Chapter 8, Problem 59P

(a)

To determine

The relation between angular acceleration and linear acceleration.

Answer to Problem 59P

The relation between angular acceleration and linear acceleration is $a=R\alpha$.

Explanation of Solution

Write the expression for linear acceleration.

$a=R\alpha$

Here, a is the linear acceleration, R is the radius and $\alpha$ is angular acceleration.

Conclusion:

Therefore, relation between angular acceleration and linear acceleration is $a=R\alpha$.

(b)

To determine

The net torque on the pulley.

Answer to Problem 59P

The net torque on the pulley is $\left(T_1-T_2\right)R$.

Explanation of Solution

Write the expression for net torque on the pulley.

${\tau }_{net}=T_{1}R-T_{2}R$

Here, ${\tau }_{net}$ is the net torque and $T_1$, $T_2$ are the tensions.

Negative sign for $T_2$ implies that the tensions are opposite in direction.

Conclusion:

Therefore, net torque on the pulley is $\left(T_1-T_2\right)R$.

(c)

To determine

Why tensions cannot be equal when masses are not equal.

Answer to Problem 59P

The blocks will not have any acceleration if their masses are equal.

Explanation of Solution

If the masses are equal, the tensions are equal. As a result, the net torque will be zero as seen from (b). This will not cause any acceleration to the masses. Hence, only when the masses are unequal, the pulley will have angular acceleration causing the blocks to accelerate.

(d)

To determine

The acceleration and tensions.

Answer to Problem 59P

The tensions are $T_1=m_1\left(g-a\right)$, $T_2=m_2\left(g+a\right)$ and acceleration is $a=\frac{\left(m_1-m_2\right)g}{m_1+m_2+\frac{M}{2}}$.

Explanation of Solution

The free body diagram of the system is given in figure 1.

Write the expression for the state of equilibrium for block 1.

$m_{1}g-T_1=m_{1}a$

Here, $m_1$ is the mass of block 1, g is the acceleration due to gravity and a is the acceleration.

Re-arrange the above equation to get $T_1$.

$T_1=m_1\left(g-a\right)$

Write the expression for the state of equilibrium for block 2.

$T_2-m_{2}g=m_{2}a$

Here, $m_2$ is the mass of block 2, g is the acceleration due to gravity and a is the acceleration.

Re-arrange the above equation to get $T_2$.

$T_2=m_2\left(g+a\right)$

Equate the expressions for torque.

$\left(T_1-T_2\right)R=\frac{1}{2}M{R}^2\alpha$

Here, M is the mass of the pulley.

Use the expressions for $\alpha$,$T_1$ and $T_2$ in the above expression.

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

Re-arrange the above equation to get a.

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

Conclusion:

Therefore, tensions are $T_1=m_1\left(g-a\right)$, $T_2=m_2\left(g+a\right)$ and acceleration is $a=\frac{\left(m_1-m_2\right)g}{m_1+m_2+\frac{M}{2}}$.

(e)

To determine

The acceleration of the blocks using Newton’s equation of motion.

Answer to Problem 59P

The acceleration of the blocks using Newton’s equation of motion is $\frac{\left(m_1-m_2\right)g}{m_1+m_2+\frac{M}{2}}$.

Explanation of Solution

Refer Example 8.2

Write the expression for speed.

$v=\sqrt{\frac{2\left(m_1-m_2\right)gh}{m_1+m_2+\frac{I}{R^2}}}$

Here, h is height and I is the moment of inertia.

Write the expression for moment of inertia.

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

Replace I by $\frac{1}{2}M{R}^2$ in the expression for v.

$\begin{array}{c}v=\sqrt{\frac{2\left(m_1-m_2\right)gh}{m_1+m_2+\frac{\frac{1}{2}M{R}^2}{R^2}}}\\ =\sqrt{\frac{2\left(m_1-m_2\right)gh}{m_1+m_2+\frac{M}{2}}}\end{array}$

Write the expression for a using Newton’s equation of motion.

$a=\frac{v^2}{2h}$

Substitute th expression for v in the above equation.

$\begin{array}{c}a=\frac{{\left(\sqrt{\frac{2\left(m_1-m_2\right)gh}{m_1+m_2+\frac{M}{2}}}\right)}^2}{2h}\\ =\frac{\left(m_1-m_2\right)g}{m_1+m_2+\frac{M}{2}}\end{array}$

Conclusion:

Therefore, acceleration of the blocks using Newton’s equation of motion is $\frac{\left(m_1-m_2\right)g}{m_1+m_2+\frac{M}{2}}$.