A coin rests 18.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.420. The turntable starts from rest at t = 0 and rotates with a constant angular acceleration of 0.650 rad/s2. (a) After 3.00 s, what is the angular velocity of the turntable? (b) At what speed will the coin start to slip? (c) After what period of time will the coin start to slip on the turntable?
To solve these problems, we can use principles of rotational motion and the concept of static friction. In each step, I'll explain the relevant equations and how to use them to find the answers.
(a) To find the angular velocity of the turntable after 3.00 s, we can use the equation:
θ = θ0 + ω0t + (1/2)αt^2,
where θ is the angular displacement, θ0 is the initial angular displacement, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.
Since the turntable starts from rest (ω0 = 0), the equation simplifies to:
θ = (1/2)αt^2.
We know the initial angular displacement is zero (θ0 = 0) and the angular acceleration is given as 0.650 rad/s^2. Plugging in these values, we have:
θ = (1/2)(0.650 rad/s^2)(3.00 s)^2.
Evaluating this expression, we find that θ = 2.925 rad. Since the coin is 18.0 cm from the center of the turntable, the angular displacement can be related to the linear displacement using the formula: θ = s/r, where s is the linear displacement and r is the distance from the center of rotation. Rearranging this equation, we have: s = θ * r.
Substituting the values, we get: s = (2.925 rad)(0.18 m) = 0.5265 m.
The linear displacement s is equal to r times the arc length, given by s = rθ. Therefore, we have: s = (0.18 m)(2.925 rad) = 0.5265 m.
So, after 3.00 s, the turntable has rotated by 2.925 radians.
(b) The coin will start to slip when the maximum static friction force is exceeded. The maximum static friction force can be calculated using the equation:
f_s,max = μ_s * N,
where f_s,max is the maximum static friction force, μ_s is the coefficient of static friction, and N is the normal force.
The normal force is equal to the weight of the coin, which is given by:
N = mg,
where m is the mass of the coin and g is the acceleration due to gravity.
We don't have the mass of the coin, so let's assume it's 1.00 kg for this example.
N = (1.00 kg)(9.81 m/s^2) = 9.81 N.
Now, we can calculate the maximum static friction force:
f_s,max = (0.420)(9.81 N) = 4.10 N.
The maximum static friction force represents the force at which the coin is at the verge of slipping.
(c) To find the period of time after which the coin starts to slip, we need to determine the angular acceleration of the turntable when the maximum static friction force is reached.
The torque acting on the coin is given by:
τ = Iα,
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
For a point mass rotating about an axis passing through one end, the moment of inertia is given by:
I = (1/3) * m * r^2,
where m is the mass of the coin, and r is the distance from the center of rotation.
Plugging in the values, we get:
I = (1/3) * (1.00 kg) * (0.18 m)^2 = 0.018 kg·m^2.
Since the torque is caused by the maximum static friction force, we can write:
τ = f_s,max * r,
where r is the distance from the center of rotation as before.
Substituting the given values, we have:
τ = (4.10 N) * (0.18 m) = 0.738 N·m.
Using the equation τ = I * α, we can solve for α:
α = τ / I = (0.738 N·m) / (0.018 kg·m^2) = 41.0 rad/s^2.
Now, we can find the time it takes for the coin to start slipping using the equation:
α = α_0 + ω_0 * t,
where α is the angular acceleration, α_0 is the initial angular acceleration (zero in this case), ω_0 is the initial angular velocity (zero in this case), and t is the time.
Rearranging the equation, we get:
t = α / α_0 = α / 0 = α/0 = infinity.
Since the initial angular acceleration is zero and remains zero, the coin will never start to slip on the turntable.