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?
(a) ω₃=ε•t₃ =0.65•3 =1.95 rad/s
(b) ma=F(fr)
centripetal acceleration a=v²/R =ω²R
F(fr) =μmg
m ω²R= μmg
ω=sqrt(μg/R)=sqrt(0.42•9.8/0.18)=4.78 rad/s
(c) ω =ε•t
t= ω/ε=4.78/ 0.65=7.35 s
To answer these questions, we need to use the equations of rotational motion and consider the forces acting on the coin.
(a) After a period of time t, the angular velocity of the turntable can be determined using the formula:
Angular velocity (ω) = Initial angular velocity + Angular acceleration * time
Since the turntable starts from rest, the initial angular velocity is 0. Therefore, the formula simplifies to:
ω = Angular acceleration * time
Substituting the given values, we have:
ω = 0.650 rad/s² * 3.00 s
= 1.95 rad/s
Therefore, the angular velocity of the turntable after 3.00 seconds is 1.95 rad/s.
(b) The coin will start to slip when the maximum static friction force between the coin and the turntable surface is exceeded. The maximum static friction force can be calculated using the formula:
Maximum static friction force (F_max) = Coefficient of static friction * Normal force
In this case, the normal force is equal to the weight of the coin, which is given by:
Weight (W) = Mass * Acceleration due to gravity
Since the coin is at rest, the net torque acting on it is zero. At the point of slipping, the friction force acts as the only torque. Therefore, the maximum static friction force is balanced by the torque equation:
F_max * radius = Moment of inertia * Angular acceleration
Simplifying the equation and substituting the given values, we get:
F_max * 0.180 m = Mass * radius² * Angular acceleration
Using Newton's second law for rotational motion, Mass = Moment of inertia / radius², we can rewrite the equation as:
F_max * 0.180 m = Moment of inertia * Angular acceleration
Substituting the given value of radius and the formula for moment of inertia of a disc (I = (1/2) * Mass * radius²), we have:
F_max * 0.180 m = (1/2) * Mass * radius² * Angular acceleration
The mass of the coin cancels out, giving us:
F_max * 0.180 m = (1/2) * radius² * Angular acceleration
The maximum static friction force can be determined by rearranging the equation:
F_max = (1/2) * radius * Angular acceleration
Finally, we substitute the given values:
F_max = (1/2) * 0.180 m * 0.420 rad/s²
= 0.0378 N
Therefore, the maximum static friction force between the coin and the turntable is 0.0378 N.
(c) To determine when the coin starts to slip on the turntable, we need to consider the maximum static friction force and compare it to the actual static friction force. At the point when the actual static friction force reaches the maximum static friction force (i.e., the point of slipping), the coin will start to slip.
The actual static friction force is given by:
Actual static friction force (F_actual) = Coefficient of static friction * Normal force
Using the same reasoning as in part (b), the normal force is equal to the weight of the coin, given by:
Weight (W) = Mass * Acceleration due to gravity
Substituting the given values, we have:
F_actual = 0.420 * Mass * Acceleration due to gravity
To find the acceleration due to gravity, we use the value of 9.8 m/s².
Finally, to solve for the time at which the coin starts to slip, we set the actual static friction force equal to the maximum static friction force, and solve for time:
0.420 * Mass * 9.8 m/s² = 0.0378 N
From the equation, we can see that the mass cancels out, so this expression is true regardless of the mass of the coin. We can then solve for time:
0.420 * 9.8 m/s² = 0.0378 N
Simplifying, we find:
4.116 = 0.0378 N
Therefore, the coin will start to slip on the turntable after approximately 4.116 seconds.