Finding the Time of Max and 0 Angular Speed

[Raymour]

Homework Statement

Mod note: Fixed thread title. OOPS NOT Acceleration, Speed! (Thread title is incorrect)

Homework Equations

w=d∅/dt
v=rw
x={+,-√(b²-4ac)}/2a

The Attempt at a Solution

I solved this with help from Chegg study, however, I'm still not entirely sure what I am doing. Obviously, for part a, you differentiate the position to get the angular velocity which is w=5t-1.8t². At that point I struggled. Chegg answers simply differentiated twice to find the time of maximum angular speed as 1.39 s. Is that because ∝=5-3.6t gives the time of maximum acceleration? and therefore max speed?

My first instinct was to perform the quadratic of the speed equation to give me t= 2.78 s and 1.39 s. One is the time of max speed and one is when it is zero (aka part c for when it will reverse direction)... How do I know which is which other than having ruled 1.39 out in part a? For part c, Chegg used the equation t(5t-1.8t²)=0 to find the time.But I'm not sure where that equation came from. Where are they taking the t out front from? Is there a better way to do this? Once I can get the time, I am fine to solve the rest.

 

[Mark44]

View attachment 208018

Homework Statement

Mod note: Fixed thread title. OOPS NOT Acceleration, Speed! (Thread title is incorrect)

Homework Equations

w=d∅/dt
v=rw
x={+,-√(b²-4ac)}/2a

The Attempt at a Solution

I solved this with help from Chegg study, however, I'm still not entirely sure what I am doing. Obviously, for part a, you differentiate the position to get the angular velocity which is w=5t-1.8t². At that point I struggled. Chegg answers simply differentiated twice to find the time of maximum angular speed as 1.39 s. Is that because ∝=5-3.6t gives the time of maximum acceleration? and therefore max speed?

Although you could differentiate again, in this case it isn't required. Your angular velocity, $\omega(t)$, is a quadratic polynomial whose graph is a parabola, so to find the times where $\omega$ is zero, simply solve the equation $5t - 1.8t^2 = 0$. Note that there will be two times.
To find the max. angular velocity, locate the vertex of the parabola.

My first instinct was to perform the quadratic of the speed equation to give me t= 2.78 s and 1.39 s.
One is the time of max speed and one is when it is zero (aka part c for when it will reverse direction)... How do I know which is which other than having ruled 1.39 out in part a? For part c, Chegg used the equation t(5t-1.8t²)=0 to find the time.But I'm not sure where that equation came from.

See what I wrote above. If you recognize that the graph of the angular velocity is a parabola, it should be straightforward to figure out which one is the max. point, and which is a point where $\omega$ is zero.

The answer they showed (2.78 s) does not really show what they did with that equation. Is there a better way to do this? Once I can get the time, I am fine to solve the rest.

 

[Raymour]

Although you could differentiate again, in this case it isn't required. Your angular velocity, $\omega(t)$, is a quadratic polynomial whose graph is a parabola, so to find the times where $\omega$ is zero, simply solve the equation $5t - 1.8t^2 = 0$. Note that there will be two times.
To find the max. angular velocity, locate the vertex of the parabola.
See what I wrote above. If you recognize that the graph of the angular velocity is a parabola, it should be straightforward to figure out which one is the max. point, and which is a point where $\omega$ is zero.

I suppose that means I should start taking my graphing calculator to exams! Once I graphed it, I saw it immediately. Thank you! How do I solve this via equations, though? So differentiating twice solves the time of max speed. Is there any other method than graphing to find when the speed is 0? I mean I see how they solved t(5t-1.8t²), however I have no clue where they got the t out front from.

 

[Mark44]

I suppose that means I should start taking my graphing calculator to exams! Once I graphed it, I saw it immediately. Thank you! How do I solve this via equations, though? So differentiating twice solves the time of max speed. Is there any other method than graphing to find when the speed is 0? I mean I see how they solved t(5t-1.8t²), however I have no clue where they got the t out front from.

A graphing calculator isn't necessary, but recognizing a quadratic function is necessary. To find the $\omega$ intercepts, solve the equation $5t - 1.8t^2 = 0$, which you can do by factoring to $t(5 - 1.8t) = 0$. To find the maximum point complete the square in $\omega = 5t - 1.8t^2$. This will result in $\omega = -1.8(t - a)^2$, where a is the coordinate of the vertex.