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) (b) At what speed will the coin start to slip? c) (c) After what period of time will the coin start to slip on the turntable?
w = alpha t = .65*3 = 1.95 radians/second
friction force = .42 * m * g = 4.12 m
Ac = w^2 R
centripetal force required = m w^2 R
m w^2 R = 4.12 m if slip is outward
Note I am ignoring tangential slip due acceleration, check later
w^2 = 4.12/R = 4.12/.18 = 22.9
w = 4.78 rad/s at spin out
4.78 = alpha t
t = 4.78/.65 = 7.36 seconds
go back and check tangential slip to make sure it is negligible
tangential acceleration = alpha R
= .65 * .18 = .117 meters/s^2
that is tiny compared to g so forget it
To solve this problem, we can use the equations of rotational motion.
(a) To find the angular velocity of the turntable after 3.00 seconds, we can use the following equation:
ω = ω0 + αt
Where ω is the final angular velocity, ω0 is the initial angular velocity (which is 0 in this case since the turntable starts from rest), α is the angular acceleration, and t is the time.
Plugging in the values, we have:
ω = 0 + (0.650 rad/s²) * (3.00 s)
ω = 1.95 rad/s
So, the angular velocity of the turntable after 3.00 seconds is 1.95 rad/s.
(b) To find the speed at which the coin will start to slip, we need to consider the maximum static friction force that can be exerted on the coin. 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.
In this case, the normal force can be calculated as the weight of the coin:
N = m * g
Where m is the mass of the coin and g is the acceleration due to gravity.
Given that the coin rests on the turntable, the normal force is equal to the weight of the coin:
N = m * g = (mass of the coin) * (acceleration due to gravity)
Next, we can calculate the maximum static friction force:
f_s(max) = μ_s * N
Then, the force causing the coin to slip is the centripetal force, which can be calculated as:
f_c = m * r * ω²
Where f_c is the centripetal force, m is the mass of the coin, r is the distance between the coin and the center of the turntable, and ω is the angular velocity of the turntable.
Since the coin is on the verge of slipping, the static friction force is equal to the centripetal force:
f_s(max) = f_c
Equating the two equations:
μ_s * N = m * r * ω²
We can substitute the expression for N:
μ_s * (m * g) = m * r * ω²
Simplifying the equation:
μ_s * g = r * ω²
Solving for the speed ω:
ω = sqrt(μ_s * g / r)
Plugging in the given values:
ω = sqrt((0.420) * (9.8 m/s²) / (0.18 m))
ω ≈ 5.24 rad/s
Therefore, the speed at which the coin will start to slip is approximately 5.24 rad/s.
(c) Finally, to calculate the period of time it takes for the coin to start slipping on the turntable, we can use the equation:
t = ω / α
Where t is the time, ω is the angular velocity, and α is the angular acceleration.
Plugging in the values:
t = (1.95 rad/s) / (0.650 rad/s²)
t ≈ 3.00 s
Therefore, after approximately 3.00 seconds, the coin will start to slip on the turntable.
To solve these questions, we need to understand the concept of static friction and its relationship to rotational motion.
(a) To find the angular velocity of the turntable after 3.00 s, we can use the equation for angular acceleration:
ω = ω₀ + αt
where:
ω is the final angular velocity,
ω₀ is the initial angular velocity (which is zero in this case since the turntable starts from rest),
α is the angular acceleration, and
t is the time.
Plugging in the values we have, we get:
ω = 0 + (0.650 rad/s²) * (3.00 s)
ω = 1.95 rad/s
Therefore, the angular velocity of the turntable after 3.00 s is 1.95 rad/s.
(b) To find the speed at which the coin will start to slip, we need to consider the maximum static friction force between the coin and the turntable. The formula for static friction is:
f_friction ≤ μ_s * N
where:
f_friction is the friction force,
μ_s is the coefficient of static friction, and
N is the normal force (which is equal to the weight of the coin in this case).
The weight of the coin can be calculated using the formula:
weight = mass * gravitational acceleration (g)
Since we don't have the mass of the coin, we need to know the gravitational acceleration, which is approximately 9.8 m/s².
Assuming the coin is on a horizontal turntable, the normal force (N) acting on the coin will be equal to its weight:
N = weight = mass * g
Now, we can substitute this value into the equation for static friction:
f_friction ≤ μ_s * weight
The speed at which the coin will start to slip is when the static friction force reaches its maximum value, which is equal to the product of the coefficient of static friction and the normal force:
f_max = μ_s * weight
To find the speed, we can use the formula for centripetal force:
f_centrifugal = mass * v² / R
Since the coin is on the turntable, the centrifugal force is acting towards the center of the turntable,
f_centrifugal = f_max
Therefore,
mass * v² / R = μ_s * weight
Since we don't know the mass of the coin, we can cancel it from both sides of the equation:
v² / R = μ_s * g
Solving for v, we get:
v = sqrt(μ_s * g * R)
Plugging in the given values, we have:
v = sqrt(0.420 * 9.8 m/s² * 0.18 m)
v = 0.619 m/s
Therefore, the speed at which the coin will start to slip is approximately 0.619 m/s.
(c) To find the period of time after which the coin starts to slip, we need to find the angular velocity at which the coin starts slipping and then calculate the corresponding time.
Using the formula for the angular velocity we found in part (a), we can rearrange the equation to solve for time:
ω = ω₀ + αt
Rearranging to solve for t:
t = (ω - ω₀) / α
Plugging in the given values, we have:
t = (1.95 rad/s - 0 rad/s) / 0.650 rad/s²
t = 3.00 s
Therefore, after 3.00 s, the coin will start to slip on the turntable.