Question

The figure shows a graph of the angular velocity of a rotating wheel as a function of time. Although not shown in the graph, the angular velocity continues to increase at the same rate until t = 8.6 s. What is the angular displacement of the wheel from 0 to 8.6 s?

Answer 1

Answer:

θ = 33.54 rad

Explanation:

This is a circular motion exercise

         θ= θ₀ + w₀ t + ½ α t²

suppose that for t = 0 the body is at its initial point θ₀ = 0 and from the graph we see that the initial angular velocity w₀ = -9.0 rad / s

we look for the angular acceleration,

         α = $\frac{\Delta w}{ \Delta t}$

from the graph taken two points

         α = $\frac{0 - (-9.0)}{3.0 - 0}$

         α = 3 rad / s²

we substitute in the first equation

            θ = 0 -9 t + ½ 3 t²

the displacement is requested for t = 8.6 s

           θ = = -9  8.6 + 3/2  8.6²

           θ = 33.54 rad

Answer 2

The angular displacement of the wheel from 0 to 8.6s is 33 radians.

Given to us

t = 8.6s

Acceleration of the wheel

We know acceleration is the ratio of velocity and time, therefore, the slope of the velocity-time graph will give us acceleration, therefore,

At point t=3, ω =  0

At point t = 5, ω = 6

Acceleration = slope of the Velocity-time graph = 3 rad/sec²

Angular displacement

Using the equation,

$\theta = \omega_0t+\dfrac{1}{2}\alpha t^2$

SUbstitute values,

$\theta = (-9.0\times 8.6)+\dfrac{1}{2}(3\times 8.6^2)\\\theta = 33\rm\ radians$

Hence, the angular displacement of the wheel from 0 to 8.6s is 33 radians.

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