Chapter 15.1, Problem 15.19P

(a)

To determine

Find the velocity and acceleration of point B just before the power is turned off.

Answer to Problem 15.19P

The velocity and acceleration of point B just before the power is turned off are $\underset{\_}{20.9\text{ft/s}}$ and $\underset{\_}{1,316{\text{ft/s}}^2}$.

Explanation of Solution

Given information:

The initial speed $\left({\omega }_0\right)$ of sprocket wheel and the chain is $600\text{rpm}\left(\text{Counterclockwise}\right)$.

The radius (r) of the wheel is $4\text{in}\text{.}$

The time taken (t) by the wheel and the chain coast to come to rest is $4\text{s}$.

Consider the motion as uniformly decelerated motion.

Calculation:

Show the position of point A and B as shown in Figure 1.

Calculate the velocity of the point B just before the power is turned off using the relation:

$v_0=r{\omega }_0$

Substitute $4\text{in}\text{.}$ for r and $600\text{rpm}$ for ${\omega }_0$.

$\begin{array}{c}{v}_0=\left(4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\times \left(600\text{rpm}\times \frac{2\pi }{60}\right)\\ =0.333\times 62.83\\ =20.9\text{ft/s}\end{array}$

Thus, the velocity of the point B just before the power is turned off is $\underset{\_}{20.9\text{ft/s}}$.

Calculate the uniform angular acceleration $\left(\alpha \right)$ of the wheel using the relation:

$\omega ={\omega }_0+\alpha t$ (1)

The value of angular velocity $\left(\omega \right)$ is zero at time $t=4\text{s}$.

Substitute 0 for $\omega$, $600\text{rpm}$ for ${\omega }_0$, and $4\text{s}$ for t in Equation (1).

$\begin{array}{c}0=\left(600\text{rpm}\times \frac{2\pi }{60}\right)+4\alpha \\ =62.83+4\alpha \\ \alpha =\frac{62.83}{4}\\ =-15.708{\text{rad/s}}^2\end{array}$

$\alpha =15.708{\text{rad/s}}^2\left(\text{Clockwise}\right)$

Consider just before power is turned off.

The angular acceleration $\left(\alpha \right)$ immediately after the power off is zero.

Consider the normal and tangential component of the acceleration at point B are denoted by ${\left({\text{a}}_B\right)}_n$ and ${\left({\text{a}}_B\right)}_t$.

Calculate the tangential component of acceleration of the point B using the relation:

${\left({\text{a}}_B\right)}_t=r\alpha$ (2)

Substitute $0$ for $\alpha$ and $4\text{in}\text{.}$ for r in Equation (2).

$\begin{array}{c}{\left({\text{a}}_B\right)}_t=0\times 4\\ =0\end{array}$

Calculate the normal component of acceleration of the point B using the relation:

${\left({\text{a}}_B\right)}_n=r{\omega }_0^2$ (3)

Substitute $4\text{in}\text{.}$ for r and $600\text{rpm}$ for ${\omega }_0$ in Equation (3).

$\begin{array}{c}{\left({\text{a}}_B\right)}_n=4\times {\left(600\times \frac{2\pi }{60}\right)}^2\\ =4\times {62.83}^2\\ =15,791\text{in}{\text{./s}}^2\times \left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)\\ =1,316{\text{ft/s}}^2(\leftarrow )\end{array}$

Neglect the effect of tangential acceleration as it is small.

Thus, the acceleration of point B just before the power is turned off is $\underset{\_}{1,316{\text{ft/s}}^2}$.

(b)

To determine

Find the velocity and acceleration of point B just after 2.5 s.

Answer to Problem 15.19P

The velocity and acceleration of point B just after 2.5 s are $\underset{\_}{7.85\text{ft/s}}$ and $\underset{\_}{185.1{\text{ft/s}}^2\left(\text{Acting at }\beta \text{=1}\text{.6}°\text{Counterclockwise below the horizontal}\right)}$.

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

Calculate the uniform angular velocity $\left(\omega \right)$ of the wheel using the relation:

$\omega ={\omega }_0+\alpha t$ (4)

The value of angular velocity $\left(\omega \right)$ is zero at time $t=4\text{s}$.

Substitute $-15.708\text{rad/s}$ for $\alpha$, $600\text{rpm}$ for ${\omega }_0$, and $4\text{s}$ for t in Equation (4).

$\begin{array}{c}\omega =\left(600\text{rpm}\times \frac{2\pi }{60}\right)+2.5\times \left(-15.708\text{rad/s}\right)\\ =62.83-39.27\\ \omega =23.56\text{rad/s}\left(\text{Counterclockwise}\right)\end{array}$

Consider the time $t=2.5\text{s}$.

Calculate the velocity of the point B using the relation:

$v_B=r\omega$

Substitute $23.56\text{rad/s}$ for $\omega$ and $4\text{in}\text{.}$ for r.

$\begin{array}{c}{v}_B=4\text{in}\text{.}\times 23.56\text{rad/s}\\ \text{=94}\text{.24}\text{in}\text{./s}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =7.85\text{ft/s}\end{array}$

Thus, the velocity of point B just after 2.5 s is $\underset{\_}{7.85\text{ft/s}}$.

Show the components of acceleration as shown in Figure 2.

Refer Figure 2.

Consider the normal and tangential component of the acceleration at point B are denoted by ${\left({\text{a}}_B\right)}_n$ and ${\left({\text{a}}_B\right)}_t$.

Calculate the tangential component of acceleration of the point B using the relation:

${\left({\text{a}}_B\right)}_t=r\alpha$ (5)

Substitute $15.708{\text{rad/s}}^2$ for $\alpha$ and $4\text{in}\text{.}$ for r in Equation (5).

$\begin{array}{c}{\left({\text{a}}_B\right)}_t=15.708\times 4\\ =62.83\text{in}{\text{./s}}^2(\uparrow )\end{array}$

Calculate the normal component of acceleration of the point B using the relation:

${\left({\text{a}}_B\right)}_n=r{\omega }^2$ (6)

Substitute $4\text{in}\text{.}$ for r and $23.56\text{rad/s}$ for $\omega$ in Equation (6).

$\begin{array}{c}{\left({\text{a}}_B\right)}_n=4\times {23.56}^2\\ =2,221\text{in}{\text{./s}}^2\times \left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)(\leftarrow )\\ =185.1{\text{ft/s}}^2(\leftarrow )\end{array}$

Calculate the angle $\beta$ using the relation:

$\begin{array}{l}\tan \beta =\frac{{\left({\text{a}}_B\right)}_t}{{\left({\text{a}}_B\right)}_n}\\ \tan \beta =\frac{62.83\text{in}{\text{./s}}^2}{2,220\text{in}{\text{./s}}^2}\\ \beta ={\tan }^{-1}\left(0.0283\right)\\ \beta =1.6°\end{array}$

Neglect the effect of tangential acceleration as it is small.

Thus, the acceleration of point B just after 2.5 s is $\underset{\_}{185.1{\text{ft/s}}^2\left(\text{Acting at }\beta \text{=1}\text{.6}°\text{Counterclockwise below the horizontal}\right)}$.