Question

Figure shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{}\mathbf{kg}$, and its center is initially at XY coordinates $\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}$; the block has mass , and its center is initially at XY coordinates $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{m}\mathbf{)}$the mass of the cord and pulley is 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 acceleration of the center of mass of the cart–block system? (b) What is the velocity of the com as a function of time ? (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.

Short answer

(a) The acceleration of the center of mass of the cart-block system in unit vector notation is${\overrightarrow{a}}_{com}=(2.35\stackrel{⏜}{i}-1.57\stackrel{⏜}{j})\frac{m}{s^2}$.

(b) The velocity of the center of mass as a function of time is ${\overrightarrow{v}}_{com}=(2.35\stackrel{⏜}{i}-1.57\stackrel{⏜}{j})t\frac{m}{s}$.

(c) Sketch the path taken by the center of mass: The Path taken by the center of mass is straight.

(d) The angle between the path and the x-axis is$\theta =-34°$.

Step-by-step solution

Listing the given quantities:

The mass of the cart is $m_1=0.600\mathrm{kg}$.

The XY coordinates of the mass $m_1$is $(x,y)=(-0.500,0)m$.

The mass of the block is $m_2=0.400kg$.

The XY coordinates of the mass $m_1$is $(x,y)=(0,-0.100)m$.

Understanding the concept of Newton’s laws and center of mass:

Use the concept of Newton’s second law and find the acceleration of the system. By using the motion of the system’s center of mass, you can find the acceleration and velocity of the center of mass. You can find the direction of the center of mass by using the trigonometry equation.

Formulae:

Newton’s second law of motion is,

$F_{net}=ma$

Here, is the net force, and is the mass, and is the acceleration.

Therefore,

$$\begin{gathered} M\stackrel{\rightharpoonup }{a_{cm}}=m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}+\cdots +m_n\overrightarrow{a} \\ \overrightarrow{V_{cm}}=\int \overrightarrow{a_{cm}}dt \\ tan\theta =\frac{v_y}{v_x} \end{gathered}$$

(a) Calculations of acceleration of center of mass of the cart-block system

You can apply Newton’s second law for the cart system, then

$T=m_{1}a$ ….. (1)

Here, T is the tensile force and$m_1$ is the mass of the cart.

For the block system,

$T-m_{2}g=-m_{2}a$ ….. (2)

Here, $m_2$ is the mass of the block and g is the acceleration due to gravity having a value $9.8\frac{m}{s^2}$.

From equation (1) equation (2), you have

$$\begin{gathered} m_{1}a-m_{2}g=-m_{2}a \\ (m_1+m_2)a=m_{2}g \end{gathered}$$

$$\begin{gathered} a=\frac{m_{2}g}{\left(m_1+m_2\right)} \\ =\frac{0.400\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{\left(0.600\mathrm{kg}+0.400\mathrm{kg}\right)} \\ =3.92\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The cart is accelerating along positive x-axis and the block is accelerating in the downward direction. By using the motion of system’s center of mass, you can find center of mass of the acceleration of the system as

$$\begin{gathered} M\overrightarrow{a_{cm}}=m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}+...+m_n\overrightarrow{a} \\ \overrightarrow{a_{cm}}==\frac{m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}}{m_1+m_2} \end{gathered}$$

$$\begin{gathered} \overrightarrow{a_{cm}}=\frac{0.600\mathrm{kg}\times \left(3.92\mathrm{m}/{\mathrm{s}}^2\right)\stackrel{⏜}{\mathrm{i}}+0.400\mathrm{kg}\times \left(-3.92\mathrm{m}/{\mathrm{s}}^2\right)\stackrel{⏜}{\mathrm{j}}}{\left(0.600\mathrm{kg}+0.400\mathrm{kg}\right)} \\ =\frac{2.532\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\stackrel{⏜}{\mathrm{i}}-1.568\mathrm{kg}.\times \mathrm{m}/{\mathrm{s}}^2\stackrel{⏜}{\mathrm{j}}}{\left(1.00\mathrm{kg}\right)} \\ =\left[\left(2.35\right)\stackrel{⏜}{i}+\left(-1.57\right)\stackrel{⏜}{j}\right]m/s^2 \end{gathered}$$

Hence, the acceleration of the center of mass of the cart-block system in unit vector notation is $\stackrel{\rightharpoonup }{a_{com}}=\left[\left(2.35\right)\stackrel{⏜}{i}+\left(-1.57\right)\stackrel{⏜}{j}\right]\frac{m}{s^2}$.

(b) The velocity of center of mass as function of time:

Calculate the velocity of the center of mass as a function of time:

Integrating the equation of the acceleration of center of mass of the system as

$$\begin{gathered} \overrightarrow{v_{cm}}=\int \overrightarrow{a_{cm}}dt \\ =\int \left[\left(2.35\right)\stackrel{⏜}{i}-\left(1.57\right)\stackrel{⏜}{j}\right]m/s^{2}dt \\ =\left[\left(2.35\right)\stackrel{⏜}{i}-\left(1.57\right)\stackrel{⏜}{j}\right]\int dt \\ =\left[\left(2.35\right)\stackrel{⏜}{i}-\left(1.57\right)\stackrel{⏜}{j}\right]tm/s \end{gathered}$$

(c) The sketch of the path taken by the center of mass  :

The ratio of the y-component to the x-component of the velocity does not change with the time. This ratio gives the velocity vector. The path taken by the center of mass is straight.

(d) Calculations of angle between the path and the x- axis

The angle between the path and the x-axis:

The angle between the path and x –axis is,

$$\begin{gathered} \tan \theta =\frac{v_y}{v_x} \\ =\frac{-1.57m/s}{2.35} \\ =-34° \end{gathered}$$

Hence, the path of the center of mass is straight at a downward angle of $34°$.