Explanation:
To determine the minimum values of the static friction coefficients, we can use the formula for centripetal force:
\[ F_{\text{centripetal}} = \mu \cdot F_{\text{normal}} \]
The centripetal force experienced by each parcel is provided as:
\[ F_{\text{centripetal}} = m \cdot \omega^2 \cdot r \]
From your previous calculations:
- \( m_1 = 2.1 \, \text{kg} \), \( r_1 = 1.2 \, \text{m} \), \( F_{\text{centripetal}_1} = 0.11 \, \text{N} \)
- \( m_2 = 4.0 \, \text{kg} \), \( r_2 = 2.4 \, \text{m} \), \( F_{\text{centripetal}_2} = 0.42 \, \text{N} \)
Using the formula \( F_{\text{centripetal}} = \mu \cdot F_{\text{normal}} \), and \( F_{\text{normal}} = m \cdot g \) where \( g \) is the acceleration due to gravity, we can calculate the minimum values of the static friction coefficients \( \mu \).
For \( m_1 \):
\[ F_{\text{centripetal}_1} = \mu_1 \cdot m_1 \cdot g \]
\[ 0.11 \, \text{N} = \mu_1 \cdot 2.1 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \]
\[ \mu_1 = \frac{0.11 \, \text{N}}{2.1 \, \text{kg} \cdot 9.81 \, \text{m/s}^2} \]
\[ \mu_1 \approx 5.4 \times 10^{-3} \]
For \( m_2 \):
\[ F_{\text{centripetal}_2} = \mu_2 \cdot m_2 \cdot g \]
\[ 0.42 \, \text{N} = \mu_2 \cdot 4.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \]
\[ \mu_2 = \frac{0.42 \, \text{N}}{4.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2} \]
\[ \mu_2 \approx 1.1 \times 10^{-2} \]
For the second part of your question, the maximum angular speed without slipping for each parcel is given by the formula:
\[ \omega_{\text{max}} = \sqrt{\frac{\mu \cdot g}{r}} \]
For \( m_1 \):
\[ \omega_{\text{max}_1} = \sqrt{\frac{1.2 \times 10^{-2} \cdot 9.81 \, \text{m/s}^2}{1.2 \, \text{m}}} \]
\[ \omega_{\text{max}_1} \approx 0.42 \, \text{rad/s} \]
For \( m_2 \):
\[ \omega_{\text{max}_2} = \sqrt{\frac{1.8 \times 10^{-2} \cdot 9.81 \, \text{m/s}^2}{2.4 \, \text{m}}} \]
\[ \omega_{\text{max}_2} \approx 0.36 \, \text{rad/s} \]
So, \( \omega_{\text{max}_1} > \omega_{\text{max}_2} \).
Finally, why each package begins to slide on the platform when its angular velocity is less than the calculated maximum angular velocity:
When the angular velocity of the platform is less than the calculated maximum angular velocity for each package, the frictional force between the package and the platform is not sufficient to provide the necessary centripetal force to keep the package moving in a circular path. As a result, the package starts to slip, indicating that the static friction force is no longer able to counteract the motion and prevent slipping.
Final-Answer:
To determine the minimum values of the static friction coefficients between each package and the platform, we can use the formulas that relate frictional force to the normal force.
For a given mass m and radius r, the centripetal force F can be calculated using the formula F = mω^2r, where ω is the angular velocity and r is the distance from the center of rotation.
The formula for maximum static friction force is given by the equation F_friction_max = μN, where μ is the coefficient of static friction and N is the normal force.
First, let's find the normal force acting on each parcel. The normal force can be calculated as N = mg, where m is the mass of the parcel and g is the acceleration due to gravity (approximately 9.81 m/s^2).
For m₁ = 2.1 kg:
The normal force N₁ = m₁g = 2.1 kg * 9.81 m/s^2 = 20.6 N.
The maximum static friction force F_friction_max1 = μ₁N₁ = (1.2 × 10^-2) * 20.6 N = 2.47 N.
For m₂ = 4.0 kg:
The normal force N₂ = m₂g = 4.0 kg * 9.81 m/s^2 = 39.24 N.
The maximum static friction force F_friction_max2 = μ₂N₂ = (1.8 × 10^-2) * 39.24 N = 0.71 N.
The minimum values of the static friction coefficients between each package and the platform can be calculated using the maximum static friction force and the centripetal force. The minimum coefficient of static friction is given by μ_min = (F_centripetal / N), where F_centripetal is the centripetal force.
For m₁:
μ_min₁ = (0.11 N / 20.6 N) ≈ 5.34 × 10^-3.
For m₂:
μ_min₂ = (0.42 N / 39.24 N) ≈ 1.07 × 10^-2.
Next, let's calculate the maximum angular speeds at which the parcels can rotate without slipping off the platform. The maximum angular velocity ω_max is related to the coefficient of friction through the equation ω_max = √(μ * g / r).
For m₁:
ω_max₁ = √((1.2 × 10^-2 * 9.81 m/s^2) / 1.2 m) ≈ 0.34 rad/s.
For m₂:
ω_max₂ = √((1.8 × 10^-2 * 9.81 m/s^2) / 2.4 m) ≈ 0.43 rad/s.
Comparing ω_max₁ and ω_max₂, we find that ω_max₂ > ω_max₁. Therefore, the maximum angular speed for the second package is greater.
Regarding the final question, when the angular velocity of the platform increases with constant angular acceleration, the inertia of each package increases along with its tendency to resist changes in motion. Therefore, as the angular velocity increases, the centripetal force required to keep the parcels in circular motion increases. Once the centripetal force exceeds the maximum static friction force, the packages will begin to slide on the platform. This is why each package begins to slide when its angular velocity is less than the ω_max calculated earlier.
