Question

Help please ? http://oi67.tinypic.com/2cctjyv.jpg

Answer 1

@CShrix

Answer 2

\[\huge F_c \propto \frac{1}{r}\]
Now matter what we do, we must make this statement true. Now let's look at what we're being told\[\huge F_c \propto \frac{1}{0.25r}\]
In order to make our first statement true, what must we multiply \(F_c\) in our second equation?

Answer 3

What do you mean exactly?

Answer 4

What I'm doing is I'm solely looking at how the radius affects the centripetal force. Nothing else. Why? Because the question says so! It's asking about how radius affects the centripetal force. So we must satisfy the proportionality relationship that I gave in the first statement. As you can see, all coefficients are equal to 1. Well, then it tells us that our radius has been reduced by a factor of 1/4. So our coefficients are messed up and we have to "re-equalize" both sides so that the coefficients match 1. The coefficient that ends up in front of the centripetal force will be our answer.

So, we know that\[\huge F_c \propto \frac{1}{r}\]But now we're given that\[\huge F_c \propto \frac{1}{\frac{1}{4}r} \propto \frac{4}{r}\]What must we multiply \(F_c\) by so that we can reduce all of our coefficients back to 1 (and to get our true statement back the way it should be!)?

Answer 5

We should multiply by 4?

Answer 6

Yes! Here is how we can check our work!
\[\huge \cancel{4}F_c \propto \frac{\cancel{4}}{r} \implies F_c \propto \frac{1}{r}\]

Answer 7

Okay so then what would we do with the .25?

Answer 8

Nothing, we already accounted for it when we multiplied the left side by 4.
0.25=1/4. The 1/4 in the denominator became the 4 on the numerator. Basic algebraic techniques with fractions.

Answer 9

Alright. I was over thinking it then.

Answer 10

Yep! X)

Now for the second question, replace v with (3v) and do the same process. Let me know what you get!

Answer 11

Okay so it's 3^2/ r then?

Answer 12

I'll get you started here...
We know that we're trying to observe centripetal force and how velocity affects it. Therefore, we should isolate those two variables specifically. In contrast to our previous statements, we now create a new relationship:
\[\huge F_c \propto v^2\]Now substitute v with (3v) and follow a similar process.

Answer 13

Okay it's 9

Answer 14

Yep! X)

\[\huge F_c \propto 9v^2 \implies \cancel{9}F_c \propto \cancel{9}v^2\]As you can see, in order get the same relationship that the equation given in the question tells us, we have to multiply the force by 9! So in other words, if we triple the velocity, then our force increases by a factor of 9!

Answer 15

**Assuming everything else remains constant!

Answer 16

If both velocity and radius are simultaneously changing, we have a much bigger and more complicated problem at hand.

Answer 17

Right. Okay so
A car is driving around a curve that can be approximated as being circular. What direction does the centripetal force point?

Answer 18

I know static friction is involved since the tires are not moving.

Answer 19

The centripetal force always points inward (hence the term 'centripetal'). However, it's a fictitious force. There's no actual centripetal force that acts upon a car. It's more like a "category" of a force or a trait of a force. In the case where a car is driving in a circle, the friction acts as the centripetal force.
http://www.batesville.k12.in.us/physics/phynet/mechanics/circular%20motion/an_unbanked_turn.htm

Answer 20

Okay so it has to go inward so that the car will actually turn

Answer 21

Correct!

Answer 22

Thanks!

Answer 23

You are very welcome X)