A coin is placed 17.0 from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 33 is reached and the coin slides off.What is the coefficient of static friction between the coin and the turntable?

forcefriction=forcecentripetal

mu*mg=m*v^2/r

At this point, I don't know what your 33 is? m/s? rpm?

oops sorry, 33 rpm and 17 cm

I have a very similar problem due on mastering physics in an hour.. I came here to get help but now 30 minutes later I'm actually going to be giving help :). Hopefully it wont be too late depending on your time zone.

u= coefficient of static friction
force of friction = u*mg

IM NOT THAT SMART SO THIS MIGHT BE WRONG BUT HOPEFULLY IT WILL HELP.. I ALSO SUCK AT SIG FIGS SO ITS PROBABLY SLIGHTLY OFF.

set force of friction = to centripetal force

so u*m*g = m*v^2/r

m cancels

so u*g = v^2/r.
r = 17cm = .17m

the circumference of the circle = 2pi*r
which roughly = 1.068m (only 2 sig figs tho)

convert 33 rpm to rps to get .55 rps.

so the object move .55 rotations in one second so it moves .55*1.068m in one second. so it moves .5874m/s reduce this to 2 sig figs to get .59m/s

use the equation we formed earlier :
u*g = v^2/r.

u= v^2/rg
u= (.59m/s)^2/((.17m)(9.80m/s^2))
u=.21

THANK YOU!!!!!!!

TY! TY!

To determine the coefficient of static friction between the coin and the turntable, we can use the centripetal force required to keep the coin in place before it slides off.

First, let's calculate the centripetal force using the formula:

Centripetal Force (F) = Mass (m) × Radius (r) × Angular Velocity (ω)^2

Given that the radius of the turntable is 17.0 cm (0.17 m), the angular velocity is 33 rad/s, and the mass of the coin is unknown (m).

Now, let's assume the mass of the coin is represented by m.

The gravitational force acting on the coin is given by:

Gravitational Force (Fg) = Mass (m) × Gravity (g)

The frictional force between the coin and the turntable is equal to the centripetal force required to keep the coin in place, which is:

Frictional Force (Ff) = Static Frictional Coefficient (μs) × Normal Force (N)

Since the coin is placed on the turntable, the normal force is equal to the gravitational force acting on the coin:

Normal Force (N) = Mass (m) × Gravity (g)

Since the coin is at rest, the gravitational force is equal to the frictional force:

Gravitational Force (Fg) = Frictional Force (Ff)

Therefore, we can set the equations equal to each other:

Mass (m) × Gravity (g) = Static Frictional Coefficient (μs) × Mass (m) × Gravity (g)

Simplifying:

μs = (Gravity (g)) / (Gravity (g))

Since gravity cancels out, the coefficient of static friction (μs) is equal to 1.

Therefore, the coefficient of static friction between the coin and the turntable is 1.