A coin is placed 11.0 cm 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 31 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?
forcefriction=mg*mu
centripetalforce= mw^2r
w= 31*2PIr/60 in rad/sec
set them equal, solve for mu.
i got 3.64 but that is wrong because it is too large
I didn't get that, I get on the order of .1
Check your work, or post it here.
mu*9.8=(((30/60)(2pi11))^2)/11
mu=11.8
To find the coefficient of static friction between the coin and the turntable, we can use the concept of centripetal force.
When the coin is not sliding, the static friction between the coin and the turntable provides the centripetal force required to keep the coin moving in a circle.
To start, let's calculate the speed of the coin when it slides off the turntable.
Given:
Distance from the axis, r = 11.0 cm = 0.11 m
Speed at which the coin slides off, v = 31 rpm
First, let's convert the speed from rpm to m/s:
v = 31 rpm * (2π radian/min) * (1 min/60 s) = 3.25 π m/s
Using the formula for centripetal force, Fc = m * v^2 / r, we can find the centripetal force required to maintain circular motion at this speed.
Assuming the mass of the coin, m, cancels out when dividing by the mass, we can write:
Fc = v^2 / r
Substituting the given values:
Fc = (3.25 π m/s)^2 / 0.11 m = 31.25 π N
The centripetal force required is 31.25 π N.
Now, let's calculate the maximum static frictional force that can be exerted on the coin before it starts sliding.
The maximum static frictional force, Fsf, can be expressed as:
Fsf = μs * Normal force
Where:
μs is the coefficient of static friction
Normal force is the force exerted on the coin due to gravity, which is equal to the weight of the coin, Fg = m * g
Given that the coin remains fixed on the turntable until it slides off, the maximum static frictional force is equal to the centripetal force at the point when the coin starts sliding.
Thus, we can equate Fsf to Fc:
Fsf = μs * Fg = Fc
Substituting the values:
μs * (m * g) = 31.25 π N
The mass of the coin, m, cancels out, so we are left with:
μs * g = 31.25 π N
Finally, we can solve for the coefficient of static friction, μs:
μs = (31.25 π N) / g
The value of g, the acceleration due to gravity, is approximately 9.8 m/s^2.
Calculating:
μs = (31.25 π N) / (9.8 m/s^2) ≈ 9.95
Therefore, the coefficient of static friction between the coin and the turntable is approximately 9.95.