19.8 Pulley in rolling  (Page 3/3)

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

The directions of angular and linear accelerations are as shown in the figure. Now, considering pulley and blocks, we have three equations as :

(i) Rotation of pulley :

$$\begin{array}{l}(T_2-T_1)xR=I\alpha =\frac{1}{2}xM{R}^2\alpha \end{array}$$

$$\begin{array}{l}\Rightarrow \alpha =\frac{2(T_2-T_1)}{MR}\end{array}$$

and

(ii) Translation of $m_1$ :

(iii) Translation of m2 :

Putting values of " $T_1$ " and " $T_2$ " from equations 7 and 8, in the expression of acceleration (equation - 6), we have :

Reaaranging, $$\begin{array}{l}\Rightarrow a_C(M+2m_1+2m_2)=2g(2m_2-m_1)\end{array}$$

$$\begin{array}{l}\Rightarrow a_C=\frac{2g(2m_2-m_1)}{(M+2m_1+2m_2)}\end{array}$$

Putting values of " $a_C$ " from equation - 9, we have :

$$\begin{array}{l}\Rightarrow T_1=m_{1}g+m_{1}x\frac{2g(2m_2-m_1)}{(M+2m_1+2m_2)}\end{array}$$

Rearranging, we have :

$$\begin{array}{l}\Rightarrow T_1=\frac{m_{1}g(M+4m_2)}{(M+2m_1+2m_2)}\end{array}$$

Similarly,

$$\begin{array}{l}\Rightarrow T_2=\frac{m_{2}g(M+4m_1)}{(M+2m_1+2m_2)}\end{array}$$

A mass-less pulley connects rolling motion of a disk to the translation of a block

Example 3

Problem : In the “pulley – blocks” arrangement, a string is wound over a circular cylinder of mass “M”. The string passes over a mass-less pulley as shown in the figure. The other end of the string is attached to a block of mass “m”. If the cylinder is rolling on the surface towards right, then find accelerations of the cylinder and block.

Solution : In this question, string passes over a mass-less pulley. It means that the tension in the string through out is same. Since string is attached to the top of the cylinder, the tension force acts tangentially to the cylinder in rolling. In this case, the friction is acting in forward direction (as the cylinder tends to slide backward).

The linear acceleration of the top position and center of mass are different. As such, linear accelerations of the cylinder and blocks are different. Let the accelerations of cylinder and block be “ $a_1$ “and “ $a_2$ ” respectively.

(i) Rolling of cylinder :

The magnitude acceleration of the rolling cylinder (i.e. its COM) is :

Now, we need to relate the acceleration of the cylinder to that of string (and hence that of block attached to it). For convenience, we consider only magnitudes here. Now, we know that the magnitude of the velocity of the top point of the cylinder is related to that of COM as :

$$\begin{array}{l}{v}_T=2v_C\end{array}$$

Differentiating with respect to time, the magnitude of the acceleration of the string and the block, attached to it, is :

(ii) Translation of cylinder :

Applying Newton’s second law for translation of the cylinder,

(iii) Rotation of cylinder :

Applying Newton’s second law for rotation,

$$\begin{array}{l}(T-f_S)xR=I\alpha =\frac{1}{2}xM{R}^2\alpha \end{array}$$

$$\begin{array}{l}\Rightarrow \alpha =\frac{2(T-f_S)}{MR}\end{array}$$

$$\begin{array}{l}\Rightarrow \alpha MR=2T-f_S\end{array}$$

For rolling, we use relation of accelerations as given by equation - 11 :

$$\begin{array}{l}\Rightarrow 2T-2f_S=M{a}_1\end{array}$$

Putting values of " $f_S$ " from equation - 13,

$$\begin{array}{l}\Rightarrow 2T-2(M{a}_1-T)=M{a}_1\end{array}$$

(iv) Translation of block :

The free body diagram of the block is shown in the figure. Here,

$$\begin{array}{l}mg-T=m{a}_2\end{array}$$

Substituting value of "T" from equation – 14,

$$\begin{array}{l}m{a}_2-mg=T=\frac{3M{a}_1}{4}\end{array}$$

Substituting this value of " $a_1$ " in terms of " $a_2$ " from equation – 11 and solving for" $a_2$ " ,

$$\begin{array}{l}\Rightarrow a_2=\frac{8mg}{(8m+3M)}\end{array}$$

As the magnitude of " $a_1$ " is given by :

$$\begin{array}{l}\Rightarrow a_1=\frac{a_2}{2}=\frac{4mg}{(8m+3M)}\end{array}$$