)/(M + I/r^2)
V = √(2*1.5*9.8*1.2)/(1.5 + 0.006/0.1^2)
V = √(35.28)/(2.1)
V = 4.09 m/s

b)
Mgh = 1/2*MV^2 + 1/2*Iw^2 + 1/2*Kx^2
35.28 = 12.546 + 5.01843 + 250x^2
x = √(17.71557)/(250)
x = 0.26 m

So what did I do wrong?

 

[BvU]

Hello Zach, welcome to PF

A clear and concise post. Not all that verbose on the attempt at solution, but it looks as if you know what you are doing.
So just a quick hint and you'll be fine, I hope:

I see that you fill in 1.5 kg for M. But it's not just the cart body that's losing potential energy ! Nor is it just the cart body that has kinetic energy from linear motion: a moving wheel has kinetic energy also if it doesn't rotate !

And in b) the problem formulation "when the cart comes to rest" should help you see what you can do to improve that answer

I sense a little snag there: If your problem statement means the spring is along the incline, the cart will lose some more potential energy, so it becomes a bit nastier equation than when the spring is horizontal.

 

[ZachAW]

I see that you fill in 1.5 kg for M. But it's not just the cart body that's losing potential energy ! Nor is it just the cart body that has kinetic energy from linear motion: a moving wheel has kinetic energy also if it doesn't rotate !.

so M is the mass of the moving system, which means M is 1.5 + 4(0.3) = 2.7 Kg
V = √(2*2.7*9.8*1.2)/(2.7 + 0.006/0.1^2)
V = √(63.504)/(3.3)
V = 4.38 m/s
And for part b I think you should see the diagram for better understanding

 

[BvU]

Sure helps ! So it's the easy way. Now think of $v_{\rm final}$ and $\omega_{\rm final}$