Question

A 10 6 kg space-station in the form of a huge wheel is rotating at 0.2 rad/s. For purposes of this problem, you may consider it's center of mass to be at rest, and that all the mass is concentrated in the rim, which is 100 m from the center. A large meteor (5 x 105 kg) is speeding towards it at 700 m/s, along a line that would first hit the rim and then go through the center if it kept on going. The meteor hits the rim of the station, where it sticks. (A) What was the moment of inertia of the station before being hit? (B) What was the angular momentum of the station before being hit? (C) How fast is the center of mass of the station plus meteor travelling after collision? (D) How fast is the station now rotating after collision?

Answer 1

Answer:

a) I = $5*10^9kg*m^2$

b) L = $10^9$ (kg*m^2)/s

c) $V_s$= 233.3m/s

d) $W_s$= 0.1 rad/s

Explanation:

a) We know that:

I= $\frac{1}{2}mR^2$

where I is the moment of inertia, m the mass and R the radius. So, replacing values, we get:

I= $\frac{1}{2}(10^6kg)(100m)^2$

I = $5*10^9kg*m^2$

b)  We know that:

L = IW

where L is the angular momentum, I the moment of inertia and W the angular velocity. So, replacing values, we get:

L = $(5*10^9)(0.2rad/s)$

L = $10^9$ (kg*m^2)/s

c) Using the conservation of the linear momentum:

$P_i = P_f$

so:

$M_mV_m = M_sV_s$

where $M_m$ is the mass of the meteor, $V_m$ is the velocity of the meteor, $M_s$ is the mass of the meteor and the space-station after the collition and $V_s$ is the velocity of the meteor and the space-station after the collition. So, replacing values, we get:

$(5*10^5kg)(700m/s) = (5*10^5+10^6)V_s$

Solving for $V_s$:

$V_s$= 233.3m/s

d) Using the conservation of the angular momentum:

$L_i = L_f$

so:

$I_aW_a = I_sW_s$

where $I_a$ is the moment of inertia of the station, $W_a$ is the angular velocity of the station, $I_s$ is the moment of inerta of the meteor and the space-station after the collition and $W_s$ is the angular velocity of the meteor and the space-station after the collition. So, replacing values, we get:

$I_aW_a = (I_a + MR^2)W_s$

$10^9 = (5*10^9+(5*10^5(100^2)W_s$

solving for $W_s$:

$W_s$= 0.1 rad/s