A coin rests 12.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.350. The turntable starts from rest at t = 0 and rotates with a constant angular acceleration of 0.749 rad/s2.
After what period of time will the coin start to slip on the turntable?
To find the period of time at which the coin starts to slip on the turntable, we can use the concept of static friction and the equation for angular acceleration.
The static friction force acts to prevent the coin from slipping on the turntable as long as it's below the maximum value determined by the coefficient of static friction. When the maximum static friction force is reached, the coin will start to slip.
We can calculate the maximum static friction force using the equation:
F_max = μ * N
where F_max is the maximum static friction force, μ is the coefficient of static friction, and N is the normal force.
Since the coin is on a turntable, the normal force is equal to the weight of the coin, which can be calculated by multiplying the mass of the coin (assuming its mass is known) by the acceleration due to gravity (9.8 m/s^2).
Once we have the maximum static friction force, we can set it equal to the product of the mass of the coin and its radial acceleration:
F_max = m * r * α
where m is the mass of the coin, r is the distance of the coin from the center of the turntable, and α is the angular acceleration of the turntable.
Now we can set these two equations equal to each other and solve for time (t):
μ * m * g = m * r * α
To find the period of time at which the coin starts to slip, we can rearrange this equation and solve for t:
t = α / (μ * g)
Plugging in the values given in the question:
r = 12.0 cm = 0.12 m (converting cm to m)
μ = 0.350
α = 0.749 rad/s^2
Using the acceleration due to gravity g = 9.8 m/s^2, we can now calculate the period of time:
t = (0.749 rad/s^2) / (0.350 * 9.8 m/s^2) ≈ 0.0234 s
Therefore, after approximately 0.0234 seconds, the coin will start to slip on the turntable.