## Question:

The flywheel shown has a radius of 20 in. a weight of 250 lbs, and a radius of gyration of 15 in. A 30-lb block A is attached to a wire that is wrapped around the flywheel, and the system is released from rest. Neglecting the effect of friction, determine (a) the acceleration of block A, (b) the speed of block A after it has moved 5 ft.

## Step-by-step

$\sum { { M }_{ B } } =\sum { { \left( { M }_{ B } \right) }_{ eff } } :\left( { m }_{ A }g \right) r=\overline { I } \alpha +\left( { m }_{ A }a \right) r\\$

${ m }_{ A }gr={ m }_{ F }{ k }^{ 2 }\left( \frac { a }{ r } \right) +{ m }_{ A }ar\\$

$a=\frac { { W }_{ A } }{ { m }_{ A }+{ m }_{ F }{ \frac { k }{ r } }^{ 2 } }$

(a) Acceleration of $a=\frac { \left( 30Ib \right) }{ \left( \frac { 30Ib }{ 32.2ft/{ s }^{ 2 } } \right) +\left( \frac { 250Ib }{ 32.2ft/{ s }^{ 2 } } \right) { \left( \frac { 15in. }{ 20in. } \right) }^{ 2 } } =5.6615ft/{ s }^{ 2 }$

$or { a }_{ A }=5.66ft/{ s }^{ 2 }\downarrow$

(b) Velocity of A ${ v }_{ A }^{ 2 }=({ { v }_{ A }) }_{ 0 }^{ 2 }+2{ a }_{ A }s$

For $s=5 ft { v }_{ A }^{ 2 }=0+2\left( 5.6615in./{ s }^{ 2 } \right) \left( 5ft \right)$

$\\ \\ =56.6154f{ t }^{ 2 }/{ s }^{ 2 }$

$\\ \\ { v }_{ A }=7.5243ft/s$

or ${ v }_{ A }=7.52ft/s$