$7$ to $9)$ To solve these problems, you needed to obtain the following data as youreadtheproblem: M $=50\mathrm{.}0$kg$,$i $=0\mathrm{.}0$rad, $$f $=?$rad,

$$i $=0\mathrm{.}0$m$/$s$,$f $=5\mathrm{.}00$rad$/$s$,=?$rad$/$s$2,$ t $=30\mathrm{.}0$s$,$

$=0\mathrm{.}5$M $$R$2$

$7)$ The Moment of Inertia of the Disk is easily calculated as follows:

R $=10\mathrm{.}0$m$,$andobjectisaDISK $$ I

Disk

$=0\mathrm{.}5$M $$R$2=(0\mathrm{.}5)(50\mathrm{.}0$kg$)(10\mathrm{.}0$m$)2=2,500$kg m$2$

$8)$ To solve this problem, you must look at the data and the three Equations of Circular Motion $($ECM$).$ Doing this, you see the problem can be solved using ECM #$2$ in the following manner:

$=+$t $5\mathrm{.}00=0\mathrm{.}0+(30\mathrm{.}0)=5\mathrm{.}0030\mathrm{.}0=0\mathrm{.}167$rad$/$s$2$ fi

Note: The Positive Angular Acceleration indicates the Disk is Accelerating. This tells us how quickly the disk is Spinning Up to a higher Angular Velocity.

$9)$ Whenever you see the word Net Torque, you should think of the following formula: $$ net $=$ I Disk $$ net $.$ Using this formula, you get the following:

I

Disk

$=(2,500$kg m$2)(0\mathrm{.}167$rad$/$s$2)=418$m N

$10)$ The Moment of Inertia for this System of Particles about the Z$-$axis

is calculated as follows:

I$=$m$$x$2=(20\mathrm{.}0)(0\mathrm{.}0)2+(10\mathrm{.}0)(3\mathrm{.}0)2+(40\mathrm{.}0)(6\mathrm{.}0)2+(30\mathrm{.}0)(10\mathrm{.}0)2$

$=4,530$kg m$2$