Question

Assume you are on abig wheel at fair ground vehicles 1rev every 8 sec .The operator of the wheel decoded to stop the wheel by applying brakes.The brake produce on acceleration of 0.1rad per sec 2,your seat os 4 m from the axis of rotation

a,what is angular velocity?
b,what os your velocity befor the brakes are applied?
c,how long does it take the big wheel to stop?
d,how many revolution does in the wheel make is slowing down?

Answer 1

[AI-generated answer — Explain Bot]

To answer these questions, we need to use the formulae related to angular velocity, acceleration, and time. Let's go through each question step by step:

a) Angular velocity (ω) is the change in angle (θ) per unit of time (t). The formula to calculate angular velocity is ω = θ / t. In this case, the wheel completes 1 revolution (which is equal to 2π radians) every 8 seconds, so the angular velocity is:

ω = (2π radians) / (8 seconds)
= π / 4 radians per second

Therefore, the angular velocity is π/4 radians per second.

b) Velocity (v) is the distance traveled (s) per unit of time (t). The formula to calculate velocity is v = s / t. In this case, your seat is 4 meters from the axis of rotation, and the wheel completes 1 revolution (which is equal to the circumference of the circle) every 8 seconds. Using these values, we can calculate the velocity:

v = (2π * 4 meters) / (8 seconds)
= π / 2 meters per second

Therefore, your velocity before the brakes are applied is π/2 meters per second.

c) To calculate how long it takes the big wheel to stop, we need to use the formula of acceleration (a) and time (t). The formula to calculate time is t = (ωf - ωi) / a, where ωf is the final angular velocity (which is 0 because the wheel stops) and ωi is the initial angular velocity (which is π/4 radians per second). Substituting the given values, we get:

t = (0 - π/4) / 0.1 rad/sec^2
= - π/4 / 0.1 rad/sec^2
= - 2.5 π seconds

Since time cannot be negative in this scenario, we take the absolute value:

t = 2.5 π seconds

Therefore, it takes 2.5 π seconds for the big wheel to stop.

d) The number of revolutions the wheel makes while slowing down can be calculated using the formula:

θ = ωi * t + 0.5 * a * t^2

Since the final angular velocity is 0, we can rearrange the formula to solve for θ:

θ = ωi * t

Substituting the values:

θ = (π/4 radians per second) * (2.5 π seconds)
= 2.5 π^2/4 radians

Therefore, the wheel makes 2.5 π^2/4 radians or approximately 1.96 revolutions while slowing down.