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

Answer:

a) v= 2.1 m/s

b) ω = 0.807 rad/s

Explanation

Conceptual analysis :

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

$a_{c} =\frac{v^{2} }{r}$ 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):

$a_{c} =\frac{v^{2} }{r}$

$1.7 =\frac{v^{2} }{2.6}$

$v^{2} =1.7*2.6 = 4.42$

$v=(\sqrt{4.42})\frac{m}{s}$

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