A 140 g coin sits on a horizontally rotating
turntable. The turntable makes one revolution each 2 s. The coin is located 12 cm from
the axis of rotation of the turntable.
The acceleration of gravity is 9.81 m/s
What is the frictional force acting on the
coin?
Answer in units of N.
The coin will slide off the turntable if it is
located more than 28 cm from the axis of
rotation.
What is the coefficient of static friction?
Well, it seems like the poor coin is having a spin of its life! To find the frictional force acting on the coin, we can use the centripetal force formula, which is given by:
Fc = m * (v^2 / r)
Where:
- Fc is the centripetal force
- m is the mass of the coin (140g = 0.14kg)
- v is the tangential velocity of the coin (which is the circumference of the circle divided by the time taken to complete one revolution)
- r is the distance from the axis of rotation (12cm = 0.12m)
So, let's calculate the velocity first:
v = (2 * π * r) / T
Where:
- v is the velocity
- r is the distance from the axis of rotation (0.12m)
- T is the time taken to complete one revolution (2s)
Plugging in the values:
v = (2 * π * 0.12) / 2
v ≈ 0.377 m/s
Now, let's calculate the centripetal force:
Fc = 0.14 * (0.377^2 / 0.12)
Fc ≈ 0.157 N
So, the frictional force acting on the coin is approximately 0.157 N.
Now, let's tackle the coefficient of static friction. The maximum distance from the axis of rotation that the coin can be located before sliding off is given as 28 cm = 0.28 m.
The maximum static frictional force can be calculated using the formula:
Fs(max) = μs * N
Where:
- Fs(max) is the maximum static frictional force
- μs is the coefficient of static friction
- N is the normal force (which is equal to the weight of the coin)
Since the coin is not sliding off, the static frictional force must be equal to the centripetal force we calculated earlier.
Therefore:
μs * N = Fc
μs * (m * g) = Fc
μs = (Fc) / (m * g)
Let's calculate it:
μs = (0.157) / (0.14 * 9.81)
μs ≈ 0.113
So, the coefficient of static friction is approximately 0.113.
Now, I hope the coin doesn't flip out with all these calculations!
To determine the frictional force acting on the coin, we need to consider the centripetal force required to keep the coin in circular motion.
First, let's calculate the angular velocity (ω) of the turntable. We know that the turntable completes one revolution in 2 seconds, so we can find ω using the formula:
ω = 2π / T
where T is the period of rotation. Plugging in the values, we get:
ω = 2π / 2 = π rad/s
Next, we can calculate the centripetal acceleration (ac) using the formula:
ac = ω^2 * r
where r is the distance of the coin from the axis of rotation. Plugging in the values, we get:
ac = (π rad/s)^2 * 0.12 m = π^2 * 0.12 m/s^2
Now, to find the centripetal force (FC) acting on the coin, we can use the formula:
FC = m * ac
where m is the mass of the coin. Plugging in the values, we get:
FC = 0.140 kg * π^2 * 0.12 m/s^2
Now, we know that the frictional force (Ff) between the coin and the turntable provides the necessary centripetal force to keep the coin in circular motion. Therefore:
Ff = FC
Plugging in the calculated value for FC, we get:
Ff = 0.140 kg * π^2 * 0.12 m/s^2
Now, we can calculate the numerical value for Ff:
Ff ≈ 0.166 N (rounded to three decimal places)
So, the frictional force acting on the coin is approximately 0.166 N.
Regarding the coefficient of static friction (μs), we can use the information given about the maximum displacement at which the coin will slide off the turntable. When the coin is at the threshold of sliding off, the frictional force equals the maximum static frictional force. Therefore:
Ff = μs * N
where N is the normal force. Since the weight of the coin is acting vertically downwards, the normal force is equal to the force of gravity acting on the coin:
N = m * g
Plugging in the values, we get:
N = 0.140 kg * 9.81 m/s^2
Now, we can substitute the calculated value for N into the equation:
Ff = μs * (0.140 kg * 9.81 m/s^2)
Since the maximum displacement is given to be 28 cm (0.28 m), we can use this information to find the maximum static coefficient of friction (μs). At the threshold of sliding, the maximum static friction force equals the centripetal force required to keep the coin in circular motion:
Ff = FC
Plugging in the known values, we get:
μs * (0.140 kg * 9.81 m/s^2) = 0.140 kg * π^2 * 0.28 m/s^2
Canceling out the mass term:
μs * 9.81 m/s^2 = π^2 * 0.28 m/s^2
Now, we can isolate μs:
μs = (π^2 * 0.28 m/s^2) / 9.81 m/s^2
Calculating the numerical value for μs, we get:
μs ≈ 0.282 (rounded to three decimal places)
So, the coefficient of static friction between the coin and the turntable is approximately 0.282.