Question

A 250-g weight is tied to a piece of thread wrapped around a spool, which is suspended in such a way that it can rotate freely. When the weight is released, it accelerates toward the floor as the thread unwinds.

250-g weight is tied to a piece of thread wrapped around a spool, which is suspended in such a way that it can rotate freely. When the weight is released, it accelerates toward the floor as the thread unwinds. Assume that the spool can be treated as a uniform solid cylinder of radius R = 3 cm and mass Ms= 100 g. Find the tension in the thread and the magnitude of the acceleration of the weight as it descends. Assume the thread has negligible mass and does not slip or stretch as it unwinds.

Answer 1

Concepts and reason

The concepts required to solve the given problem are Newton’s second law of motion, torque, moment of inertia of a solid cylinder about an axis passing through its center and relation of angular acceleration with linear acceleration.

Initially, write the equation of motion for the weight tied with the spool by using Newton’s second law of motion and by knowing net force on it. Then, find the tension in the thread in terms of acceleration and mass of the spool by taking expression for torque on the spool due to tension in the thread by using relation of angular acceleration with linear acceleration. After that, calculate the acceleration of the weight by solving these two equations. Finally, calculate the tension in the thread by using acceleration of the weight and mass of spool.

Fundamentals

According to Newton’s second law of motion, the net force ${F_{{\rm{net}}}}$ on an object is equal to product of mass m and acceleration a of the object.

${F_{{\rm{net}}}} = ma$

The magnitude of torque $\tau$ on an object by applying force F on the object at a distance r from the pivot is given by following expression.

$\tau = Fr$

Torque $\tau$ is also defined as the product of moment of inertia I and angular acceleration $\alpha .$

$\tau = I\alpha$

The moment of inertia I of a solid cylinder of radius R and mass M about an axis passing through its center is given by following expression.

$I = \frac{1}{2}M{R^2}$

The relation of angular acceleration $\alpha$ of a rotating object of radius R with linear acceleration a of the object is given by following expression.

$\alpha = \frac{a}{R}$

Draw the following free body diagram for weight.

Here, T is the tension, R is the radius of the spool, ${M_s}$ is the mass of the spool, g is the acceleration due to gravity and ${m_w}$ is the mass of the weight.

From the above free body diagram, the net force ${F_{{\rm{net}}}}$ on the weight is given by following expression.

${F_{{\rm{net}}}} = {m_w}g - T$

Assume the acceleration of the system is a.

From Newton’s second law of motion, the net force on the weight is equal to product of mass ${m_w}$ and acceleration a of the system.

${F_{{\rm{net}}}} = {m_w}a$

Substitute ${m_w}a$ for ${F_{{\rm{net}}}}$ in the above equation ${F_{{\rm{net}}}} = {m_w}g - T.$

${m_w}a = {m_w}g - T$

The torque $\tau$ on the spool due to tension T in the thread is given as follows:

$\tau = TR$

Torque $\tau$ is defined as the product of moment of inertia of the spool I and angular acceleration $\alpha$

$\tau = I\alpha$

Substitute $\frac{1}{2}{M_s}{R^2}$ for I, and $\frac{a}{R}$ for $\alpha$ in the above equation.

$\begin{array}{c}\\\tau = \left( {\frac{1}{2}{M_s}{R^2}} \right)\left( {\frac{a}{R}} \right)\\\\ = \frac{{{M_s}Ra}}{2}\\\end{array}$

Substitute TR for $\tau$ in the above equation.

$\begin{array}{c}\\TR = \frac{{{M_s}Ra}}{2}\\\\T = \frac{{{M_s}a}}{2}\\\end{array}$

The equation of motion for weight is given by following expression.

${m_w}a = {m_w}g - T$

The tension in the thread is given by following expression.

$T = \frac{{{M_s}a}}{2}$

Substitute $\frac{{{M_s}a}}{2}$ for T in the above equation ${m_w}a = {m_w}g - T$ to solve for a.

$\begin{array}{c}\\{m_w}a = {m_w}g - \frac{{{M_s}a}}{2}\\\\a\left( {{m_w} + \frac{{{M_s}}}{2}} \right) = {m_w}g\\\\a = \frac{{{m_w}g}}{{{m_w} + \frac{{{M_s}}}{2}}}\\\end{array}$

Substitute 250 g for ${m_w},$ 100 g for ${M_s}$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for g in the above equation to solve for acceleration.

$\begin{array}{c}\\a = \frac{{\left( {250\,{\rm{g}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)}}{{250\,{\rm{g}} + \frac{{100\,{\rm{g}}}}{2}}}\\\\ = 8.2\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

The tension T in the thread is given by following expression.

$T = \frac{{{M_s}a}}{2}$

Substitute 100 g for ${M_s},$ and $8.2\,{\rm{m/}}{{\rm{s}}^2}$ for a in the above equation.

$\begin{array}{c}\\T = \frac{{\left( {100\,{\rm{g}}} \right)\left( {\frac{{1\,{\rm{kg}}}}{{1000\,{\rm{g}}}}} \right)\left( {8.2\,{\rm{m/}}{{\rm{s}}^2}} \right)}}{2}\\\\ = 0.41\,{\rm{N}}\\\end{array}$

Ans:

The acceleration of the weight is $8.2\,{\rm{m/}}{{\rm{s}}^2}.$

The tension in the thread is 0.41 N.