Chapter 8, Problem 14P

(a)

To determine

The angular acceleration.

Answer to Problem 14P

The angular acceleration is $\text{1}\text{.45}\times {\text{10}}^{\text{-4}}{\text{ rad/sec}}^{\text{2}}$

Explanation of Solution

Given:

  $\begin{array}{l}{\text{ω}}_{\text{initial}}\text{ = 0 rpm = 0 rad/sec}\\ {\text{ω}}_{\text{final}}\text{ = 1 rpm = 280}\times \left(\text{2π rad / revolution}\right)\times \text{(1/60) rad / sec = 0}\text{.1047 rad/sec}\\ \text{time (t) = 12 minutes = 720 seconds}\end{array}$

Formula used:

The angular acceleration is determined by the following formula,

  $\text{Angular Acceleration (α) = }\frac{{\text{ω}}_{\text{final}}-{\text{ω}}_{\text{initial}}}{\text{time}}{\text{ radian/sec}}^{\text{2}}$

Calculation:

Substitute the given values,

  $\text{α = }\frac{\text{0}\text{.1047}-\text{0}}{\text{720}}\text{ = 1}\text{.45}\times {\text{10}}^{\text{-4}}{\text{ rad / s}}^{\text{2}}$

Conclusion:

The angular acceleration $\text{ 1}\text{.45}\times {\text{10}}^{\text{-4}}{\text{ rad/sec}}^{\text{2}}$

(b)

To determine

The radial and tangential components of the linear acceleration.

Answer to Problem 14P

Radial components of the linear acceleration is $\text{7}\text{.93}\times {\text{10}}^{\text{-3}}{\text{ m/sec}}^{\text{2}}$

Tangential components of the linear acceleration is $\text{6}\text{.16}\times {\text{10}}^{\text{-4}}{\text{ m/sec}}^{\text{2}}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Radius (r) = }\frac{\text{8}\text{.5}}{\text{2}}\text{ = 4}\text{.25 m}\\ \text{Angular Acceleration (α) = 1}\text{.45}\times {\text{10}}^{\text{-4}}{\text{ rad/s}}^{\text{2}}\\ \text{Time (t) = 5 minutes = 300 seconds}\end{array}$

Formula used:

The instantaneous angular velocity is determined by the following formula,

  ${\text{Instantaneous Angular Velocity (ω}}_{\text{f}}{\text{) = ω}}_{\text{i}}\text{ + (α}\times \text{t)}$

Radial and tangential components of the linear acceleration is calculated by following formula,

  $\begin{array}{l}{\text{Radial Components (a}}_{\text{rad}}{\text{) = ω}}_{\text{f}}{}^{\text{2}}\times \text{r}\\ {\text{Tangential Components (a}}_{\text{tan}}\text{)= α}\times \text{r }\end{array}$

Calculation:

Substitute the given values,

  ${\text{ω}}_{\text{f}}\text{= 0 + (1}\text{.44}\times {\text{10}}^{\text{-4}}\times \text{300) = 0}\text{.0432 rad/s}$

Radial and tangential components are,

  $\begin{array}{l}{\text{a}}_{\text{rad}}\text{ = (0}{\text{.0432)}}^{\text{2}}\times \text{4}\text{.25 = 7}\text{.93}\times {\text{10}}^{\text{-3}}{\text{ m/s}}^2\\ {\text{a}}_{\text{tan}}\text{= 1}\text{.45}\times {\text{10}}^{\text{-4}}\times \text{ 4}\text{.25 = 6}\text{.16}\times {\text{10}}^{\text{-4}}{\text{ m/s}}^{\text{2}}\end{array}$

Conclusion:

Radial components of the linear acceleration is $\text{7}\text{.93}\times {\text{10}}^{\text{-3}}{\text{ m/sec}}^{\text{2}}$

Tangential components of the linear acceleration is $\text{6}\text{.16}\times {\text{10}}^{\text{-4}}{\text{ m/sec}}^{\text{2}}$