Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 71P from Chapter 17 from Hibbeler's Engineering Mechanics.
Problem 71P
Step-by-Step Solution
We are given the mass of the reel is $m = 400\;{\rm{kg}}$, the radius of the gyration is ${k_A} = 0.75\;{\rm{m}}$, and the radius of circular reel of cable is $r = 0.5\;{\rm{m}}$.
We are asked to determine the angular velocity when $t = 2\;{\rm{s}}$, starting from rest, if the force $P = \left( {20{t^2} + 80} \right)\;{\rm{N}}$, when t is in seconds.
Step 2
The free body diagram of the system is shown as:
We have the radius of the cable containing roll is $R = 1\;{\rm{m}}$.
We have the time is $t = 2\;{\rm{s}}$.
Step 3
The formula to calculate the mass moment of inertia of the reel about point A is,
\[{I_A} = m{\left( {{k_A}} \right)^2}\]Step 4
Substitute the values in the above expression.
\[\begin{array}{l} {I_A} = \left( {400\;{\rm{kg}}} \right){\left( {0.75\;{\rm{m}}} \right)^2}\\ {I_A} = 225\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]Step 5
The formula to calculate the angular acceleration of the reel is,
\[P\left( r \right) = {I_A}\alpha \]Step 6
Substitute the values in the above expression.
\[\begin{array}{c} \left( {20{t^2} + 80} \right)\left( {0.5} \right) = \left( {225} \right)\alpha \\ \alpha = \left( {0.177 + 0.044{t^2}} \right)\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]Step 7
The formula to calculate the angular velocity of the reel is,
\[d\omega = \alpha dt\]Step 8
Substitute the values in the above expression.
\[\begin{array}{c} d\omega = \alpha dt\\ \int\limits_0^\omega {d\omega } = \int\limits_0^{2\;{\rm{s}}} {\left( {0.177 + 0.044{t^2}} \right)} dt\\ \left[ \omega \right]_0^\omega = \left[ {0.177t + 0.044\left( {\frac{{{t^3}}}{3}} \right)} \right]_0^{2\;{\rm{s}}}\\ \left[ {\omega - 0} \right] = \left[ {0.177\left( 2 \right) + 0.044\left( {\frac{{{2^3}}}{3}} \right) - 0} \right] \end{array}\]Step 9
On further solving, we get:
\[\begin{array}{l} \omega = \left[ {0.177\left( 2 \right) + 0.044\left( {\frac{{{2^3}}}{3}} \right) - 0} \right]\\ \omega = 0.47\;{\rm{rad/s}} \end{array}\]