Question

A 20.0 -g metal cylinder is placed on a turntable, with its center \(80.0 \mathrm{~cm}\) from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is \(\mu_{s}=0.800\). A thin, massless string of length \(80.0 \mathrm{~cm}\) 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 60.0 rpm (rotations per minute).

Short answer

Answer: The tension in the string when the angular velocity of the turntable is 60.0 rpm is 0.472 N.

Step-by-step solution

Understand the problem and convert given units

We are given the mass of the metal cylinder (m = 20.0 g), the distance from the center of the turntable to the cylinder (r = 80.0 cm), the coefficient of static friction (μs = 0.800), the length of the string (l = 80.0 cm), and the angular velocity (ω = 60.0 rpm). First, we need to convert the given units to their appropriate SI units: 1. Mass: 20.0 g = 0.020 kg (1 g = 0.001 kg) 2. Distance: 80.0 cm = 0.80 m (1 cm = 0.01 m) 3. Angular velocity: 60.0 rpm = 6.28 rad/s (1 rpm = 0.1047 rad/s) Now, we have: m = 0.020 kg r = 0.80 m μs = 0.800 l = 0.80 m (which is equal to r) ω = 6.28 rad/s

Calculate maximum static friction force

To calculate the maximum static friction force between the cylinder and the turntable's surface, we have: \(f_{s_{max}} = μ_{s} N\) Since the cylinder is in uniform circular motion, the only force acting on it in the vertical direction is the gravitational force, which is equal to the normal force (N): \(N = mg\) Therefore, the maximum static friction force can be calculated as: \(f_{s_{max}} = μ_{s}mg\) Plugging in the values we have: \(f_{s_{max}} = (0.800)(0.020\,\text{kg})(9.81\,\text{m/s}^2) = 0.157\,\text{N}\)

Calculate centripetal force

The cylinder experiences centripetal force (Fc) due to its circular motion, which can be calculated using the formula: \(F_{c} = m⋅a_{c}\) where \(a_{c}\) is the centripetal acceleration of the cylinder, which can be found using the formula: \(a_{c} = r⋅ω^2\) Substituting the values of \(r\) and \(ω\): \(a_{c} = (0.80\,\text{m}) (6.28\,\text{rad/s})^2 = 31.46\,\text{m/s}^2\) Now, we can find the centripetal force: \(F_{c} = (0.020\,\text{kg})(31.46\,\text{m/s}^2) = 0.629\,\text{N}\)

Calculate the tension in the string

Since \(F_{c} > f_{s_{max}}\), the string will experience some tension to keep the cylinder in uniform circular motion. The tension can be found using the difference between the centripetal force and the maximum static friction force: \(T = F_{c} - f_{s_{max}}\) Plugging in the values: \(T = 0.629\,\text{N} - 0.157\,\text{N} = 0.472\,\text{N}\) Hence, the tension in the string when the angular velocity of the turntable is 60.0 rpm is 0.472 N.

Key concepts

Understanding Centripetal Force

Centripetal force is a fundamental concept in physics, specifically when dealing with the motion of objects in a circular path. It is not a force outwards, but rather the inward force that keeps an object moving in a circular path at a constant speed. The essence of centripetal force lies in its direction, which is always towards the center of the circle.

In the case of the metal cylinder on the turntable, as the turntable spins, the cylinder experiences this inward force. The formula for centripetal force is given by \(F_c = m \cdot a_c\), where \(a_c\) is the centripetal acceleration, expressed as \(a_c = r \cdot \omega^2\). By increasing the angular velocity \(\omega\), the centripetal acceleration also increases, leading to a larger centripetal force required to maintain the circular motion.

When the cylinder moves at a constant speed in a circular path, it's in a state known as uniform circular motion, where the magnitude of the centripetal force remains constant even though its direction continuously changes as the cylinder circles around.

Static Friction in Rotational Systems

Static friction plays a crucial role in scenarios involving objects at rest or moving at constant speed along a surface without slipping. It's the frictional force that prevents two surfaces from sliding past each other and is described by the coefficient of static friction (\(\mu_s\)) along with the normal force (\(N\)) through the equation \(f_{s_{max}} = \mu_s N\).

In the exercise, static friction is what initially holds the cylinder in place on the turntable. The maximum static friction force is the threshold value up to which the cylinder can stick to the turntable's surface without sliding off. If the centripetal force exceeds this maximum static friction force, the cylinder will begin to move until there's a need for additional force, in this case provided by the tension in the string, to sustain its circular path. It's important to realize that as the turntable spins faster, approaching or surpassing this maximum can lead to changes in how the cylinder moves, emphasizing the delicate balance between forces in uniform circular motion.

Angular Velocity and Its Effects

Angular velocity is defined as the rate of change of the angle by which an object rotates. In simpler terms, it describes how fast an object is spinning around a central point and is usually given in units of radians per second (rad/s), though it can also be expressed in revolutions per minute (rpm) as it is in our exercise.

To convert the angular velocity from rpm to rad/s, as shown in the solution, we use the conversion factor \(1\, \text{rpm} = 0.1047\, \text{rad/s}\). With the angular velocity, we can determine how quickly an object is moving along its circular path and thus its centripetal acceleration. The higher the angular velocity, the greater the centripetal acceleration and force needed to maintain uniform circular motion.

Understanding angular velocity allows us to predict the behavior of rotating systems, like our turntable, and harness such motion in various technologies – from car wheels to planetary orbits. In the exercise, the challenge is to ensure that as angular velocity increases, the forces at play can still maintain the cylinder in stable motion without it slipping off or requiring excessive tension in the string.