Question

Far out in space, a 100,000-kg rocket and a 200,000-kg rocket are docked at opposite ends of a motionless 90m-long tunnel. The tunnel is rigid and has much less mass than the rockets. The rockets start their engines simultaneously, each providing 50,000N of constant thrust in opposite directions.A. Find the center of mass, as measured from rocket 1B. What is the moment of inertia of the system, assuming the rockets can be treated as point masses?C. Find the net torque on the system about the center of massD. What is the angular acceleration of the system?E. What is the structure's angular velocity after 30.0 seconds?

Answer 1

Answer:

a) $X_{cm} = 60m$

b) I = $5.4*10^8Kg*m^2$

c) $T= 4.5*10^6$ N*m

d) a = $8.3*10^{-3}rad/s^2$

e) W= 0.25 rad/s

Explanation:

a) we know that:

$X_{cm} = \frac{m_1x_1+m_2x_2}{m1+m2}$

where $X_cm$ is the ubicaton of the center of mass, $m_1$ the mass of the first rocket, $x_1$ its distances with the rocket 1, $m_2$ the mass of the second rocket and $x_2$ its distance with the rocket 1. So, replacing values, we get:

$X_{cm} = \frac{(100,000kg)(0)+(200,000kg)(90m)}{100,000kg+200,000kg}$

$X_{cm} = 60m$

So, the center of mass is at 60m from the rocket 1.

b) we know that:

$I = M_1R_1^2 +M_2R_2^2$

where I is the moment of inertia, $M_1$ is the mass of the rocket 1, $R_1$ its distance from the center of mass, $M_2$ the mass of the second rocket and $R_2$ the distance between the rocket 2 and the center of mass. So, replacing values, we get:

$I = (100,000kg)(60m)^2 +(200,000kg)(30m)^2$

I = $5.4*10^8Kg*m^2$

c) We know that:

T = Fr

where T is the net torque, F is the force and r is the distance between the rocket and the radius. Then:

$FR_1+FR_2 = T$

Replacing values, we get:

50,000N(60m)+50,000N(30m) = T

$T= 4.5*10^6$ N*m

d) We know that:

T = Ia

where T is the net torque, I the moment of inertia and a is the angular aceleration. So, replacing values, we get:

$4.5*10^6Nm = 5.4*10^8Kg*m^2a$

solving for a:

a = $8.3*10^{-3}rad/s^2$

e) Finally, using:

W = at

where W is the angular velocity, a is the angular aceleration and t is the time.

Then, replacing values. we get:

W = ($8.3*10^{-3}rad/s^2$)(30s)

W = 0.25 rad/s