A star has a mass of 2.40 x 1030 kg and is moving in a circular orbit about the
ID: 1449364 • Letter: A
Question
A star has a mass of 2.40 x 1030 kg and is moving in a circular orbit about the center of its galaxy. The radius of the orbit is 3.1 x 104 light-years (1 light-year = 9.5 x 1015 m), and the angular speed of the star is 1.8 x 10-15 rad/s. (a) Determine the tangential speed of the star. (b) What is the magnitude of the net force that acts on the star to keep it moving around the center of the galaxy?
The initial angular velocity and the angular acceleration of four rotating objects at the same instant in time are listed in the table that follows. For each of the objects (a), (b), (c), and (d), determine the final angular speed after an elapsed time of 5 s.
Initial angular velocity 0 Angular acceleration (a) +37.0 rad/s +6.0 rad/s2 (b) +37.0 rad/s -6.0 rad/s2 (c) -37.0 rad/s +6.0 rad/s2 (d) -37.0 rad/s -6.0 rad/s2Explanation / Answer
1) m =2.4*10^30 kg , r =3.1*10^4 light year = 2.935*10^20 m
w =1.8*10^-15 rad/s
(a) v =rw = 2.935*10^20*1.8*10^-15 = 5.283*10^5 m/s
(b) F =mv^2/r = (2.4*10^30*5.283*5.283*10^10)/(2.935*10^20)
F =2.28*10^21 N
2) t =5 s
(a) wo =37 rad/s , = 6 rad/s^2
from rotational kinematic equation
w = wo+ t
w = 37+(6*5) = 67 rad/s
(b) wo =37 rad/s , = -6 rad/s^2
from rotational kinematic equation
w = wo+ t
w = 37-(6*5) = 7 rad/s
(c) wo = -37 rad/s , = 6 rad/s^2
from rotational kinematic equation
w = wo+ t
w = -37+(6*5) = -7 rad/s
(d) wo = -37 rad/s , = -6 rad/s^2
from rotational kinematic equation
w = wo+ t
w = -37-(6*5) = -67 rad/s
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