Question

A textbook of mass 2.09kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.120m , to a hanging book with mass 2.99kg . The system is released from rest, and the books are observed to move a distance 1.30m over a time interval of 0.750s . Part A What is the tension in the part of the cord attached to the textbook? =9.66N Part B What is the tension in the part of the cord attached to the book? Take the free fall acceleration to be = 9.80m/s^2 =15.5N Part C What is the moment of inertia of the pulley about its rotation axis? Take the free fall acceleration to be = 9.80 m/s^2???? .

Answer 1

Concepts and reason

The concept is used to solve this problem is the equation of motion, free body diagram and torque.

In the given problem, the two masses are connected to the pulley. Use the equation of motion to calculate the acceleration along vertical direction. Write the dynamic equations to find the tension in the cord.

To find the moment of inertia, first find the angular acceleration. Then, use the expressions of the torque.

Fundamentals

The equation of the motion can be expressed as follows,

$s = ut + \frac{1}{2}a{t^2}$

Here, $s$ is the distance, $u$ is the initial speed, $v$ is the final speed, $a$ is the acceleration and $t$ is the time taken be object.

From newton’s second law,

$F = ma$

Here, $F$ is the force and $m$ is the mass of the object.

The relation between angular acceleration and linear acceleration can be expressed as follows,

$a = \alpha r$

Here, $\alpha$ is the angular acceleration and $r$ is the radius.

The expression for the torque can be expressed as follows,

$\vec \tau = \vec F \times \vec r$

Here, $r$ is the perpendicular distance.

(A)

Consider the free body diagram of the system as,

Calculate the value of the acceleration by using the equation of motion along vertical direction.

$y = ut + \frac{1}{2}{a_y}{t^2}$

Substitute $0{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for $u$ , $1.30{\rm{ m}}$ for $y$ and $0.75{\rm{ s}}$ for $t$ in the expression of the equation of motion.

$\begin{array}{l}\\\left( {1.30{\rm{ m}}} \right) = \left( {0{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.75{\rm{ s}}} \right) + \frac{1}{2}{a_y}{\left( {0.75{\rm{ s}}} \right)^2}\\\\\left( {1.30{\rm{ m}}} \right) = \frac{1}{2}{a_y}{\left( {0.75{\rm{ s}}} \right)^2}\\\end{array}$

Rearrange the expression to get the value of acceleration along vertical direction.

${a_y} = 4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$

Write the dynamic equation when the textbook is moving with acceleration.

${T_1} = {m_1}{a_y}$ …… (1)

Here, ${m_1}$ represents the mass of the textbook and ${a_y}$ is the vertical acceleration.

Substitute $2.09{\rm{ kg}}$ for ${m_1}$ and $4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for ${a_y}$ in the expression of ${T_1}$ .

$\begin{array}{c}\\{T_1} = \left( {2.09{\rm{ kg}}} \right)\left( {4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)\\\\ = 9.6558{\rm{ N}}\\\end{array}$

(B)

Write the dynamic equation for the book.

${m_2}g - {T_2} = {m_2}{a_y}$ …… (2)

Here, ${m_2}$ is the mass of the book, ${T_2}$ is the tension in the part of the cord attached to the book, $g$ is the acceleration due to gravity and ${a_y}$ is the acceleration along vertical direction.

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$ , $4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${a_y}$ and $2.99{\rm{ kg}}$ for ${m_2}$ in the expression of the dynamic equation of the book.

$\begin{array}{c}\\\left( {2.99{\rm{ kg}}} \right)\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right) - \left( {2.99{\rm{ kg}}} \right)\left( {4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right) = {T_2}\\\\{T_2} = 15.4882{\rm{ N}}\\\end{array}$

(C)

Find the radius.

$r = \frac{{{\rm{diameter}}}}{2}$

Substitute $0.120{\rm{ m}}$ for ${\rm{diameter}}$ in the expression of radius.

$\begin{array}{c}\\r = \frac{{\left( {0.120{\rm{ m}}} \right)}}{2}\\\\ = 0.06{\rm{ m}}\\\end{array}$

Calculate the angular acceleration of the pulley.

${a_y} = \alpha r$

Substitute $4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for ${a_y}$ and $0.06{\rm{ m}}$ for $r$ in the expression of the angular acceleration.

$\begin{array}{c}\\\left( {4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right) = \alpha \left( {0.06{\rm{ m}}} \right)\\\\\alpha = \left( {\frac{{4.62{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}{{0.06{\rm{ m}}}}} \right)\\\\ = 77{\rm{ rad}} \cdot {{\rm{s}}^{ - 2}}\\\end{array}$

Write the expression of the torque acting on the pulley.

$\tau = I\alpha$ …… (3)

Write the expression of the torque acting on the pulley.

$\tau = \left( {{T_2} - {T_1}} \right)r$ …… (4)

Use equation (3) and (4) to find the moment of inertia.

$I\alpha = \left( {{T_2} - {T_1}} \right)r$

Rearrange the equation to find the value of the moment of inertia.

$I = \frac{{\left( {{T_2} - {T_1}} \right)r}}{\alpha }$

Substitute ${\rm{15}}{\rm{.49 N}}$ for ${T_2}$ , ${\rm{9}}{\rm{.66 N}}$ for ${T_1}$ , $0.06{\rm{ m}}$ for $r$ and $137{\rm{ rad}} \cdot {{\rm{s}}^{ - 2}}$ for $\alpha$ in the expression of moment of inertia.

$\begin{array}{c}\\I = \frac{{\left( {{\rm{15}}{\rm{.49 N}} - {\rm{9}}{\rm{.66 N}}} \right)\left( {0.06{\rm{ m}}} \right)}}{{{\rm{77 rad}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = 0.0045{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

Ans: Part A

The tension in the part of the cord attached to the textbook is $9.66{\rm{ N}}$ .