Chapter 17.1, Problem 17.14P

To determine

(a)

Velocity of cylinder A.

Answer to Problem 17.14P

Velocity of cylinder A is, $v_a=4.78m/s$

Explanation of Solution

Given information:

Mass of block (m_b) = 15kg.

Mass of cylinder (m_a) = 5kg.

Radius of gyration (k) = 160mm.

Coefficient of friction between block and surface ($\mu$) = 0.2.

External force 0n the cylinder (P) = 200N.

Radius of outer pulley (r_a) = 250mm.

Radius of inner pulley (r_b) = 150mm.

Distance moved by the cylinder A (s_a) = 1m.

Calculation:

Let radius of inner pulley is r_b and radius of outer pulley is r_a.

Velocity of block B attached to the inner pulley is v_b and velocity of cylinder A is v_a.

Velocity ratio

$\frac{v_b}{v_a}=\frac{r_a}{r_b}$

Distance moved by the cylinder A is s_a and distance moved by the block B is s_b.

So, displacement ratio

$\begin{array}{l}\frac{s_b}{s_a}=\frac{r_b}{r_a}\\ \frac{s_b}{1\times 1000mm}=\frac{150mm}{250mm}\\ s_b=600mm=0.6m\end{array}$

For initial condition; Angular velocity is zero. So, initial kinetic energy

E₁=0

Final kinetic energy is the summation of kinetic energy of cylinder A, kinetic energy of block B and kinetic energy of double pulley.

$\begin{array}{l}{E}_2=\frac{1}{2}{m}_a{v}_a{}^2+\frac{1}{2}{m}_b{v}_b{}^2+\frac{1}{2}{I}_c{\omega }_c{}^2\\ E_2=\frac{1}{2}{m}_a{v}_a{}^2+\frac{1}{2}{m}_b{\left(\frac{r_b}{r_a}{v}_a\right)}^2+\frac{1}{2}\left(m_c{k}^2\right){\left(\frac{v_a}{r_a}\right)}^2\\ E_2=\frac{1}{2}\times 5\times v_a{}^2+\frac{1}{2}\times 15{\left(\frac{0.150}{0.250}{v}_a\right)}^2+\frac{1}{2}\left(15\times {0.16}^2\right){\left(\frac{v_a}{0.250}\right)}^2\\ E_2=\frac{1}{2}\left(5+5.4+6.14\right)v_a{}^2\\ E_2=\frac{1}{2}\times 16.54\times v_a{}^2\\ E_2=8.27v_a{}^{2}J\end{array}$

Equlibrium force in normal direction.

$\begin{array}{l}{N}_B=m_{B}g\cos {30}^o\\ N_B=15\times 9.81\times \cos {30}^o\\ N_B=127.44N\end{array}$

Frictional force

$\begin{array}{l}f=\mu N_B\\ f=0.2\times 127.44\\ f=25.487N\end{array}$

Work done by the system

$\begin{array}{l}W=P{s}_a+m_{a}g{s}_a-f{s}_b-m_{b}g{s}_b\sin {30}^o\\ W=200\times 1+5\times 9.81\times 1-25.497\times 0.6-15\times 9.81\times 0.6\times \sin {30}^o\\ W=200+49.05-15.292-44.145\\ W=189.213J\end{array}$

Substitute the value of E₁, E₂ and W in work energy equation.

$\begin{array}{l}{E}_1+W=E_2\\ 0+189.213=8.27v_a{}^2\\ v_a{}^2=\frac{189.213}{8.27}\\ v_a{}^2=22.92\\ v_a=4.78m/s\end{array}$

To determine

(b)

Total distance moved by block B.

Answer to Problem 17.14P

Total distance moved by block B is s = 1.936 m.

Explanation of Solution

Given information:

Mass of block (m_b) = 15kg.

Mass of cylinder (m_a) = 5kg.

Radius of gyration (k) = 160mm.

Coefficient of friction between block and surface ($\mu$) = 0.2.

External force 0n the cylinder (P) = 200N.

Radius of outer pulley (r_a) = 250mm.

Radius of inner pulley (r_b) = 150mm.

Distance moved by the cylinder A (s_a) = 1m.

Calculation:

Velocity of block B

Velocity of block B attached to the inner pulley is v_b and velocity of cylinder A is v_a.

$\begin{array}{l}\frac{v_b}{v_a}=\frac{r_a}{r_b}\\ \frac{v_b}{4.78}=\frac{0.150}{0.250}\\ v_b=2.87m/s\end{array}$

Angular velocity of the pulley

$\begin{array}{l}{\omega }_c=\frac{v_a}{r_a}\\ {\omega }_c=\frac{4.78}{0.150}\\ {\omega }_c=19.15rad/s\end{array}$

Kinetic energy after cylinder A strike the ground is the summation of kinetic energy of block B and kinetic energy of double pulley.

$\begin{array}{l}{E}_3=\frac{1}{2}{m}_b{v}_b{}^2+\frac{1}{2}{I}_c{\omega }_c{}^2\\ E_3=\frac{1}{2}{m}_a{v}_a{}^2+\frac{1}{2}{m}_b{\left(\frac{r_b}{r_a}{v}_a\right)}^2+\frac{1}{2}\left(m_c{k}^2\right){\left(\frac{v_a}{r_a}\right)}^2\\ E_3=\frac{1}{2}\times 15{\left(\frac{0.150}{0.250}{v}_a\right)}^2+\frac{1}{2}\left(15\times {0.16}^2\right){\left(\frac{v_a}{0.250}\right)}^2\\ E_3=\frac{1}{2}\left(5.4+6.14\right)v_a{}^2\\ E_3=\frac{1}{2}\times 11.54\times v_a{}^2\\ E_3=5.77\times {4.78}^2\\ E_3=132.31J\end{array}$

Kinetic energy after the system comes to rest is zero.

E₄=0

Work done by the system

$\begin{array}{l}W=-f{s}_b{}^{\text{'}}-m_{b}g{s}_b\text{'}\sin {30}^o\\ W=-25.497\times s_b{}^{\text{'}}-15\times 9.81\times s_b{}^{\text{'}}\times \sin {30}^o\\ W=-99.60s_b{}^{\text{'}}J\end{array}$

Substitute the value of E₁, E₂ and W in work energy equation

$\begin{array}{l}{E}_1+W=E_2\\ 132.31-99.60s_b{}^{\text{'}}=0\\ s_b{}^{\text{'}}=\frac{132.31}{99.60}\\ s_b{}^{\text{'}}=1.336m\end{array}$

Total distance moved by the block B is

$\begin{array}{l}s=s_b+s_b{}^{\text{'}}\\ s=0.6+1.336\\ s=1.936m\end{array}$