Question

Suppose a 57 g mass is attached to one end of a spring and spun in a circle on a horizontal frictionless stace. The spring's unstretched length is 10 cm and

its spring constant is 29 N/m. What angular velocity, in rpm, will stretch the spring to a length of 12 cm?
Express your answer in revolutions per minute.

Answer 1

The angular velocity that will stretch the spring to a length of 12 cm is approximately -2.959 rpm.

Step-by-step explanation:

To find the angular velocity that will stretch the spring to a length of 12 cm, we can use the equation:

F = -kx

Where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.

Since the spring is being spun in a circle, the force applied to the spring is given by:

F = mω²r

Where m is the mass, ω is the angular velocity, and r is the radius of the circle.

Plugging in the given values, we have:

mω²r = -kx

Solving for ω, we get:

ω = -√[(k/m) * (x/r)] = -√[(29 N/m) / (57 g / 0.1 m)]

Converting to rpm, we divide by 2π radians per minute to get:

ω = -√[(29 N/m) / (57 g / 0.1 m)] * (1 min / 2π radians)

which is approximately -2.959 rpm.