A 6.80 g coin is placed 10.0 cm from the center of a turntable. The coin has a coefficient of static friction of 0.80 and a coefficient of kinetic friction of 0.60 with the turntable. The turntable's angular velocity slowly speeds up until the coin starts to slide off. Assuming that the local acceleration due to gravity is -9.80 m/s2?, calculate the angular speed (in rev/min) where the coin slides off.
mw^2r>mg*mustatic
w^2>g*.8/.1
w>sqrt(9.8*.8/.1) in rad per sec
Now in rev, min, w=sqrt(9.8*.8/.1)*60/2PI
To solve this problem, we need to understand the forces acting on the coin and how they affect its motion.
1) First, let's calculate the net force acting on the coin. The force causing the coin to slide off the turntable is the maximum static friction force, given by:
F_friction = μ * N
where μ is the coefficient of static friction and N is the normal force.
The normal force is equal to the weight of the coin, which can be calculated as:
N = m * g
where m is the mass of the coin and g is the acceleration due to gravity.
In this case, the mass of the coin is given as 6.80 g. To convert grams to kilograms (kg), divide by 1000:
m = 6.80 g / 1000 = 0.0068 kg
The acceleration due to gravity is given as -9.80 m/s^2, taking the negative value because it is acting downwards. Let's substitute the values into the equation:
N = 0.0068 kg * (-9.80 m/s^2) = -0.06664 N
Note that the negative sign indicates that the normal force is acting in the opposite direction to the acceleration due to gravity.
Now, we can calculate the maximum static friction force:
F_friction = 0.80 * (-0.06664 N) = -0.053312 N
2) Next, let's examine the forces that cause the coin to remain on the turntable as it rotates. When the turntable rotates, there is a centrifugal force acting away from the center of rotation. This force can be calculated using the equation:
F_centrifugal = m * r * ω^2
where r is the radius (distance from the center of the turntable to the coin) and ω is the angular velocity.
In this case, the radius r is given as 10.0 cm, which can be converted to meters by dividing by 100:
r = 10.0 cm / 100 = 0.10 m
Now we can substitute the values into the equation:
F_centrifugal = 0.0068 kg * 0.10 m * ω^2
3) At the point where the coin just starts to slide off, the net force acting on the coin will be zero:
F_net = F_centrifugal - F_friction = 0
Substituting the values we calculated earlier, we have:
0.0068 kg * 0.10 m * ω^2 - (-0.053312 N) = 0
4) Now we can solve the equation for ω (angular velocity), which will give us the value at which the coin starts to slide off. Rearranging the equation:
0.0068 kg * 0.10 m * ω^2 = 0.053312 N
ω^2 = 0.053312 N / (0.0068 kg * 0.10 m)
ω^2 = 7838.82 rad^2/s^2
Taking the square root of both sides:
ω = √(7838.82 rad^2/s^2) = 88.54 rad/s
5) Finally, we can convert the angular velocity to revolutions per minute (rev/min). Since 1 revolution is equal to 2π radians, and 1 minute is equal to 60 seconds, we have:
ω (rev/min) = (88.54 rad/s) * (1 rev / 2π rad) * (60 s / 1 min) = 267.27 rev/min
Therefore, the angular speed at which the coin slides off the turntable is approximately 267.27 rev/min.