Rotational problem

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Homework Statement


During a certain time interval, the angular position of a swinging door is described by θ = 4.92 + 10.7t + 2.07t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.
first at t=0s
then at t=3.10s
finding for each time interval [/B]
θ=
ω=
α=


Homework Equations


dθ/dt
dω/dt [/B]


The Attempt at a Solution


so finding the angular position for t=0 i just took the 4.92 because both other parts are multiplied by time which here was 0 hence 4.92+0+0 and for t=3.10 just entered that time into that equation for θ and got 57.98 which was correct. When i tried to find ω at t=0 because ω=Δθ/Δt and you cant have denominator as 0 i thought they were trying to say that it had no angular speed at t=o which seems to make sense, but the answer wasnt 0 and same thing for angular acceleration.
then for t=3.10
ω=Δθ/Δt and got 4.56 rad/s which was also wrong and so was the angular acceleration.
am i meant to be using instantaneous angular speed and acceleration? if so how do i do that?[/B]

 

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  • #2
collinsmark
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Homework Statement


During a certain time interval, the angular position of a swinging door is described by θ = 4.92 + 10.7t + 2.07t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.
first at t=0s
then at t=3.10s
finding for each time interval [/B]
θ=
ω=
α=


Homework Equations


dθ/dt
dω/dt [/B]


The Attempt at a Solution


so finding the angular position for t=0 i just took the 4.92 because both other parts are multiplied by time which here was 0 hence 4.92+0+0 and for t=3.10 just entered that time into that equation for θ and got 57.98 which was correct. When i tried to find ω at t=0 because ω=Δθ/Δt and you cant have denominator as 0 i thought they were trying to say that it had no angular speed at t=o which seems to make sense, but the answer wasnt 0 and same thing for angular acceleration.[/B]
You shouldn't simply divide by t. The answer is a bit more involved than that.
then for t=3.10
ω=Δθ/Δt and got 4.56 rad/s which was also wrong and so was the angular acceleration.
am i meant to be using instantaneous angular speed and acceleration? if so how do i do that?
Yes, I interpret the problem for asking for the instantaneous angular velocity and angular acceleration.

Do you know how to find the instantaneous slope of a curved function?

Edit: do you know another way of expressing [tex] \lim_{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t} \ ?[/tex]
 
  • #3
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You shouldn't simply divide by t. The answer is a bit more involved than that.

Yes, I interpret the problem for asking for the instantaneous angular velocity and angular acceleration.

Do you know how to find the instantaneous slope of a curved function?
Dont you take a line tangent to the point and find its gradient ??
 
  • #4
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Dont you take a line tangent to the point and find its gradient ??
Essentially, yes.

Do you know the branch of mathematics that allows you to easily find the tangent on a curved line (and it's gradient, pretty much all in one step)?
 
  • #5
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Essentially, yes.

Do you know the branch of mathematics that allows you to easily find the tangent on a curved line?
are we talking about derivatives? I am familiar with deriving functions and integrating functions but i am not sure how i could apply the first derivative to find the instantaneous slope of a curved function. hopefully you mean derivatives ?? haha
 
  • #6
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are we talking about derivatives? I am familiar with deriving functions and integrating functions but i am not sure how i could apply the first derivative to find the instantaneous slope of a curved function. hopefully you mean derivatives ?? haha
Yes, you take the derivative. The derivative of function is its slope. Taking the derivative of a function with respect to time gives the rate of change of the function.

What is the rate of change of angular position?

What is the rate of change of that? :wink:
 
  • #7
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Yes, you take the derivative. The derivative of function is its slope. Taking the derivative of a function with respect to time gives the rate of change of the function.

What is the rate of change of angular position?

What is the rate of change of that? :wink:
is the rate of change of the angular position just the angular velocity ??
and the rate of change of that would just be angular acceleration wouldn't it ??
 
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  • #8
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is the rate of change of the angular position just the angular velocity ??
and the rate of change of that would just be angular acceleration wouldn't it ??
Correct and correct! :smile:
 
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  • #9
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Correct and correct! :smile:
okay great so how do i now apply this to the question ? im still a bit confused on how i would go about solving the question
 
  • #10
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okay great so how do i now apply this to the question ? im still a bit confused on how i would go about solving the question
Well, let's start with the angular velocity (we can come back to the angular acceleration when we're finished with that).

You know that the angular velocity, [itex] \omega [/itex], is the rate of change of the angular position, [itex] \theta [/itex]. And you know that for a given function [itex] F(t) [/itex], the rate of change of that function is the derivative of that function with respect to time, [itex] \frac{d}{dt} \{F(t) \} [/itex].

So how do you express [itex] \omega (t) [/itex] in terms of [itex] \theta(t) [/itex]?
 
  • #11
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Well, let's start with the angular velocity (we can come back to the angular acceleration when we're finished with that).

You know that the angular velocity, [itex] \omega [/itex], is the rate of change of the angular position, [itex] \theta [/itex]. And you know that for a given function [itex] F(t) [/itex], the rate of change of that function is the derivative of that function with respect to time, [itex] \frac{d}{dt} \{F(t) \} [/itex].

So how do you express [itex] \omega (t) [/itex] in terms of [itex] \theta(t) [/itex]?
would it just be the first derivative of θ(t)??
 
  • #12
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would it just be the first derivative of θ(t)??
Yes. That is correct! :smile:
 
  • #13
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Yes. That is correct! :smile:
so in this main question im trying to solve, when they ask for ω i solve that by finding the answer to the first derivative of θ??
 
  • #14
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so in this main question im trying to solve, when they ask for ω i solve that by finding the answer to the first derivative of θ??
Yes, that is correct.
 
  • #15
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Yes, that is correct.
just to make sure,
θ= 4.92+10.7+2.07t2
ω = dθ/dt = 10.07+4.14t
α = dω/dt = d2 θ/d2t = 4.14
right or ??
 
  • #16
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just to make sure,
θ= 4.92+10.7+2.07t2
ω = dθ/dt = 10.07+4.14t
α = dω/dt = d2 θ/d2t = 4.14
right or ??
You missed a t in your theta, but otherwise yes, I think you have the right idea. :woot:
 
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