Can someone help me solve this?

Question

Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is || = 180 km/s and the orbital period of each is 11.1 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)

 _____ solar masses

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 810 N and a radius of 1.51 m. A child applies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?

 _____ J

Answer 1

First Question:

This is an interesting problem. As far as we know, there's only one force acting on each of the stars. Using our general Second law equation and the equation of centripetal motion, we can get the following.

F_g = ma ; a = v²/r

F_g = (mv²)/r

Also, we can use Newton's law of gravitation to find an equation for the force being exerted by one body on the other.

F_Ng = (Gm₁m₂)/(r²)

suppose m₁ = m₂ = M

also note that r is not from planet to center, but planet to next planet, so we substitute 2r for r.

F_Ng = (GM²)/(2r)²

We can set these equations equal to each other.

(GM²)/(2r)² = (Mv²)/r

M = (4v²r)/G

r can be found with the velocity and the time it takes for one rotation. r=(vT)/(2π)

M = (2v³T)/(πG) = ( 2*(180x10³ m/s)³ (11.1 days) (86400 s/day) ) / (π * 6.67x10^-11 Nm²/kg²)

M ≈ 5.338x10³¹ kg

So the mass of each star is M = 5.338x10³¹ kg.

Second Question:

This is an inertia question. The moment of inertia for a disk is denoted by I = 0.5mR² , while the equation for torque on a moment is τ = I*α and the equation for torque from force is τ=F*r*sinθ. We can use τ = F*r because the force is being applied at the tangent.

The force being applied is 49.0N at the radius of 1.51m which gives a torque of

τ =(49N)*(1.51m)

τ = 73.99Nm

Now let's find the moment for the merry go round.

I = 0.5mR²

But we don't have the mass, we have the weight. Let's convert.

F = ma; m = (810N)/(9.81m/s²) ≈ 82.569kg

Plug this value in to the inertia equation with our radius of 1.51m.

I = 0.5*(82.569kg)*(1.51m)² = 94.133kg m²

Okay, cool. We have the moment of inertia, and the torque being applied. Let's put them together in the torque equation to find the angular acceleration.

τ = I*α

73.99Nm = 94.133kg m² * α

α = 0.786 rad/s²

Let's use the equation for rotational motion to find the speed we'll be at after 2.95 seconds.

ω= ω₀+α*t 

ω = 0 rad/s + 0.786 rad/s² * 2.95s = 2.32 rad/s

Kinetic energy for angular motion equation:

KE = 0.5*I*ω²

Substitute known values:

KE = 0.5*(94.133kg m²)* (2.32 rad/s) = 109 J

Hope this helps!

Answer 2

Let

v = 180 km/s be the orbital speed,

T = 11.1 days be the period,

M be the mass of each star (to be found),

r be the distance of each star from their center of gravity,

G = 6.674×10^-11 m³kg^-1s^-2 be the gravitational constant.

First we can calculate r from the given period -- each star travels around the circle in time T.

T = 2πr/v

r = vT/2π

Now we calculate the mass M.

For the stars to travel around in a circle, their centripetal acceleration,

v²/r,

must equal their gravitational acceleration,

GM/(2r)².

After setting those two quantities equal, we get

M = 4v²r/G = 2v³T/πG

Plugging in actual values

M = 2 x 180³ km³/s³ 11.1 day / (π × 6.674×10^-11 m³kg^-1s^-2)

= 6.17×10¹⁷ km³/s³ day m^-3 kg s²

= 6.17×10¹⁷ x 10⁹ m³/s³ x 86400 s m^-3 kg s²

= 53.3×10³⁰ kg

That is over 26 solar masses.