Rotational Kinematics Question Homework Help needed

[rmarkatos]

Our sun rotates in a circular orbit about the center of the Milky Way Galaxy. The radius of the orbit is 2.2x10^20 m, and the angular speed of the sun is 1.2x10^-15 rad/s. (a) What is the tangential speed of the sun? (b) How long (in years) does it take for the sun to make one revolution around the center?

(a) for part a all i did was VT=r(w) w=omega

(b) for this part since we know the constant speed is 1.2x10^-15 rad/s and since its one revolution theta should equal 2(pi) radians so am i supposed to use w(omega)= 0(theta)/t(time)

is this correct?

 

[rmarkatos]

thanks for the help everyone.......

 

[kyle8921]

Yes, your approach is correct. To find the tangential speed, you can use the formula VT = rω, where VT is the tangential speed, r is the radius of the orbit, and ω is the angular speed. Plugging in the values given, we get VT = (2.2x10^20 m)(1.2x10^-15 rad/s) = 2.64x10^5 m/s. This is the speed at which the sun is moving tangentially around the center of the Milky Way Galaxy.

For part (b), we can use the formula ω = θ/t, where ω is the angular speed, θ is the angle of rotation, and t is the time taken. As you correctly stated, for one revolution, θ = 2π radians. So, we can rearrange the formula to get t = θ/ω = (2π rad)/(1.2x10^-15 rad/s) = 5.24x10^15 seconds. To convert this to years, we can divide by the number of seconds in a year, which is approximately 3.15x10^7 seconds. This gives us the time taken for one revolution as approximately 166 billion years. This is a very long time, considering the current estimated age of the universe is around 13.8 billion years.