Figure 9-44 shows an arrangement with an air track, in which...

Figure $9-44$ shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass $m_{1}=0.600 \mathrm{kg}$ and its center is initially at $x y$ coordinates $(-0.500$ $\mathrm{m}, 0 \mathrm{m} ) ;$ the block has mass $m_{2}=0.400 \mathrm{kg},$ and its center is initially at $x y$ coordinates $(0,-0.100 \mathrm{m}) .$ The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. In unit-vector notation, what is the acceleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function of time $t ?$ (c) Sketch the path taken by the com. (d) If the path is curved, determine whether it bulges upward to the right or downward to the left, and if it is straight, find the angle between it and the $x$ axis.

Figure $9-44$ shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass
$m_{1}=0.600 \mathrm{kg}$ and its center is initially at $x y$ coordinates $(-0.500$
$\mathrm{m}, 0 \mathrm{m} ) ;$ the block has mass $m_{2}=0.400 \mathrm{kg},$ and its center is initially at
$x y$ coordinates $(0,-0.100 \mathrm{m}) .$ The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move
until the cart hits the pulley. The friction between the cart and the air
track and between the pulley and its axle is negligible. In unit-vector notation, what is the acceleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function
of time $t ?$ (c) Sketch the path taken by the com. (d) If the path is
curved, determine whether it bulges upward to the right or downward
to the left, and if it is straight, find the angle between it and the $x$ axis.

Key Concepts

Center of Mass

The center of mass of a system is the weighted average position of all the mass in the system. It acts as if all the mass were concentrated at that point when analyzing the external forces and motion. Calculating the acceleration or velocity of the center of mass involves summing all external forces over the total mass, independent of the internal forces between the components.

Newton's Second Law for a System

Newton's second law states that the net external force acting on a system equals the total mass of the system times the acceleration of its center of mass. This principle applies to multi-object systems as well, meaning that when internal forces cancel, only external forces determine the motion of the center of mass.

System Analysis and Internal Forces Cancellation

When analyzing a system made up of multiple interconnected parts, it is important to identify which forces are internal and which are external. Internal forces (such as tension in a cord connecting two objects) do not affect the acceleration of the center of mass because they cancel out according to Newton’s third law. This simplifies the analysis by allowing one to consider only the net external forces.

Vector Representation and Unit Vectors

Expressing physical quantities like displacement, velocity, or acceleration in unit-vector notation allows for a clear representation of their direction and magnitude in a chosen coordinate system. This method is crucial for problems involving motion in multiple dimensions, where the decomposition of vectors into components simplifies the analysis.

Kinematics Under Constant Acceleration

Kinematics deals with the motion of objects and is particularly streamlined under conditions of constant acceleration. In such scenarios, the equations of motion provide direct relations between displacement, velocity, time, and acceleration, enabling the derivation of velocity as a function of time and the prediction of the trajectory of the center of mass.

Transcript

00:01 In this question, we have an attract arrangement.

00:04 So there's a card is connected to a hanging block by a cord.

00:08 M1 is 0 .6 kg, m2 is 0 .4 kg.

00:12 And the whole system is released from rest.

00:16 So we want to find the acceleration of the center of mass of the card block system.

00:23 The velocity of cm as a function of time.

00:26 We need to sketch the path and find the angle.

00:29 Angle of the displacement between the x -axis.

00:41 So to solve this problem, to find the acceleration of the center of mass, so one thing we can do is first to find a magnitude of acceleration of the system.

01:05 So this is equal to the total effect.

01:11 So for this whole system is like the point 4 kg is pulling both m1 and m2.

01:18 So the force that is pulling the two blocks will be 0 .4 would be m2g and divide by m1 plus m2.

01:28 And then you put in the numbers 0 .4 times 9 .8, divide by 0 .4 plus 0 .6.

01:36 And then you get 3 .92 meters per second square.

01:41 Okay, and then to find the acceleration of the center of mass, acm, okay.

01:48 Okay, so we know that a1, this is a1, and then this is a2, okay, and they have the same or many tube, okay, just different direction.

02:03 So the formula for acm is m1, a1, plus m2, a2, divide by m1 plus m2...

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