Question

A bead slides without friction around a loop–the–loop (see figure below). The bead is released from rest at a height h = 3.80R.

What is its speed at point A? (Use the following as necessary: the acceleration due to gravity g, and R.)
V=__

How large is the normal force on the bead at point A if its mass is 4.10 grams?
magnitude=__
Direction=downward

Answer 1

You can use energy principles, right?

Answer 2

Yeah, I'm okay with it.. I'm stuck on that ^ "R"..

Answer 3

It's just a constant, treat it like one.

Answer 4

Think of it like a unit, like meters.

Answer 5

The height goes from 3.80R to 2R

Answer 6

Figure out the change in energy.

Answer 7

I got it! sqrt(3.6gR) (: yay!

Answer 8

\[
KE = \Delta PE = m\times g\times (3.80-2)R = 1.80mgR
\]\[\begin{split}
KE = \frac{1}{2}mv^2 = 1.80mgR &\implies mv^2 = 3.60mgR \\
&\implies v = \sqrt{3.60gR}

\end{split}
\]

Answer 9

Now you can use \[
a= \frac{v^2}{R} \implies F= ma = \frac{mv^2}{R}
\]

Answer 10

uhm^ I'm lost now

Answer 11

Do you know about \[
a = \frac{v^2}{R}
\]

Answer 12

The inward acceleration \(a\) of a body moving in a circle of radius \(R\) with velocity \(v\).

Answer 13

3.60gR/R?

Answer 14

Yeah, the Rs cancel each other out.

Answer 15

To change that acceleration into Force, you just multiply the mass in.

Answer 16

3.60mg?

Answer 17

if its mass is 4.10 grams?

Answer 18

144.8

Answer 19

You want to convert grams to kilos I think.

Answer 20

14.48