How Is Effective Weight Calculated in a Spinning Carnival Ride?

[Violagirl]

Homework Statement

On a carnival ride, a girl of mass m stands in a cylindrical cage of radius R. The cage is spun about its cylindrical axis so that her speed is v. What is her effective weight?

Homework Equations

w^e= mg-ma

w_x^e = mv/R², w^e_y = -mg

The Attempt at a Solution

This is an example problem in my book and I'm not sure how they put everything together to solve it. Upon presenting the above forumulas, it says to next use the Pythagorean theorem for it to get:

w^e = m √g² + (v^2/R)2

Then goes on to say that the girl's effective weight is greater than mg.
I'm not sure how they set up their formula into the Pythagorean formula from the formulas above. Any clarification would be appreciated.

 

[BruceW]

mg-ma is only the effective weight when the acceleration and gravity are in the same direction (or with sign of 'a' reversed for opposite directions). But in this problem, gravity and acceleration are not parallel or anti-parallel.

Think about the centripetal and gravitational forces. Which directions will they be in? From this, you can see why they use Pythagoras' formula.

 

[Violagirl]

mg-ma is only the effective weight when the acceleration and gravity are in the same direction (or with sign of 'a' reversed for opposite directions). But in this problem, gravity and acceleration are not parallel or anti-parallel.

Think about the centripetal and gravitational forces. Which directions will they be in? From this, you can see why they use Pythagoras' formula.

Oh! Ok, that makes more sense, I didn't understand in reading my textbook, thanks for clarifying! So then, they each form their own vectors in coming together to produce a triangle, which is why Pythagorean Thorem is needed. Thank you!

 

[BruceW]

yep, that's it :)

 

[Nickie]

To solve this problem, we first need to understand what is meant by "effective weight". Effective weight is the apparent weight experienced by an object when it is accelerating or moving in a circular motion. In this case, the girl is moving in a circular motion due to the spinning of the cylindrical cage.

To calculate the effective weight, we can use the formula: we = mg - ma, where m is the mass of the girl, g is the acceleration due to gravity, and a is the centripetal acceleration (which is the acceleration towards the center of the circle).

We can also use the formula for centripetal acceleration: a = v^2/R, where v is the speed of the girl and R is the radius of the cylindrical cage.

Using these formulas, we can rewrite the equation for effective weight as: we = mg - mv^2/R.

Now, to use the Pythagorean theorem, we need to find the components of the girl's weight in the x and y directions. The x-component is given by wxe = mv/R^2, and the y-component is given by wey = -mg.

Using the Pythagorean theorem, we can find the magnitude of the effective weight (we) as: we = √(wxe^2 + wey^2).

Substituting the values for wxe and wey, we get: we = √(mv^2/R^2 + mg^2).

We can simplify this to: we = m √(v^2/R^2 + g^2).

Therefore, the formula given in the attempt at a solution is correct. The girl's effective weight is greater than her actual weight (mg) because she is experiencing an additional force due to the centripetal acceleration.

Hope this helps clarify the solution for you!