A coin is placced 12.0 cm from the axis of a roatating turntable of variable speed. When the speed of the turntable is slowly incraeased, the coin remains fixed on the turntable until a rate of 50 rpm is reached at which point the coin slides off. What is the coefficent of static friction between the turntable and the coin?
ok it was rpm not mph
so then in order to do one round it would take .83 seconds
Us = (rg)^-1 (t^-1 2 pi r)^2
i got .70
how do i do this
First of all, convert rpm to radians per second, omega (w).
50 rpm = (50 rev/min)*(2 pi rad/rev)*(1 min/60 s) = 5.236 radians/s
When sliding begins, the static friction force equals the centripetal force. That means:
(mus)*M g = M r w^2
mus is the static friction coefficient. M cancels out.
mus = r w^2/g
r = 0.12 m.
Solve for mus
To find the coefficient of static friction between the turntable and the coin, we can use the equation:
Us = (rg)^-1 (t^-1 2 pi r)^2
Where:
- Us is the coefficient of static friction
- r is the distance of the coin from the axis of rotation (12.0 cm or 0.12 m)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken for one rotation (0.83 seconds)
Now let's plug in the values:
Us = (0.12 m * 9.8 m/s^2)^-1 * (0.83 s^-1 * 2 * π * 0.12 m)^2
Us = (1.176 m^2/s^2)^-1 * (2π * 0.0996 m)^2
Us = 0.85 * 0.3543 m^2/s^2
Us ≈ 0.3013
Therefore, the coefficient of static friction between the turntable and the coin is approximately 0.3013.
To calculate the coefficient of static friction between the turntable and the coin, we can use the following formula:
Us = (r * g)^-1 * (t^-1 * 2π * r)^2
Where:
- Us is the coefficient of static friction
- r is the distance of the coin from the axis of rotation (12.0 cm or 0.12 m)
- g is the acceleration due to gravity (approximately 9.8 m/s²)
- t is the time taken for one complete rotation in seconds (0.83 s)
Let's plug in the values and calculate:
Us = (0.12 m * 9.8 m/s²)^-1 * (0.83 s^-1 * 2π * 0.12 m)^2
Us = (1.176 m²/s²)^-1 * (0.988 rad/s * 0.12 m)^2
Us = 0.849 * (0.118 m/s)^2
Us = 0.849 * 0.013924 m²/s²
Us ≈ 0.011825
Therefore, the coefficient of static friction between the turntable and the coin is approximately 0.011825.