A 15.0-g metal cylinder is placed on a turntable, with its center 73 cm from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is μs = 0.79. A thin, massless string of length 73 cm connects the center of the turntable to the cylinder, and initially, the string has zero tension in it. Starting from rest, the turntable very slowly attains higher and higher angular velocities, but the turntable and the cylinder can be considered to have uniform circular motion at any instant. Calculate the tension in the string when the angular velocity of the turntable is 55 rpm (rotations per minute).
Help! I've changed cm to m and grams to kg and plugged it in many times and cannot attain the correct answer. Can someone explain this to me please?
You don't need to give me the answer, I would like an explanation.
To find the tension in the string when the angular velocity of the turntable is 55 rpm, we can use the concept of centripetal force.
First, convert the given measurements to SI units. The radius, which is 73 cm, should be converted to meters by dividing by 100. The mass of the cylinder, which is given as 15.0 grams, should be converted to kilograms by dividing by 1000.
The radius of the turntable, r, is given as 0.73 m (since 73 cm = 0.73 m).
The mass of the cylinder, m, is given as 0.015 kg (since 15.0 g = 0.015 kg).
The coefficient of static friction, μs, is given as 0.79.
Now, let's proceed with the steps to calculate the tension in the string:
1. Determine the acceleration of the cylinder on the turntable. Since the cylinder is in uniform circular motion, its acceleration is given by centripetal acceleration, ac = ω^2 * r, where ω is the angular velocity in radians per second.
To convert 55 rpm to radians per second, use the conversion factor 1 rotation = 2π radians:
ω = (55 rpm) * (2π radians / 1 minute) * (1 minute / 60 seconds) = 5.76 radians per second (approximately).
Thus, ac = (5.76 radians per second)^2 * 0.73 m = 22.33 m/s^2 (approximately).
2. Calculate the maximum static friction force between the cylinder and the turntable. This can be done using the equation F_static_max = μs * N, where N is the normal force exerted on the cylinder.
In circular motion, the normal force N is equal to the weight of the cylinder. Thus, N = mg, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, the maximum static friction force is F_static_max = 0.79 * (0.015 kg) * (9.8 m/s^2) = 0.11 N (approximately).
3. Determine the tension in the string. At any instant, the tension in the string is equal to the centripetal force acting on the cylinder, which is provided by the static friction force. Therefore, the tension T in the string is also equal to F_static_max.
Therefore, the tension in the string when the angular velocity is 55 rpm is T = 0.11 N (approximately).
Note: It's important to ensure that all the units are consistent throughout the calculations. Double-check your unit conversions and calculations to make sure they are accurate.
To solve this problem, you need to analyze the forces acting on the metal cylinder.
First, let's consider the forces acting radially towards the center of the turntable. These forces include the tension in the string and the static friction force between the cylinder and the turntable's surface.
At any instant, the static friction force must be equal to or less than the maximum static friction force, which is given by:
Ffs(max) = μs * N
where μs is the coefficient of static friction and N is the normal force acting on the cylinder.
The normal force N can be determined by considering the weight of the cylinder and the tension in the string.
The weight of the cylinder can be calculated as:
W = m * g
where m is the mass of the cylinder and g is the acceleration due to gravity.
The tension in the string can be determined using centripetal force:
T = m * ω^2 * R
where T is the tension, m is the mass of the cylinder, ω is the angular velocity in radians per second, and R is the radius of the turntable.
By equating these forces, we can solve for the tension in the string:
T + Ffs(max) = W
Substituting the expressions for T and Ffs(max), we get:
m * ω^2 * R + μs * N = m * g
Now, substituting the given values:
m = 0.015 kg (given mass of the cylinder)
ω = 2π * 55/60 rad/s (convert 55 rpm to radians per second)
R = 0.73 m (given radius of the turntable)
μs = 0.79 (given coefficient of static friction)
g = 9.8 m/s^2 (acceleration due to gravity)
We can now plug in these values and solve for N:
(0.015 kg) * (2π * 55/60 rad/s)^2 * (0.73 m) + (0.79) * N = (0.015 kg) * (9.8 m/s^2)
Solving for N, we find:
N ≈ 0.342 N
Finally, substituting this value of N into our equation for tension, we get:
T + (0.79) * 0.342 N = (0.015 kg) * (9.8 m/s^2)
Simplifying this equation, we can solve for T:
T ≈ 0.139 N
So, the tension in the string when the angular velocity of the turntable is 55 rpm is approximately 0.139 N.
It's important to note that the calculations might vary slightly depending on the rounding of intermediate values, but this should give you a general idea of the approach to solving such a problem.