# 19.8 Pulley in rolling  (Page 2/3)

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## Examples

In this section, we work out with few representative examples to illustrate the application of rolling motion in the case of pulley :

• Pulley or disk of finite mass is translating
• Pulley or disk of finite mass is stationary
• A mass-less pulley connects rolling motion of a disk to the translation of a block.

## Pulley or disk of finite mass is translating

Example 1

Problem : A long rope of negligible mass is wrapped many times over a solid cylinder of mass "m" and radius "R". Other end of the rope is attached to a fixed ceiling and the cylinder is then let go at a given instant. Find the tension in the string and acceleration of the cylinder.

Solution : The rolling of the cylinder is guided on the rope. We select a coordinate system in which positive y-direction is along the downward motion as shown in the figure.

As discussed earlier, we set out to write three equations and solve for the required quantity. Three equations for rolling motions are (i) Newton's second law for translation (ii) Newton's second law for rotation and (iii) Equation of accelerated rolling.

From the application of Newton's second law for translation in y - direction :

$\begin{array}{l}\sum {F}_{y}=\mathrm{mg}-T=m{a}_{C}\end{array}$

From the application of Newton's second law for rotation :

$\begin{array}{l}\tau =TR=I\alpha \end{array}$

$\begin{array}{l}T=\frac{I\alpha }{R}\end{array}$

Note that force due to gravity passes through center of mass and does not constitute a torque to cause angular acceleration. Also, we do not consider sign for angular quantities as we shall be using only the magnitudes here . Now, from relation of linear and angular accelerations (equation of accelerated rolling), we have :

$\begin{array}{l}{a}_{C}=\alpha R\end{array}$

Putting the value of "α" from above in the equation - 4, we have :

$\begin{array}{l}T=I\frac{{a}_{C}}{{R}^{2}}\end{array}$

The moment of inertia for solid cylinder about its axis is,

$\begin{array}{l}I=\frac{m{R}^{2}}{2}\end{array}$

Hence,

$\begin{array}{l}⇒T=\frac{m{R}^{2}{a}_{C}}{2{R}^{2}}\\ ⇒T=\frac{m{a}_{C}}{2}\end{array}$

Substituting the expression of tension in the equation of force analysis in y-direction (equation - 3), we have :

$\begin{array}{l}\mathrm{mg}-\frac{m{a}_{C}}{2}=m{a}_{C}\\ ⇒g-\frac{{a}_{C}}{2}={a}_{C}\\ ⇒{a}_{C}\left(1+\frac{1}{2}\right)=g\\ ⇒{a}_{C}=\frac{2g}{3}\end{array}$

Putting this value in the expression of "T" (equation - 6), we have :

$\begin{array}{l}⇒T=\frac{m{a}_{C}}{2}=\frac{mg}{3}\end{array}$

## Pulley or disk of finite mass is stationary

Example 2

Problem : In the “pulley – blocks” arrangement shown in the figure. The masses of the pulley and two blocks are “M”, “ ${m}_{1}$ ” and “ ${m}_{2}$ " respectively. If there is no slipping between pulley and rope, then find (i) acceleration of the blocks and (ii) tensions in the string.

Solution : The blocks have different masses (hence weights). The net force on the pulley due to difference in weight accelerates the pulley in rotation. Let the tensions in the string be “ ${T}_{1}$ “ and “ ${T}_{2}$ “. Since the rope is inextensible, the different points on the rope has same acceleration as that of the rolling i.e. ${a}_{C}$ . Let ${m}_{2}>{m}_{1}$ .

The magnitude of torque on the pulley (we neglect the consideration of direction) is :

$\begin{array}{l}T=\left({T}_{2}-{T}_{1}\right)xR\end{array}$

The pulley is fixed to the ceiling. As such, it is constrained not to translate in response to the net vertical force on it. Instead, the net vertical force causes translational acceleration of the string and the masses attached to the string. The length of string released from the pulley is equal to the distance covered by a point on the rim of the pulley. It means that the linear acceleration of the string is related to angular acceleration of the pulley by the equation of accelerated rolling. Hence,

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