Acceleration of center of mass of cart-block system

• esoteric deviance
In summary: CM equations are a good start. You don't have Fnet given, adn you will not be able to find it unless you find the tension in the string. You can do that, but you don't have to. The two masses share a common distance of motion, which means their velocities and accelerations must have the same magnitude. Those magnitudes can be found without actually calculating the tension. You know the directions. From your CM position equations you can come up with the CM velocity and CM acceleration equations. So what you're saying is:(change in d1) = (change in d2), v1 = v2, and a1 = a
esoteric deviance
http://www.fileden.com/files/2006/11/15/381656/My%20Documents/problem%2013.jpg
The figure above shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block.
The cart has mass m1 = 0.600 kg, and its center is initially at xy coordinates (-0.500 m, 0 m).
The block has mass m2 = 0.400 kg, and its center is initially at xy coordinates (0, 0.100 m).
The cart is released from rest, and both cart and block move until the cart hits the pulley.
The friction between the cart and air track and between the pulley and its axle is negligible as is the mass of the cord and pully.

• In unit-vector notation, what is the acceleration of the center of mass of the cart-block system?
• What is the velocity of the com as a function of time t?

so..i know that acom = Fnet/M and that i probably have to find acom,x and acom,y, but I'm not sure and if i do need to solve for those, how do i find the Fnet?

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I think you are making it too complicated. The internal forces will cancel, by the 3rd law, right? Fnet would just be the external forces.

Dorothy

Still not sure what to do, sorry :(.

The only forces are the tension on the cord, and gravity, as far as I can tell. The tension on the cord is an internal force, which just cancels out by Newton's third law. So Fnet has got to have something to do with gravity, it would seem.

so if the only external force acting on the system is gravity, then
Fnet,system = Mg = (m1 + m2)g = 9.8?

esoteric deviance said:
so if the only external force acting on the system is gravity, then
Fnet,system = Mg = (m1 + m2)g = 9.8?
There is another external force besides gravity. You don't need to find it to do the problem, but you could. It is the force of the pulley acting on the string. You can continue to find the motions of the individual masses and then find the motion of the CM from those. But you have too many g in your equation, and something is missing.

Yeah, I guess I'm pretty lost. For some reason, the more I try to think about this problem, the more confused I seem to become :(.
I hate to ask this, but might you be willing to give me a step-by-step explanation of how to solve it?

I don't know if they have any significance (or if they are even correct), but here are some values I've come up with so far:
• x1 = -0.5, y1 = 0
• x2 = 0, y2 = -0.1
• xcom,sys = (m1x1 + m2x2)/(M) = -0.3
• ycom,sys = (m1y1 + m2y2)/(M) = -0.04
and a couple of the equations I've been trying to work with:
• Fnet = Macom
• Macom = m1a1 + m2a2

esoteric deviance said:
Yeah, I guess I'm pretty lost. For some reason, the more I try to think about this problem, the more confused I seem to become :(.
I hate to ask this, but might you be willing to give me a step-by-step explanation of how to solve it?

I don't know if they have any significance (or if they are even correct), but here are some values I've come up with so far:
• x1 = -0.5, y1 = 0
• x2 = 0, y2 = -0.1
• xcom,sys = (m1x1 + m2x2)/(M) = -0.3
• ycom,sys = (m1y1 + m2y2)/(M) = -0.04
and a couple of the equations I've been trying to work with:
• Fnet = Macom
• Macom = m1a1 + m2a2
The initial position and CM equations are a good start. You don't have Fnet given, adn you will not be able to find it unless you find the tension in the string. You can do that, but you don't have to. The two masses share a common distance of motion, which means their velocities and accelerations must have the same magnitude. Those magnitudes can be found without actually calculating the tension. You know the directions. From your CM position equations you can come up with the CM velocity and CM acceleration equations.

so what you're saying is:
(change in d1) = (change in d2), v1 = v2, and a1 = a2?

the equations for CM of velocity and acceleration:
vcom = (m1v1 + m2v2)/(M)
acom = (m1a1 + m2a2)/(M)

i don't know what to do...if i set the velocities and accelerations equal to each other, all i end up with is vcom = v and acom = a.

esoteric deviance said:
so what you're saying is:
(change in d1) = (change in d2), v1 = v2, and a1 = a2?

the equations for CM of velocity and acceleration:
vcom = (m1v1 + m2v2)/(M)
acom = (m1a1 + m2a2)/(M)

i don't know what to do...if i set the velocities and accelerations equal to each other, all i end up with is vcom = v and acom = a.

The vectors are not equal. The magnitudes of the vectors are equal. The horizontal distance m1 moves is equal to the vertical distance m2 falls

What is the definition of "acceleration of center of mass"?

The acceleration of center of mass is a measure of how quickly the center of mass of a system is changing its velocity. It is a vector quantity that takes into account both the magnitude and direction of the change in velocity.

How is the acceleration of center of mass calculated for a cart-block system?

The acceleration of center of mass for a cart-block system is calculated by taking into account the mass of both the cart and the block, and the forces acting on them. The formula is a = (m1a1 + m2a2) / (m1 + m2), where m1 and m2 are the masses of the cart and block respectively, and a1 and a2 are their respective accelerations.

What factors can affect the acceleration of center of mass in a cart-block system?

The acceleration of center of mass in a cart-block system can be affected by the mass of the cart and block, the forces acting on them, and any external factors such as friction or air resistance. Additionally, the position and direction of the forces can also impact the acceleration of the system.

What is the relationship between net force and acceleration of center of mass?

The acceleration of center of mass is directly proportional to the net force acting on a cart-block system. This means that if the net force increases, the acceleration of the center of mass will also increase. Similarly, if the net force decreases, the acceleration of the center of mass will decrease.

How does the acceleration of center of mass relate to Newton's Second Law of Motion?

The acceleration of center of mass is directly related to Newton's Second Law of Motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In the case of a cart-block system, the net force is equal to the combined mass of the cart and block multiplied by their acceleration of center of mass.

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