Question

Question 6 (14 points)

A 3.68 kg object is tied to the end of a cord and whirled in a horizontal circle of radius 881 cm. The object completes 10.8 revolutions every 18.8
s. What is the centripetal acceleration of the object?

Answer 1

`a=114\ m/s^2`

Step-by-step explanation:

Given that,

Mass of an object, m = 3.68 kg

It is whirled in a horizontal circle of radius 881 cm or 8.81 m

The object completes 10.8 revolutions every 18.8 s.

We need to find the centripetal acceleration of the object. It is given by the formula as follows :

`a=\omega^2 r` ...(1)

Where
`\omega` is angular velocity

10.8 revolutions = 67.85 rad

`\omega=(67.85\ rad)/(18.8\ s)\\\\=3.6\ rad/s`

Put values in equation (1)

`a=3.6^2* 8.81\\\\a=114\ m/s^2`

So, the centripetal acceleration of the object is
`114\ m/s^2`.