Question

A dog sits 2.6 m from the center of a merry- go-round. a) If the dog undergoes a 1.7 m/s^2 centripetal acceleration, what is the dog's linear speed? Answer in units of m/s. b) What is the angular speed of the merry-go- round? Answer in units of rad/s.

Answer 1

a) v= 2.1 m/s

b) ω = 0.807 rad/s

Step-by-step explanation

Conceptual analysis :

The dog and the merry-go- round describes a circular motion, then, the following formulas apply :

`a_(c) =(v^(2) )/(r)` Formula (1)

v = ω *r Formula (2)

Where:

`a_(c)` : Centripetal acceleration(m/s²)

v: linear speed or tangential (m/s)

r : radius of the circle (m)

ω : angular speed ( rad/s)

Data

r= 2.6 m

`a_(c)` = 1.7 m/s²

Problem develpment

a) We replace data in the formula 1 to calculate the dog's linear speed(v):

`a_(c) =(v^(2) )/(r)`

`1.7 =(v^(2) )/(2.6)`

`v^(2) =1.7*2.6 = 4.42`

`v=(√(4.42))(m)/(s)`

v= 2.1 m/s

b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).

v = ω *r

2.1 = ω *2.6

ω = 2.1/2.6

ω = 0.807 rad/s