Shows an arrangement with an air track, in which a cart is...

Shows an arrangement with an air track, in which a cart is connceted 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 $m_{2}=0.400 \mathrm{~kg}_{\text {ind its center is ini- }}$ tially at $x y$ coordinates $(0,-0.100 \mathrm{~m})$. The mass of the cord and pullcy 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. (a) In unit-vector notation, what is the acccleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function of time $f ?$ (c) Sketch the path taken by the com. (d) If the path is curved, detcrmine 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 is the weighted average position of all the mass in a system. For a two-body system, it is calculated by taking the sum of each mass multiplied by its position and then dividing by the total mass. This concept simplifies problems in mechanics because the motion of the entire system can be analyzed by tracking the center of mass, especially in the absence of external forces or when they are easily accounted for.

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 multiplied by the acceleration of its center of mass. This crucial principle allows us to analyze the motion of a multi-object system by focusing on how external forces influence the center of mass, regardless of the internal forces between components.

Vector Decomposition and Unit-Vector Notation

Vector decomposition involves breaking a vector into its components along mutually perpendicular axes, typically using unit vectors such as i-hat and j-hat. This process is essential in mechanics for clearly expressing quantities like velocity, acceleration, and force in two or three dimensions, which then facilitates the analysis of motion in a coordinate system.

Kinematics of Uniform Acceleration

In problems involving uniform acceleration, kinematics provides the equations to describe velocity and displacement as functions of time. When the acceleration is constant, the velocity can be expressed as the initial velocity plus the product of acceleration and time, and the displacement can be derived by integrating the velocity function, which is critical for predicting the position of the center of mass over time.

Trajectory Analysis

Trajectory analysis involves determining the path traced by the center of mass as the system evolves in time. This requires understanding both the magnitude and direction of the velocity and acceleration vectors. The shape of the trajectory (whether linear or curved) can be deduced from the constancy or variation of these vectors, and it often involves analyzing the influence of constraints and the relative orientation of the forces acting on the system.

Transcript

00:01 In this question, we are given the mass of card that is m1 equal to 0 .6 kilogram and it is lying its center is initially at minus 0 .5 meter and 0 .5 meter.

00:26 Now there is a block whose mass is given by m.

00:32 And it is equal to 0 .4 kilogram and its center is initially at 0 meter and negative 0 .1 meter.

00:50 Now what we are supposed to find over here, so we are supposed to find over here the acceleration of center of mass, velocity, of center of mass in terms of function of time t then we're supposed to find out the path taken by center of mass and the angle now how will we do that we will solve all of these parts one by one so first of all we will solve path a that is the acceleration of center of mass now we know that the net force acting on this card block system is equal to m2 times gravitation acceleration.

01:54 So we will take positive directions for m1 in the right foot direction and m2 in the downward direction.

02:07 So by doing that and using newton's second law, so by using newton's second law, we get that acceleration is equal to m2 times gravitational acceleration divided by the sum of m1 and m2.

02:28 Now by substituting the values, we'll get acceleration is equal to 0 .4 meter per second square.

02:39 Now what will we do? the second step is that in this coordinate system, the acceleration of m1 and m2 are a x a1 is the acceleration for m1 and that is positive 0 .4 meter per second square i unit vector and the acceleration for m2 is a 2 and it is equal to negative 0 .4 meter per second square j unit vector now what will be the acceleration of center of mass it would be equal to m1 xx aeration of m1 that is a 1 plus m2 a 2 divided by the sum of m 1 and m 2 so so when you'll substitute the values, you'll get acceleration of center of mass equal to 2 .35 unit vector minus 1 .57j unit vector meter per second square.

04:28 Now what is the next thing that we need to find? this is your acceleration of center of mass.

04:36 The second thing that we are required to find out is the velocity of center of mass as a function of time t...

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