Uniform circular motion problem

[J-dizzal]

Homework Statement

A police officer in hot pursuit drives her car through a circular turn of radius 356 m with a constant speed of 70.9 km/h. Her mass is 57.1 kg.What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

Homework Equations

R=v²/a, F=mv²/R, R=V₀²sin(2θ₀)/g

The Attempt at a Solution

Im don't see why my attempt at part (a) is wrong my answer is 62.2 N[/B]

 

[Simon Bridge]

How did you use the hint?
How many forces are acting on the officer?

 

[J-dizzal]

How did you use the hint?
How many forces are acting on the officer?

I guess i just found the horizontal force or the force pointing toward the center. There must also be a vertical force in the direction of the velocity, but velocity is constant.

 

[Nathanael]

There must also be a vertical force in the direction of the velocity

By "vertical" the hint means "perpendicular to the ground" (not parallel to the velocity).

 

[J-dizzal]

By "vertical" the hint means "perpendicular to the ground" (not parallel to the velocity).

so to solve i need to consider the 3rd spatial dimension?

 

[Simon Bridge]

so to solve i need to consider the 3rd spatial dimension?

... are there any forces in the third spatial dimension that could contribute to the "net force" that you are asked to find?

"horizontal" and "vertical" are sometimes, lazily, used to refer to the orientation on the page - but not always.
You have to use the context to determine what is intended.

Since the officer moves in a circle - there is no particular reason to choose "vertically on the page" to be the direction of the velocity.
You could have drawn the velocity diagonally on the page for eg. So the instructions cannot refer to the orientation on the page.

The more precise meaning of "horizontal" is "parallel to the plane of the horizon" - and "vertical" means "perpendicular to the plane of the horizon" (the word means "aligned to the high/overhead vertex").
You have correctly identified the important forces that are horizontal in this respect - in fact, the centripetal force is always horizontal so you can see how this is starting to make more sense?

 

[J-dizzal]

... are there any forces in the third spatial dimension that could contribute to the "net force" that you are asked to find?

"horizontal" and "vertical" are sometimes, lazily, used to refer to the orientation on the page - but not always.
You have to use the context to determine what is intended.

Since the officer moves in a circle - there is no particular reason to choose "vertically on the page" to be the direction of the velocity.
You could have drawn the velocity diagonally on the page for eg. So the instructions cannot refer to the orientation on the page.

The more precise meaning of "horizontal" is "parallel to the plane of the horizon" - and "vertical" means "perpendicular to the plane of the horizon" (the word means "aligned to the high/overhead vertex").
You have correctly identified the important forces that are horizontal in this respect - in fact, the centripetal force is always horizontal so you can see how this is starting to make more sense?

if the horizontal force in my diagram is the centripetal force then perpendicular to that is the velocity which would be in the vertical direction is my understanding.

 

[Nathanael]

if the horizontal force in my diagram is the centripetal force then perpendicular to that is the velocity which would be in the vertical direction is my understanding.

Whoever wrote the hint doesn't know how you drew your diagram, though. As Simon said, you could've easily (and correctly) drawn it at any angle.

At any rate, the hint is talking about the other direction (perpendicular to both velocity and acceleration). Does the seat exert a force in that direction? If so, then that force must contribute to the net force.

 

[J-dizzal]

Whoever wrote the hint doesn't know how you drew your diagram, though. As Simon said, you could've easily (and correctly) drawn it at any angle.

At any rate, the hint is talking about the other direction (perpendicular to both velocity and acceleration). Does the seat exert a force in that direction? If so, then that force must contribute to the net force.

Yes the force the officer exerts on the seat is F=ma, (57.1kg)(-1.089m/s/s)=62.2N. But this is just the opposite of the centripetal force.

 

[Nathanael]

Yes the force the officer exerts on the seat is F=ma, (57.1kg)(-1.089m/s/s)=62.2N. But this is just the opposite of the centripetal force.

Sorry, I misread and thought we were finding the force from the seat on the officer. But that is not the force I'm talking about. I said in the other (upwards) direction. Are there any forces in that direction? (The direction out of the page in your drawing.)

 

[Simon Bridge]

Yes the force the officer exerts on the seat is F=ma, (57.1kg)(-1.089m/s/s)=62.2N. But this is just the opposite of the centripetal force.

No it isn't. Unless it's a very funny corner - think about the last time you went around a corner in a car.

 

[J-dizzal]

Sorry, I misread and thought we were finding the force from the seat on the officer. But that is not the force I'm talking about. I said in the other (upwards) direction. Are there any forces in that direction? (The direction out of the page in your drawing.)

If i was in the car taking the turn the force of me on seat would be in the direction opposite the centripetal force, ignoring my weight on the seat.

 

[J-dizzal]

If i was in the car taking the turn the force of me on seat would be in the direction opposite the centripetal force, ignoring my weight on the seat.

that would be the direction going out the right side of my diagram

 

[J-dizzal]

If i was in the car taking the turn the force of me on seat would be in the direction opposite the centripetal force, ignoring my weight on the seat.

It would be a frictional force between me and the seat.

 

[Simon Bridge]

If i was in the car taking the turn the force of me on seat would be in the direction opposite the centripetal force, ignoring my weight on the seat.

... what you are saying is that only one component of the net force of you on the seat has been calculated by you.
What do you get if you do not ignore the weight?

 

[J-dizzal]

... what you are saying is that only one component of the net force of you on the seat has been calculated by you.
What do you get if you do not ignore the weight?

If the weight of me was added, then it would be a vector pointing in the -z direction?

 

[J-dizzal]

If the weight of me was added, then it would be a vector pointing in the -z direction?

But i thought the question is asking for the force on the same axis as the centripetal force. Just in the opposite direction.

 

[Simon Bridge]

What would be a vector pointing in the -z direction? Be specific!

But i thought the question is asking for the force on the same axis as the centripetal force.

... the question asks for the "net force" ... that is the vector sum of all the forces on the object.

You correctly calculated the centripetal force - but this was the wrong answer ... therefore what you thought the question was asking must be wrong. Either that or the person who designed the assignment is wrong. Pick one. If you do not follow advise I cannot help you.

 

[J-dizzal]

What would be a vector pointing in the -z direction? Be specific!

sorry, my diagram is on the xy plane x-axis being the axis parallel to the centripetal force, y-axis is perpendicular to that. positive z axis perpendicular to the xy plane in the direction coming out of the page if your looking at it. and gravity would be a vector field perpendicular to the xy plane pointing in the negative direction.

 

[Nathanael]

But i thought the question is asking for the force on the same axis as the centripetal force. Just in the opposite direction.

All the question asked for was the "net force." If they say "net force" you can be sure they want the sum of the forces in all directions.

 

[J-dizzal]

What would be a vector pointing in the -z direction? Be specific!

... the question asks for the "net force" ... that is the vector sum of all the forces on the object.

You correctly calculated the centripetal force - but this was the wrong answer ... therefore what you thought the question was asking must be wrong. Either that or the person who designed the assignment is wrong. Pick one. If you do not follow advise I cannot help you.

I don't see

All the question asked for was the "net force." If they say "net force" you can be sure they want the sum of the forces in all directions.

ok, so that would be the sum of all forces acting on the seat by the driver; mg+62N?

 

[Nathanael]

ok, so that would be the sum of all forces acting on the seat by the driver; mg+62N?

mg and 62N are vectors. You have to use vector addition.

 

[J-dizzal]

mg and 62N are vectors. You have to use vector addition.

I got the magnitude ok, but the angle I am getting is wrong, i got 90deg

 

[J-dizzal]

I got the magnitude ok, but the angle I am getting is wrong, i got 90deg

nevermind, relative to vertical let me try again.

 

[J-dizzal]

mg and 62N are vectors. You have to use vector addition.

when i try and use the dot product to find the angle between the two vectors i got 0, so then cos^-10 is 90 deg, and that would be 0 deg from vertical. i don't see what I am doing wrong here.

 

[Nathanael]

when i try and use the dot product to find the angle between the two vectors i got 0, so then cos^-10 is 90 deg, and that would be 0 deg from vertical. i don't see what I am doing wrong here.

Yes that is the angle between the two vectors. One of them is horizontal, one of them is vertical; they are perpendicular.

The question wants to know the angle between the net-force vector and the vertical. (You calculated the angle between the two components of the net force.)

 

[J-dizzal]

Yes that is the angle between the two vectors. One of them is horizontal, one of them is vertical; they are perpendicular.

The question wants to know the angle between the net-force vector and the vertical. (You calculated the angle between the two components of the net force.)

ok, i tried using θ=net force vector dotted with unit vector jhat / product of the magnitude of both vectors and got cos^-1-1 = 180deg but its wrong still

 

[Nathanael]

ok, i tried using θ=net force vector dotted with unit vector jhat / product of the magnitude of both vectors and got cos^-1-1 = 180deg but its wrong still

is jhat the unit vector in the z direction?

Show your work please.

 

[J-dizzal]

is jhat the unit vector in the z direction?

Show your work please.

 

[Nathanael]

F_T is the net force right? The magnitude of F_T is not 559.98.

Also there are 2 angles against the vertical. The problem may want you to find the smaller of the two angles.

 

[J-dizzal]

F_T is the net force right? The magnitude of F_T is not 559.98.

Also there are 2 angles against the vertical. The problem may want you to find the smaller of the two angles.

i guess i found it.

 

[J-dizzal]

F_T is the net force right? The magnitude of F_T is not 559.98.

Also there are 2 angles against the vertical. The problem may want you to find the smaller of the two angles.

thanks again Nathanael!

 

[Nathanael]

No problem. I was going to tell you about the arctan method after you found it your way. The arctan method was the first that came to my mind.
(But the arccos(a.b/(|a||b|)) method is also a very good one to know! It's far more general.)

 

[J-dizzal]

i really need to work on recognizing what is the logical next step to take with what information i have. i spend way too much time on a single problem

 

[J-dizzal]

No problem. I was going to tell you about the arctan method after you found it your way. The arctan method was the first that came to my mind.
(But the arccos(a.b/(|a||b|)) method is also a very good one to know! It's far more general.)

did you find the angle with the arccos method? what values did you use for a and b? i could only get 180degs