Scale on Top of Rollercoaster

[doug1]

Homework Statement

You take a bathroom scale onto a stand-up rollercoaster (the riders are always in a standing position). Before the ride
starts, you stand on the scale and note that it reads 65 kg. The ride starts and at the top of the first loop you are upsidedown and your scale reads 77 kg. You estimate that the loop has a radius of 15 m. How fast were you going at the top
of the loop?

Homework Equations

Fn = reading of scale

Fg = mg

ac = v^2/r

The Attempt at a Solution

I drew a FBD where the normal force and the force of gravity both point downwards towards the centre of the circle. The centripetal force equals the sum of the normal force and the force of gravity:

Fc = Fn + Fg

The problem is that I don't know what Fn is. Is it (77kg)(9.8N/kg)? If this is the case, I would then do:

Fc = (77kg)(9.8N/kg) + (9.8)(65)

I would then divide this value by 65 to get the centripetal acceleration (21.4 m/s^2).

I would then plug 21.4 m/s^2 into the equation ac = v^2/r and solve for v (about 17.9 m/s)

 

[collinsmark]

Homework Statement

You take a bathroom scale onto a stand-up rollercoaster (the riders are always in a standing position). Before the ride
starts, you stand on the scale and note that it reads 65 kg. The ride starts and at the top of the first loop you are upsidedown and your scale reads 77 kg. You estimate that the loop has a radius of 15 m. How fast were you going at the top
of the loop?

Homework Equations

Fn = reading of scale

Fg = mg

ac = v^2/r

The Attempt at a Solution

I drew a FBD where the normal force and the force of gravity both point downwards towards the centre of the circle. The centripetal force equals the sum of the normal force and the force of gravity:

Fc = Fn + Fg

The problem is that I don't know what Fn is. Is it (77kg)(9.8N/kg)?

Yes, it is.

Bathroom scales are good at measuring the normal force on a level surface (regarding the normal force acting on objects standing on the scale). That's pretty much all they're designed to do.

It's a bit tricky here only because the particular bathroom scale you are working with gives its measurement in units of kilograms -- a measure of mass, not force. The bathroom scale assumes (correctly or not) that the total acceleration has magnitude g (i.e. 9.8 m/s²). So you are correct when you equate F_n = (77 [kg])(9.8 [N/kg]).

If this is the case, I would then do:

Fc = (77kg)(9.8N/kg) + (9.8)(65)

I would then divide this value by 65 to get the centripetal acceleration (21.4 m/s^2).

I would then plug 21.4 m/s^2 into the equation ac = v^2/r and solve for v (about 17.9 m/s)

That looks right to me.*

*(There are some assumptions being made such that the radius of the loop is in regard to your own center of gravity as you move around the loop [which is not quite the radius of the track], and g = 9.8 N/kg. And there's also an assumption that the top plate of the bathroom scale has negligible mass. But regardless of all that, you're on the right track. )