Calculating Speed of Point P in Physics Homework

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Homework Help Overview

The discussion revolves around calculating the speed of point P in a physics problem involving angular motion and displacement. The original poster attempts to find the vertical velocity component of point P while having already calculated the horizontal component. The problem involves understanding the relationship between angular velocity, radius of curvature, and the coordinates of point P.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between angular velocity and linear velocity, questioning the need for the radius of curvature. There are attempts to derive expressions for the coordinates of point P in terms of the angle and to differentiate these to find velocity components. Some participants suggest using the Chain Rule to relate differentials.

Discussion Status

There is ongoing exploration of the problem with various participants providing insights and suggestions. Some guidance has been offered regarding the use of trigonometric identities and the application of the Chain Rule. The discussion reflects a collaborative effort to clarify concepts and derive necessary expressions without reaching a definitive conclusion.

Contextual Notes

Participants note the complexity of the problem and the need to clarify the relationship between the coordinates and the angle. There is also mention of the circular motion of point P, which may influence the interpretation of distances involved.

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Homework Statement


(see attachment)


Homework Equations





The Attempt at a Solution


I was able to calculate the velocity of point P in x direction (to the left). For the speed, I need to find the velocity in y direction (vertically up). Here is my attempt:
At t=0, the distance of point P from O is l, let at time t, the angle POR be ##\theta##. The displacement is ##y=l(\sin \theta -1)##. Differentiating this equation, ##\frac{dy}{dt}=l \cos \theta \cdot \frac{d \theta}{dt}##. The term dθ/dt is equal to the angular velocity but I don't know the radius of curvature of its path which I have to use in the formula ω=vr, where ω is the angular velocity, v is the velocity in x direction and r is the radius of curvature.

Any help is appreciated. Thanks!
 

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How far does P get from O after t =0?
 
tms said:
How far does P get from O after t =0?

##l \cos \theta##.
 
Pranav-Arora said:

Homework Statement


(see attachment)


Homework Equations





The Attempt at a Solution


I was able to calculate the velocity of point P in x direction (to the left). For the speed, I need to find the velocity in y direction (vertically up). Here is my attempt:
At t=0, the distance of point P from O is l, let at time t, the angle POR be ##\theta##. The displacement is ##y=l(\sin \theta -1)##. Differentiating this equation, ##\frac{dy}{dt}=l \cos \theta \cdot \frac{d \theta}{dt}##. The term dθ/dt is equal to the angular velocity but I don't know the radius of curvature of its path which I have to use in the formula ω=vr, where ω is the angular velocity, v is the velocity in x direction and r is the radius of curvature.

Any help is appreciated. Thanks!

You're overcomplicating things. Let ∠SUT = 2θ (so you're looking for what happens when θ = 45 deg).

Take O as the origin.

Let ##u_x, u_y, p_x, p_y## be (respectively), the co-ordinates of U and P at time t.

Can you find simple expressions for those in terms of l and θ? (except that ##u_y## is simply constant at zero).

Can you find ##\frac{du_x}{dt}## in terms of ##\frac{d\theta}{dt}## using Chain Rule? Hence rearrange to determine a numerical value for ##\frac{d\theta}{dt}## at the instant of interest.

Can you differentiate ##p_x, p_y## wrt t to find the instantaneous horizontal and vertical velocities of P?

Now use Pythagoras theorem to find the resultant speed.
 
Find the coordinates of P in terms of the angle. The velocity components are the time derivatives of coordinates.

ehild

Edit:Curious beat me ...
 
Curious3141 said:
You're overcomplicating things. Let ∠SUT = 2θ (so you're looking for what happens when θ = 45 deg).

Take O as the origin.

Let ##u_x, u_y, p_x, p_y## be (respectively), the co-ordinates of U and P at time t.

Can you find simple expressions for those in terms of l and θ? (except that ##u_y## is simply constant at zero).

Can you find ##\frac{du_x}{dt}## in terms of ##\frac{d\theta}{dt}## using Chain Rule? Hence rearrange to determine a numerical value for ##\frac{d\theta}{dt}## at the instant of interest.

Can you differentiate ##p_x, p_y## wrt t to find the instantaneous horizontal and vertical velocities of P?

Now use Pythagoras theorem to find the resultant speed.

Thanks a lot Curious, that solved the problem. :smile:

The velocity of P in both the directions come out to be same!
 
Pranav-Arora said:
Thanks a lot Curious, that solved the problem. :smile:

The velocity of P in both the directions come out to be same!

That's because the sine and cosine of 45 deg are the same! :wink:
 
tms said:
How far does P get from O after t =0?
Pranav-Arora said:
##l \cos \theta##.
Not the x coordinate, but the distance from O. I was just trying to point out that P moves in a circle.
 
tms said:
Not the x coordinate, but the distance from O. I was just trying to point out that P moves in a circle.

Yep, didn't notice that at all. Thanks!
 

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