Question

3. A record 0.30 cm in diameter rotates 33.5 times per minute.

a. What is its frequency?

b. What is its period?

c. What is the linear speed of a point on its rim?

d. What is the centripetal acceleration of a point on its rim?

Answer 1

Step-by-step 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,

`T=(1)/(0.55)\\\\T=1.81\ s`

(c) Linear speed of a point on its rim,

`v=r\omega\\\\v=0.15* 10^(-2)* 2\pi * 0.55\\\\v=5.18* 10^(-3)\ m/s`

(d) Centripetal acceleration,

`a=(v^2)/(r)\\\\a=((5.18* 10^(-3))^2)/(0.15* 10^(-2))\\\\a=0.017\ m/s^2`

Hence, this is the required solution.