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zrbecker
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(EDIT: SOLVED) Book did it wrong, they swapped out the final rotational speed 1.8 rad/s with 1.7 rad/s, presumably mixed up with the radius of the sphere.
A space probe coasting in a region of negligible gravity is rotating with an angular speed of 2.4 rev/s about an axis that points in its direction of motion. The spacecraft is in the form of a thin spherical shell of radius 1.7m and mass 245kg. It is necessary to reduce the rotational speed to 1.8 rev/s by firing tangential thrusters along the "equator" of the probe. What constant force must the thrusters exert if the change in angular speed is to be accomplished as the probe rotates through 3.0 revolutions? Assume the fuel ejected by the thrusters is a negligible fraction of the mass of the probe.
Angular Velocity
[tex]\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)[/tex]
Moment of Inertia
[tex]I = \frac{2}{3} mr^2[/tex]
Kinetic Energy
[tex]K = \frac{1}{2}I\omega^2[/tex]
Torque
[tex]\tau = rF\sin\theta[/tex]
The way I attempted to do this problem was to find angular acceleration first
[tex](1.8 rev/s^2) = (2.4 rev/s^2) + 2 \alpha (3 rev)[/tex]
[tex]\alpha = -0.42 rev/s^2 = -2.63 rad/s^2[/tex]
[tex]I = \frac{2}{3}(245)(1.7)^2 = 472 kg \cdot m^2[/tex]
Then I got the torque as follows
[tex]\tau = I \alpha = 472(-2.63) = -1246[/tex]
Since the force is applied in the opposite direction that the thing is rotating I get
[tex]\tau = Fr\sin\theta = -Fr[/tex]
[tex]F = \frac{-\tau}{r} = \frac{1246}{1.7} = 733N[/tex]
The book is giving 833N
It did it the following way
[tex]I = \frac{2}{3}(245)(1.7)^2 = 472 kg \cdot m^2[/tex]
[tex]W = K_f - K_i = (1/2)(472)(2\pi * 1.7)^2 - (1/2)(472)(2\pi * 2.4)^2 = -2.67 \times 10^4[/tex]
[tex]W = \tau\theta = -rF\theta [/tex]
[tex]F = \frac{-W}{r\theta} = \frac{2.67 \times 10^4}{1.7 \cdot 6\pi} = 833N[/tex]
Homework Statement
A space probe coasting in a region of negligible gravity is rotating with an angular speed of 2.4 rev/s about an axis that points in its direction of motion. The spacecraft is in the form of a thin spherical shell of radius 1.7m and mass 245kg. It is necessary to reduce the rotational speed to 1.8 rev/s by firing tangential thrusters along the "equator" of the probe. What constant force must the thrusters exert if the change in angular speed is to be accomplished as the probe rotates through 3.0 revolutions? Assume the fuel ejected by the thrusters is a negligible fraction of the mass of the probe.
Homework Equations
Angular Velocity
[tex]\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)[/tex]
Moment of Inertia
[tex]I = \frac{2}{3} mr^2[/tex]
Kinetic Energy
[tex]K = \frac{1}{2}I\omega^2[/tex]
Torque
[tex]\tau = rF\sin\theta[/tex]
The Attempt at a Solution
The way I attempted to do this problem was to find angular acceleration first
[tex](1.8 rev/s^2) = (2.4 rev/s^2) + 2 \alpha (3 rev)[/tex]
[tex]\alpha = -0.42 rev/s^2 = -2.63 rad/s^2[/tex]
[tex]I = \frac{2}{3}(245)(1.7)^2 = 472 kg \cdot m^2[/tex]
Then I got the torque as follows
[tex]\tau = I \alpha = 472(-2.63) = -1246[/tex]
Since the force is applied in the opposite direction that the thing is rotating I get
[tex]\tau = Fr\sin\theta = -Fr[/tex]
[tex]F = \frac{-\tau}{r} = \frac{1246}{1.7} = 733N[/tex]
The book is giving 833N
It did it the following way
[tex]I = \frac{2}{3}(245)(1.7)^2 = 472 kg \cdot m^2[/tex]
[tex]W = K_f - K_i = (1/2)(472)(2\pi * 1.7)^2 - (1/2)(472)(2\pi * 2.4)^2 = -2.67 \times 10^4[/tex]
[tex]W = \tau\theta = -rF\theta [/tex]
[tex]F = \frac{-W}{r\theta} = \frac{2.67 \times 10^4}{1.7 \cdot 6\pi} = 833N[/tex]
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