Respuesta :
Answer:
(a) [tex]a_{c} = 5.41\times 10^{9} m/s^{2}[/tex]
(b) [tex]a_{t} = 2.99\times 10^{- 5} m/s^{2}[/tex]
Given:
Time period of Pulsar, [tex]T_{P} = 33.085 ms == 33.085\times 10^{- 3} s[/tex]
Equatorial radius, R = 15 Km = 15000 m
Spinning time, [tex]t_{s} = 9.50\times 10^{10}[/tex]
Solution:
(a) To calculate the value of the centripetal acceleration, [tex]a_{c}[/tex] on the surface of the equator, the force acting is given by the centripetal force:
[tex]m\times a_{c} = \frac{mv_{c}^{2}}{R}[/tex]
[tex]a_{c} = \frac{v_{c}^{2}}{R}[/tex] (1)
where
[tex]v_{c} = \frac{distance covered(i.e., circumference)}{ T}[/tex]
[tex]v_{c} = \frac{2\pi R}{Time period, T}[/tex] (2)
Now, from (1) and (2):
[tex]a_{c} = R\frac({2\pi )^{2}}{T^{2}}[/tex]
[tex]a_{c} = 15000\frac{2\pi )^{2}}{(33.085\times 10^{- 3})^{2}}[/tex]
[tex]a_{c} = 5.41\times 10^{9} m/s^{2}[/tex]
(b) To calculate the tangential acceleration of the object :
The tangential acceleration of the object will remain constant and is given by the equation of motion as:
[tex]v = u + a_{t}t_{s} = 0[/tex]
where
u = [tex]v_{c}[/tex]
[tex]a_{t} = - \frac{2\pi R}{Tt_{s}}[/tex]
[tex]a_{t} = - \frac{2\pi 15000}{33.085\times 10^{- 3}\times 9.50\times 10^{10}}[/tex]
[tex]a_{t} = 2.99\times 10^{- 5} m/s^{2}[/tex]
