Question

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg kg, traveling perpendicular to the door at 14.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

Answer 1

Answer:

0.36 rad/s

Explanation:

from the question we are given the following:

width of the door (w) = 1 m

length of the door (L) = 2 m

mass of the door (Md) = 40 kg

mass of mud (Mm) = 0.7 kg

velocity of the mud (V) = 14 m/s

angular speed = ?

we can get the angular speed by equating the initial angular momentum to the final angular momentum

initial angular momentum of the mud = final angular momentum of the mud and door

m x v x \frac{r}{2} = ($\frac{1}{3} × m × w ^{2}$ + m$(\frac{r}{2}) ^{2}$) x ω

note that r is divided by 2 because the mud hit the door at its center which is half of its width

0.7 x 14 x \frac{1}{2} = ($\frac{1}{3} × 40 × 1 ^{2}$ + (0.7 x $(\frac{1}{2}) ^{2}$)) x ω

ω = $\frac{0.7 x 14 x 0.5}{ ([tex]\frac{1}{3} × 40 × 1 ^{2}$ + (0.7 x $(\frac{1}{2}) ^{2}$))}[/tex]

= 0.36 rad/s