Question

A 10 g steel marble is spun so that it rolls at 150 rpm around the inside of a vertically oriented
steel tube (coefficient of static friction is 0.8, coefficient of kinetic friction is 0.6). The tube,
shown in Figure P8.39, is 12 cm in diameter. Assume that the rolling resistance is small enough
for the marble to maintain 150 rpm for several seconds. During this time, will the marble spin in
a horizontal circle, at constant height, or will it spiral down the inside of the tube? If the marble
spins in a horizontal circle what is the magnitude of the force of friction?

Answer 1

If the marble spirals
down how much larger would the coefficient of friction need to be in order for the marble to
spin in a horizontal circle?

Answer 2

# 2 on assignment to see figures. Help needed badly :(

Answer 3

can you explain whats going on in the pictures and what we need to find, and why?

Answer 4

find N, (find V in order to find N)
if mu*N < mg it slips down as it goes around.

Answer 5

okay

Answer 6

make sense?

Answer 7

yeah, so to change my rpm in to m/s if I multiply 150 by2pi/60 I think that would change it into angular velocity, and then I can multiply that by the radius:
150*2pi/60= 12.71
12.71*6 = 94.25m/s??

Answer 8

yes.

Answer 9

uh well r needs to be in meters...

Answer 10

if you want your V in m/s

Answer 11

nice catch!

Answer 12

12.71*0.06= 0.7626m/s

Answer 13

150*2pi/60 = 5pi

Answer 14

5pi= 15.71

Answer 15

15.71*0.06= 0.942 m/s

Answer 16

yes

Answer 17

0.942^2 = 0.89
0.89*0.01= 0.0089
0.0089/0.06= 0.15N

Answer 18

looks fine. you could save yourself some calculation by just doing it symbolically..
mu*(m (omega*r)^2)/r <=mg

mu*omega^2*r <= g

Answer 19

so .8*(15.71)^2*.06 <= 9.8

Answer 20

then that means it slips down.

Answer 21

yep

Answer 22

Wow, thank you once again! :D It's greatly appreciated!

Answer 23

sure:)