Chapter 10, Problem 36P

(a)

To determine

The time interval required for $m_1$ to hit the floor.

Answer to Problem 36P

The time interval required for $m_1$ to hit the floor is $1.95\text{ s}$.

Explanation of Solution

Strings experiences tension due to different masses the Mass $m_1$ is greater than $m_2$ hence, $m_1$ moves downwards and rotating the pulley in anticlockwisedirection.

Write the expression for heavier mass.

    $T_1-m_{1}g=-m_{1}a$

Here, $T_1$ is tension for heavier mass, $m_1$ is mass of heavier block, $g$ is acceleration under gravity and $a$ is net acceleration.

Rearrange the above equation.

    $T_1=m_{1}g-m_{1}a$                                                                                                (I)

Write the expression for lighter mass.

    $T_2-m_{2}g=m_{2}a$

Here, $T_2$ is tension for lighter mass and $m_2$ is mass of lighter block.

Rearrange the above equation.

    $T_2=m_{2}a+m_{2}g$                                                                                            (II)

Write the expression for moment of inertia for the pulley as.

    $I=\frac{1}{2}M{R}^2$

Here, $I$ is moment of inertia, $M$ is the mass of pulley and $R$ is the radius of pulley.

Write the expression for torque produced in motion of the pulley.

    $\tau =I\alpha$                                                                                                        (III)

Here, $\alpha$ is angular acceleration and $\tau$ is the torque produced.

Write the expression force torque in terms of force.

    $\tau =\left(F_{\text{net}}\right)R$                                                                                                 (IV)

Here, $F_{\text{net}}$ is the net force and $R$ is the radius of pulley.

Write the expression for net force.

    $F_{\text{net}}=T_1-T_2$

Substitute $T_1-T_2$ for $F_{\text{net}}$ in equation (IV).

    $\tau =\left(T_1-T_2\right)R$

Substitute $\left(T_1-T_2\right)R$ for $\tau$ in equation (III).

    $\left(T_1-T_2\right)R=I\alpha$                                                                                           (V)

Write the expression for angular acceleration.

    $\alpha =\frac{a}{R}$

Here, $a$ is the linear acceleration.

Substitute $\frac{a}{R}$ for $\alpha$, $m_{1}g-m_{1}a$ for $T_1$ , $m_{2}a+m_{2}g$ for $T_2$ and $\frac{1}{2}M{R}^2$ for $I$ in equation (V).

    $\begin{array}{c}\left(\left(m_{1}g-m_{1}a\right)-\left(m_{2}a+m_{2}g\right)\right)R=\left(\frac{1}{2}M{R}^2\right)\left(\frac{a}{R}\right)\\ m_{1}g-m_{1}a-m_{2}a-m_{2}g=\frac{1}{2}Ma\end{array}$

Rearrange the above expression for $a$ .

    $a=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2+\frac{1}{2}M\right)}$                                                                                   (VI)

Write the expression of motion to calculate time.

    $x=ut+\frac{1}{2}a{t}^2$

Here, $x$ is the distance covered, $u$ is the initial velocity and $t$ is the time interval.

Substitute $0$ for $u$ and solve the above equation.

    $t=\sqrt{\frac{2x}{a}}$                                                                                                     (VII)

Conclusion:

Substitute $20.0\text{ kg}$ for $m_1$, $12.5\text{ kg}$ for $m_2$, $5.00\text{ kg}$ for $M$ for $9.80{\text{ m/s}}^2$ for $g$ in equation (VI).

    $\begin{array}{c}a=\frac{\left(20.0\text{ kg}\right)-\left(12.5\text{ kg}\right)}{\left(20.0\text{ kg}+12.5\text{ kg}+\frac{1}{2}\left(5.00\text{ kg}\right)\right)}\left(9.80{\text{ m/s}}^2\right)\\ =\frac{\left(7.5\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{35.0\text{ kg}}\\ =2.10{\text{ m/s}}^2\end{array}$

Substitute $4.00\text{ m}$ for $x$ and $2.10{\text{ m/s}}^2$ for $a$ in equation (VII).

    $\begin{array}{c}t=\sqrt{\frac{2\left(4.00\text{ m}\right)}{2.10{\text{ m/s}}^2}}\\ =\sqrt{3.80}\\ =1.9509\text{ s}\end{array}$

Thus, the time interval required for $m_1$ to hit the floor is $1.95\text{ s}$.

(b)

To determine

The effect onthe pulley the moment pulley is mass less.

Answer to Problem 36P

The time reduces by $3.64\%$ for $m_1$ to hit the floor.

Explanation of Solution

Write the expression for acceleration of pulley.

    $a\text{'}=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2+\frac{1}{2}M\right)}$

Here, $a\text{'}$ is new acceleration when pulley is massless.

Substitute $0$  for $M$ in above equation.

    $a\text{'}=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)}$                                                                                 (VIII)

Write the expression for Timemoving down with acceleration $a\text{'}$.

    $t\text{'}=\sqrt{\frac{2x}{a\text{'}}}$                                                                                                    (IX)

Here $t\text{'}$ is the new time.

Write the expression for percentage change in time.

    $\text{Change in time}\left(\%\right)=\frac{t-t\text{'}}{t}\times 100$                                                                (X)

Conclusion:

Substitute $20.0\text{ kg}$ for $m_1$, $12.5\text{ kg}$ for $m_2$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (VIII).

    $\begin{array}{c}a\text{'}=\frac{20.0\text{ kg}-12.5\text{ kg}}{20.0\text{ kg}+12.5\text{ kg}}\left(9.80{\text{ m/s}}^2\right)\\ =\frac{7.5\text{ kg}\left(9.80{\text{ m/s}}^2\right)}{32.5\text{ kg}}\\ =2.26{\text{ m/s}}^2\end{array}$

Substitute $4.00\text{ m}$ for $x$ and $2.2638{\text{ m/s}}^2$ for $a\text{'}$ in equation (IX).

    $\begin{array}{c}t\text{'}=\sqrt{\frac{2\left(4.00\text{ m}\right)}{2.2638{\text{ m/s}}^2}}\\ =\sqrt{3.5388}\\ =1.8798\text{ s}\end{array}$

Substitute $1.88\text{s}$ for $t\text{'}$ and $1.95\text{s}$ for $t$ in equation (X).

    $\begin{array}{c}\text{Change in time}\left(\%\right)=\frac{1.9509\text{ s}-1.8798\text{ s}}{1.9509\text{ s}}\times 100\\ =3.64\%\end{array}$

Thus, the time reduces by $3.64\%$ for $m_1$ to hit the floor.