Question

A phonograph record 30.0cm in diameter rotates 33.5 times per minute.

Answer 1

Explanation:

Given that phonograph rotates 33.5 times per minute:

Frequency of the phonograph = $33.5 min^{-1}$

Time period =$\frac{1}{frequency}=\frac{1}{33.5 min^{-1}}=0.0298 min$

time period = 0.0298 × 60 seconds =1.788 seconds

Radius r = 15 cm = 0.15 m (1 m = 100 cm)

The linear speed =$\frac{Distance}{time}=\frac{2\pi\times r}{1.788 sec}=\frac{2\times 3.14\times 0.15 m}{1.788 sec}=0.526 m/s$

The centripetal acceleration of a point on its rim:

$\frac{(\text{linear velocity})^2}{\text{radius of the rim}}=\frac{(0.526 m/s)^2}{0.15 m}=1.84 m/s^2$

Centripetal acceleration is $1.84 m/s^2$.

Answer 2

The frequency (f) is 33.5 rpm, as given, or 0.59 rev/sec. The period of rotation (T) is 1/f, so 1/0.59 or ~ 1.69 seconds. The linear speed (v) of a point on it's rim is calculated using v= 2*pi*r/T so, v= 2 * pi * (30/2) / 1.69 ==> 2 * 3.14 * 15 / 1.69 ==> 55.74 cm/s. The centripetal acceleration (a) of a point on it's rim is caluclated using a = v^2/r so a = 55.74 ^2 / 15 ==> 207.13 cm/s^2