Acceleration of Cart Rolling Down Incline

In summary: I'm a little rusty; it's been a couple of years.)Thanks,DorothyIn summary, the conversation discusses the motion of a cart rolling down an incline with two wheels, each with a mass of m and a moment of inertia of mR^2/2. The total mass of the cart and wheels is M. Using conservation of energy and assuming no friction between cart and axles, the acceleration of the cart along the incline is shown to be a = [M/(M + 2m)] g sin theta. There is some confusion about the inclusion of a rolling friction term in the solution, but it is ultimately determined to not be necessary. There is a possibility that the book may have a
  • #1
Dorothy Weglend
247
2

Homework Statement


A cart is rolling down an incline (angle to horizontal is theta). It has two wheels, each wheel has mass m and moment of inertia mR^2/2. The total mass of the cart and the two wheels is M. Using conservation of energy (assuming no friction between cart and axles), show that the acceleration of the cart along the incline is a = [M/(M + 2m)] g sin theta.


Homework Equations


K = Iw^2/2
K = mv^2/2
v = wR (w is angular velocity).
v^2 = 2as

The Attempt at a Solution



E0 = E1
Ug = 2(K_rolling) + K_trans

Letting d be the distance traveled on the incline, this gives:

Mgd sin(theta) = (Iw^2/2)2 + Mv^2/2 = Iw^2 + Mv^2/2

Using the given moment of inertia I = mR^2/2:

Mgd sin(theta) = (mR^2/2)w^2 + Mv^2/2

Mgd sin(theta) = mv^2/2 + Mv^2/2

v^2 = 2Mgd sin(theta)/(M + m)

v^2 = 2ad:

a = Mg sin(theta)/(M + m)



I can't figure out where they got the 2m from.

I've tried adjusting the height by the radius of the wheels, but that gets complicated, because there are two of them, at two different heights, and I can't believe that would lead to the formula they give.

When you all recover from New Years, I'd appreciate any suggestions :wink:
Dorothy

(This is from Serway and Jewett, btw, 10.89)
 
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  • #2
Consider motion starting from rest over distance [tex]x[/tex] along the incline:

[tex]Mgx\sin\theta = \frac{1}{2}Mv^2 + 2(\frac{1}{2}mR^2)(\frac{v}{R})^2[/tex]

[tex]2Mgx\sin\theta = (M+2m)v^2[/tex]

On the other hand

[tex]v^2 = 2ax[/tex]

[tex]2Mgx\sin\theta = (M+2m)2ax[/tex]

[tex]a = \frac{Mg}{M+2m}\sin\theta[/tex]
 
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  • #3
phucnv87 said:
Consider motion starting from rest over distance [tex]x[/tex] along the incline:

[tex]Mgx\sin\theta = \frac{1}{2}Mv^2 + 2(\frac{1}{2}mR^2)(\frac{v}{R})^2[/tex]

Ok, but I don't understand how you got that rolling friction term.

Why isn't it

[tex]2(\frac{1}{2})(\frac{1}{2}mR^2)(\frac{v}{R})^2[/tex]

Isn't the KE of rolling friction

[tex] (\frac{1}{2}){I\omega^2}[/tex]

Why would it be different in this case?

Thanks,
Dorothy
 
  • #4
The condition states that the cart is rolling, which I suppose means that it's rolling without slipping. The condition for rolling without slipping is:

[tex]v=R\omega[/tex]
 
  • #5
Dorothy, I agree with your solution. Is there a possibility that the book is wrong?
 
  • #6
thequirk said:
The condition states that the cart is rolling, which I suppose means that it's rolling without slipping. The condition for rolling without slipping is:

[tex]v=R\omega[/tex]

Hi Quirk... Yes, I understand that bit. What I don't understand is what happened to the 1/2 in Iw^2/2, or if you prefer, Iv^2/2R.
 
  • #7
Hi Hawire,

That was my thought, yes, that the book was wrong. But I've thought that before (smile) and it wasn't. Just want to make sure I understand all of this.
 

What is acceleration?

Acceleration is the rate at which the velocity of an object changes over time. It is a vector quantity, meaning it has both magnitude and direction. It can be calculated by dividing the change in velocity by the change in time.

How does an object's mass affect its acceleration?

According to Newton's second law of motion, the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. This means that a heavier object will have a smaller acceleration than a lighter object when the same force is applied.

What factors affect the acceleration of a cart rolling down an incline?

The acceleration of a cart rolling down an incline is affected by the incline angle, the mass of the cart, and the force of gravity. The steeper the incline, the greater the acceleration. The greater the mass of the cart, the slower the acceleration. And the greater the force of gravity, the faster the acceleration.

How is the acceleration of a cart rolling down an incline calculated?

The acceleration of a cart rolling down an incline can be calculated using the formula a = gsinθ, where a is the acceleration, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the incline. This formula assumes that there is no friction present.

How is the acceleration of a cart rolling down an incline affected by friction?

Friction can act in the opposite direction of the motion of the cart, slowing down its acceleration. The amount of friction depends on the surface of the incline and the surface of the cart. A rougher incline or a cart with less friction on its wheels will experience a greater decrease in acceleration due to friction.

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