The mass of a star is 1.89·1031 kg and it performs one rotation in 28.9 days. Find its new period (in /days) if the diameter suddenly shrinks to 0.93 times its present size. Assume a uniform mass distribution before and after.
To find the new period of rotation for the star, we need to consider the conservation of angular momentum. The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. In this case, we can assume that the mass distribution is uniform, so the moment of inertia can be calculated as I = (2/5)MR^2, where M is the mass of the star and R is the radius.
First, we need to find the present radius of the star. Since the diameter suddenly decreases to 0.93 times its present size, the radius will also decrease by the same factor. Let's denote the present radius as R.
New radius = 0.93 * R
Next, we can find the present moment of inertia. Since the mass distribution is uniform, we can use the formula I = (2/5)MR^2.
Present moment of inertia = (2/5) * M * R^2
Then, we need to find the present angular velocity. The period of rotation is given as 28.9 days, which is the time taken for one complete rotation.
Present angular velocity ω = 2π / T = 2π / 28.9 (in radians/day)
Now, let's calculate the new mass of the star. The mass density of the star remains the same, so mass can be calculated using the formula M = density * volume. Since the mass distribution is uniform, the volume will decrease by the cube of the change in diameter.
New mass = M * (0.93)^3
Finally, we can find the new period of rotation using the conservation of angular momentum formula L = Iω.
New period (T') = 2π / ω' = 2π / (L / I') = 2π / [(2/5) * (New mass) * (New radius)^2 * ω]
Replace the values of new mass, new radius, and present angular velocity into the equation to find the new period of rotation.
I omegai = I omegaf
I for a solid sphere is 2/5 mr^2