Question

Compare the equation for a line through the origin, y = mx, to the equation Fc= mv^2/r , and explain what the slope of the graph of Fc vs. v2 represents. Remember that the m in the equation for a line represents the slope, and the m in the centripetal force equation represents the mass of the stopper.

I have no idea what they are asking with comparing the slopes

Answer 1

@Luigi0210

Answer 2

\(\color{#000000 }{ \displaystyle y=mx }\)
A line with a positive slope \(\color{#000000 }{ \displaystyle m }\), that
goes through the origin. Sometimes,
this would be referred to as, \(\tt"\)direct
variation\(\tt"\). The \(\color{#000000 }{ \displaystyle x }\) is proportional to
\(\color{#000000 }{ \displaystyle y }\), and the proportion constant is \(\color{#000000 }{ \displaystyle m }\).

\(\color{#000000 }{ \rm F_c=\frac{\large m\times v^2}{\large r} }\)
The equation of the Cetripetal Force,
of some particle that moves along a
circle.

where,
\(\color{#000000 }{ \displaystyle {\rm F_c} }\) Centripetal Force
\(\color{#000000 }{ \displaystyle \rm m }\) Mass
\(\color{#000000 }{ \displaystyle \rm v }\) Velocity of the partical
\(\color{#000000 }{ \displaystyle \rm r }\) Radius of the circle
\(\color{#000000 }{ \displaystyle \rm A_{\large c} }\) Centripetal acceleration

The cerntripetal acceleration is given by
\(\color{#000000 }{ \rm A_{\large c}=\frac{\Large v^2}{\Large r} }\)

So, alternatively, you can write
\(\color{#000000 }{ \rm F_{\large c}=m\times A_{\large c} }\)

And that is also a line that shows
proportion between cetripetal force
and centripetal acceleration.
(This force equation is just like any other \(F=ma\))