A dog sits 3.0 m from the center of a merry-

go-round.
a) If the dog undergoes a 1.2 m/s2 cen-
tripetal 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.

Use these formulas, which you should know:

Acceleration = V^2/R = R w^2

To solve these problems, we need to understand the relationship between linear speed, angular speed, centripetal acceleration, and the radius of rotation.

Let's start with part (a):

a) The centripetal acceleration of an object moving in a circular path is given by the formula:

a = v^2 / r

where:
a is the centripetal acceleration,
v is the linear speed,
and r is the radius of rotation.

Rearranging this formula to solve for v, we get:

v = √(a * r)

Now we can substitute the given values. The radius of rotation is 3.0 m, and the centripetal acceleration is 1.2 m/s^2. Plugging these values into the equation, we get:

v = √(1.2 * 3.0)
v = √3.6
v ≈ 1.8974 m/s (rounded to 4 decimal places)

Therefore, the dog's linear speed is approximately 1.8974 m/s.

Moving on to part (b):

b) The angular speed (ω) of an object moving in a circular path is defined as the angle (θ) traveled per unit of time. It is related to the linear speed (v) and radius of rotation (r) by the equation:

v = ω * r

To find the angular speed, we need to rearrange this equation:

ω = v / r

We already found the linear speed (v) in part (a) as 1.8974 m/s, and the radius of rotation (r) is given as 3.0 m. Plugging these values into the equation, we get:

ω = 1.8974 m/s / 3.0 m
ω ≈ 0.6325 rad/s (rounded to 4 decimal places)

Therefore, the angular speed of the merry-go-round is approximately 0.6325 rad/s.