Calculating Time Elapsed for a Wheel to Come to a Stop with Constant Retardation

Imagine a wheel rotating with an initial angular velocity of 30 revolutions per second (rev/s). The wheel ultimately comes to rest after completing 60 revolutions due to a constant retardation. In this article, we will calculate the time elapse before the wheel stops. We will use concepts from rotational motion and physics equations to solve this problem.

Given Data

Initial angular velocity, ω0 30 rev/s Final angular velocity, ω 0 rev/s (since the wheel comes to rest) Angular displacement, θ 60 rev

Converting Revisions to Radians

To apply the equations of rotational motion, we first need to convert the angular displacement from revolutions to radians.

θ 60 rev × 2π rad/rev 120π rad

Applying the Rotational Motion Equations

We use the following equation from rotational motion to find the angular acceleration α

ω2 ω02 - 2αθ

Rewriting the equation to solve for α gives us:

α (ω2 - ω02) / (2θ)

Substituting the given values (ω 0, ω0 30 rev/s, and θ 120π rad), we get:

α (0 - 30^2) / (2 × 120π) -900 / (240π) -15π / 4 rad/s^2

Calculating Time Elapse

Next, we use the equation for rotational motion to find the time t needed for the wheel to come to a stop:

ω ω0 αt

Since ω 0, we can rearrange the equation as:

t (ω - ω0) / α

Substituting the values, we get:

t (0 - 30 × 2π) / (-15π / 4) (60π / 15π) 16 seconds

Final Answer: The time elapsed before the wheel stopped is 16 seconds.

Alternative Approach

Note that another method to solve this problem involves using the rotational kinematic relations. Here, we can calculate the retardation directly and then find the time:

From the kinematic equation: ω2 - ω02 2αθ, we can solve for α as:

α (ω2 - ω02) / (2θ) -900 / 240 -7.5 rev/s2

Using another kinematic relation: ω ω0 αt, we can find t as:

t (ω - ω0) / α (-30) / (-7.5) 4 seconds

[Answer]: T 4 seconds