Question

A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 times before coming to rest. Find the constant angular acceleration of the centrifuge (in rad/s2). Consider the direction of the initial angular velocity to be the positive direction, and include the appropriate sign in your result.

Answer 1

Answer:

Explanation:

Given:

Initial angular velocity, wi = 3650 rev/min

Converting from rev/min to rad/sec,

3650 rev/min × 2pi rad/1 rev × 1 min/60 secs

= 382.23 rad/s

Final angular velocity, wf = 0 rad/s (comes to rest)

Angular displacement, Ѳ = 46 times

= 46 revs

= 46 rev × 2pi rad/1 rev

= 289.03 rad

Using equation of angular motion,

wf^2 = wi^2 + 2 × ao × Ѳ

Where,

ao = angular acceleration

0 = 382.23^2 + 2 × 289.03 × ao

Solving for ao,

ao = -252.75 rad/s^2

Answer 2

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!