Explanation:
The diameter of a record, d = 0.3 cm
Radius, r = 0.15 cm
It rotates 33.5 times per minute.
(a) Frequency, f = 33.5 rotation/minute
= (33.5/60) rotation/second
= 0.55 rotation/second
(b) Time period,
T = 1/f
So,
[tex]T=\dfrac{1}{0.55}\\\\T=1.81\ s[/tex]
(c) Linear speed of a point on its rim,
[tex]v=r\omega\\\\v=0.15\times 10^{-2}\times 2\pi \times 0.55\\\\v=5.18\times 10^{-3}\ m/s[/tex]
(d) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(5.18\times 10^{-3})^2}{0.15\times 10^{-2}}\\\\a=0.017\ m/s^2[/tex]
Hence, this is the required solution.
