Linear & Angular Velocity Related Rates

In summary: From implicitly differentiating with respect to time, I got the equation sec^2(Θ)*dΘ/dt = dx/dt Since we know that dr/dt(rate of change of revolution of beacon) = 3 revs/min, I used dimensional analysis to get that dΘ/dt at...the point (##P##,##x##) where the spotlight is hitting the shore.the point (##P##,##x##) where the spotlight is hitting the shore.
  • #1
Michele Nunes
42
2

Homework Statement


A rotating beacon is located 1 kilometer off a straight shoreline (see figure). If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be moving to a viewer who is 1/2 kilometer down the shoreline?

Homework Equations

The Attempt at a Solution


Okay so I used the equation w = 2πf where w is the angular velocity and f is the frequency of rotation. I implicitly differentiated with respect to time, then plugged in 3 rev/min for df/dt, then converted to km/hr, and got 360π km/hr. The answer is 450π km/hr. I think my units are off somewhere or I'm using the wrong equation but I'm not sure.
 

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  • #2
Michele Nunes said:

Homework Statement


A rotating beacon is located 1 kilometer off a straight shoreline (see figure). If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be moving to a viewer who is 1/2 kilometer down the shoreline?

Homework Equations

The Attempt at a Solution


Okay so I used the equation w = 2πf where w is the angular velocity and f is the frequency of rotation. I implicitly differentiated with respect to time, then plugged in 3 rev/min for df/dt, then converted to km/hr, and got 360π km/hr. The answer is 450π km/hr. I think my units are off somewhere or I'm using the wrong equation but I'm not sure.

What was the distance you used in the conversion? Can you type out your work step by step?
 
  • #3
Student100 said:
What was the distance you used in the conversion? Can you type out your work step by step?
d/dt[w] = d/dt[2πf]
dw/dt = 2π(df/dt)
dw/dt = 2π(3 rev/min) = 6π km/min = 360π km/hr
 
  • #4
Michele Nunes said:
d/dt[w] = d/dt[2πf]
dw/dt = 2π(df/dt)
dw/dt = 2π(3 rev/min) = 6π km/min = 360π km/hr

What about the distance from the shore to the light house? How are you going from revs to km?
 
  • #5
I honestly don't know how to relate the units of revolutions to kilometers, that's what I'm not understanding, like they give me the frequency of revolutions of the beacon but I don't know what to do with that to end up getting kilometers/hour.
 
  • #6
Michele Nunes said:
I honestly don't know how to relate the units of revolutions to kilometers, that's what I'm not understanding, like they give me the frequency of revolutions of the beacon but I don't know what to do with that to end up getting kilometers/hour.

First use your trig relationship to find the distance from the observer on the shore to the beacon house.
 
  • #7
Student100 said:
First use your trig relationship to find the distance from the observer on the shore to the beacon house.
Okay it's sqrt(5)/2 but what do I do with it? I've looked up angular velocity equations and they all have like 1 too many variables that I don't have enough information on.
 
  • #8
Let the lighthouse be ##Q##, let ##P## be the point on the shore nearest the lighthouse, let ##x## be the distance from ##P## to where the spotlight hits the shore, and ##\theta## be the angle of rotation from ##QP##. What you need to calculate is ##\frac{dx}{dt}##. You don't need the distance from the observer on shore to the beacon house to work this problem.
 
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  • #9
LCKurtz said:
Let the lighthouse be ##Q##, let ##P## be the point on the shore nearest the lighthouse, let ##x## be the distance from ##P## to where the spotlight hits the shore, and ##\theta## be the angle of rotation from ##QP##. What you need to calculate is ##\frac{dx}{dt}##. You don't need the distance from the observer on shore to the beacon house to work this problem.
Okay I got it now, thank you!
 
  • #10
Michele Nunes said:
Okay I got it now, thank you!
If you don't mind, could you tell me how you ended up solving this problem? I was trying it myself but I'm not sure if I have the right equation to differentiate or if I'm using the rate of the rotating beacon light correctly.
 
  • #11
Thewindyfan said:
If you don't mind, could you tell me how you ended up solving this problem? I was trying it myself but I'm not sure if I have the right equation to differentiate or if I'm using the rate of the rotating beacon light correctly.

Show us what you did if you want advice or help.
 
  • #12
LCKurtz said:
Show us what you did if you want advice or help.
Okay so based off your guidance in the last post, I basically sorted out what was given and what we want. You mentioned that the hypotenuse or the length of the beam isn't necessary for this problem so I decided to use the equation tan(Θ) = (x)/1
From implicitly differentiating with respect to time, I got the equation sec^2(Θ)*dΘ/dt = dx/dt
Since we know that dr/dt(rate of change of revolution of beacon) = 3 revs/min, I used dimensional analysis to get that dΘ/dt at that moment is 6π rad/min, but then I realized I definitely went wrong somewhere with my related equation since I would still need to know the length of the beam of light and even when I did, I got a pretty small number. I'm stuck on exactly how you have to deal with the dr/dt and what the best equation to relate these rates would be for this specific problem.

Thanks in advance!
 
  • #13
Thewindyfan said:
From implicitly differentiating with respect to time, I got the equation sec^2(Θ)*dΘ/dt = dx/dt
Since we know that dr/dt(rate of change of revolution of beacon) = 3 revs/min, I used dimensional analysis to get that dΘ/dt at that moment is 6π rad/min
You are almost there.
Hint: ##1+\tan^²(\theta)=\sec²(\theta)##
 
  • #14
Thewindyfan said:
Okay so based off your guidance in the last post, I basically sorted out what was given and what we want. You mentioned that the hypotenuse or the length of the beam isn't necessary for this problem so I decided to use the equation tan(Θ) = (x)/1
From implicitly differentiating with respect to time, I got the equation sec^2(Θ)*dΘ/dt = dx/dt
Since we know that dr/dt(rate of change of revolution of beacon) = 3 revs/min,

That would be ##\frac{d\theta}{dt}##, and it is the rate of revolution, not the "rate of change" of it. And don't forget the final units requested in the problem are kilometers/hour for ##\frac{dx}{dt}##.

I used dimensional analysis to get that dΘ/dt at that moment is 6π rad/min, but then I realized I definitely went wrong somewhere with my related equation since I would still need to know the length of the beam of light and even when I did, I got a pretty small number. I'm stuck on exactly how you have to deal with the dr/dt and what the best equation to relate these rates would be for this specific problem.

Thanks in advance!
 
Last edited:
  • #15
Samy_A said:
You are almost there.
Hint: ##1+\tan^²(\theta)=\sec²(\theta)##
Wow I can't believe I haven't thought of using this identity for these types of related rates problems! Thanks for the tip.

LCKurtz said:
That would be ##\frac{d\theta}{dt}##, and it is the rate of revolution, not the "rate of change" if it. And don't forget the final units requested in the problem are kilometers/hour for ##\frac{dx}{dt}##.

Ah so I just forgot to have ##\frac{d\theta}{dt}## in the proper units before plugging in the numbers into the differentiated equation! Thank you for pointing that out! I haven't seen a related rates problem like this so I guess I was overthinking about it.
 

Related to Linear & Angular Velocity Related Rates

1. What is the difference between linear and angular velocity?

Linear velocity is the rate of change of an object's position in a straight line, while angular velocity is the rate of change of an object's angular position around a fixed point. Linear velocity is measured in units such as meters per second, while angular velocity is measured in units such as radians per second.

2. How are linear and angular velocity related?

Linear and angular velocity are related through the formula v = rw, where v is linear velocity, r is the distance from the object to the fixed point, and w is the angular velocity. This relationship shows that an object's linear velocity is directly proportional to its angular velocity and the distance from the fixed point.

3. Can you give an example of a real-life application of linear and angular velocity related rates?

One example is the speedometer in a car. The linear velocity of the car is related to the angular velocity of the car's wheels through the radius of the wheels. As the car's wheels rotate, the speedometer measures the linear velocity of the car and displays it in units of kilometers per hour or miles per hour.

4. What is the role of calculus in understanding linear and angular velocity related rates?

Calculus is essential in understanding linear and angular velocity related rates because it allows us to calculate the rate of change of these quantities. By taking derivatives and using the chain rule, we can find the relationship between linear and angular velocity and how they change over time.

5. How does the direction of linear and angular velocity affect their relationship?

The direction of linear and angular velocity affects their relationship in that they are always perpendicular to each other. This means that when an object's linear velocity is changing, its angular velocity will also change in a different direction, and vice versa. This concept is known as the right-hand rule, where the direction of the fingers represents the direction of the linear velocity, and the direction of the thumb represents the direction of the angular velocity.

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