Calculating Speed of Point P in Physics Homework

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The discussion focuses on calculating the speed of point P in a physics problem involving angular motion. The user successfully determined the x-direction velocity but needed assistance with the y-direction velocity and the radius of curvature. A suggestion was made to simplify the problem by using trigonometric relationships and the Chain Rule to find expressions for the coordinates of point P. Ultimately, the user found that the velocities in both directions were equal due to the properties of a 45-degree angle, confirming that point P moves in a circular path. The conversation highlights the importance of applying fundamental physics concepts and trigonometry in solving motion problems.
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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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