Question

7&8

7. Antares is a star in Scorpius constellation. Its radius is 3.0AU and its mas is about 12 times the mass of the sun. Antares orbital speed is 20.7 km/sec. Eventually after a couple million years Antares will be converted to a neutron star. Its radius will be 4.3*10-5 AU. Find its orbital speed after it becomes neutron star? Find your answer by rpm. Moment of inertia of sphere is = 2/5 MR2 - 8. A rotor disk in your car's wheel with radius of 30.0 cm and mass of 15.0 kg rotates with 800 rpm and it slows down to 50 rpm in 10 second. Find its angular acceleration? (b) Number of revolutions during this period of time.(C) The required force to do this action during this period of time.

Answer 1

7) From the principle of conservation of angular momentum,

$I_1\omega_1=I_2\omega_2$

I's denotes the moment of inertia of the star before and after transformation and denote the angular velocities

S

Using the moment of inertia of a sphere we have

2 2 MRſwi 5 MRWE

$or,\ \omega_2=\left ( \frac{R_1}{R_2} \right )^2\omega_1$

R1 and R2 are the radius before and after transformation

Initial orbital speed is 20.7 km/s and radius is 3 AU

So, converting this into angular speed we have

wi velocity radius 20.7 x 103 rad/s = 4.6 x 10-8 rad/s 3 x 1.5 x 1011

Putting the values,

$\omega_2=\left ( \frac{3}{4.3\times10^{-5}} \right )^2\times4.6\times10^{-8}\ rad/s$

Converting this into orbital speed we have

$v=223.9\times4.3\times10^{-5}\times1.5\times10^{11}\ m/s=1444186\times10^3\ m/s$

8) Acceleration is deined as the rate of change of velocity

Velocities are

$\omega_1=800\ rpm=800\times\frac{2\pi}{60}\ rad/s=26.67\pi\ rad/s$

$\omega_2=800\ rpm=50\times\frac{2\pi}{60}\ rad/s=1.67\pi\ rad/s$

a) Hence, acceleration is

$\alpha=\frac{\omega_1-\omega_2}{time}=\frac{26.67\pi-1.67\pi}{10}\ rad/s^2=7.85\ rad/s^2$

b) Distance traversed in 10 s is

$\theta=\omega_1t+\frac{1}{2}\alpha t^2$

$=\left (26.67\pi\times10-\frac{1}{2}\times7.85\times10^2 \right )\ rad=445.36\ rad$

Hence, number of revolution is

$\frac{445.36}{2\pi}=70.88\approx 71$

c) Linear acceleration is

$a=\frac{30}{100}\times7.85\ m/s^2=2.355\ m/s^2$

So, force is

$F=15\times2.355\ N=35.325\ N$