Question

a uniform, solid sphere with a mass of 3.5kg is pivoted around a frictionless axis through its center, and is in contact at its surface with a flat circular disk (the center of the sphere lies in the same plane as the disk). the disk is uniform with a mass of 1.4kg, and it pivots around an axis parallel to the spheres axis. there is string wound around the disk, and this is pulled with a tension 6.6n, spinning both the disk and the sphere, as their contact point does not slip. find the friction force between the disk and the sphere. provide at least two decimal places

Explanation:

Step1: Calculate the moment of inertia for the disk and the sphere

The moment of inertia \(I\) for a solid disk is \(I_{disk}=\frac{1}{2}mr^{2}\), and for a solid sphere is \(I_{sphere}=\frac{2}{5}mr^{2}\). Let the radius of the disk be \(r_{d}\) and of the sphere be \(r_{s}\). Since there is no slipping at the contact point, \(r_{d}\alpha_{d}=r_{s}\alpha_{s}\). Let \(r_{d} = r_{s}=r\) (for simplicity of ratio calculation, as the radius will cancel out).
The torque \(\tau\) on the disk is \(\tau_{d}=T\times r - f\times r\) (where \(T = 6.6N\) is the tension and \(f\) is the friction force), and \(\tau_{d}=I_{d}\alpha_{d}\). So \(\tau_{d}=(T - f)r=\frac{1}{2}m_{d}r^{2}\alpha_{d}\), which gives \(\alpha_{d}=\frac{2(T - f)}{m_{d}r}\).
The torque on the sphere is \(\tau_{s}=f\times r\), and \(\tau_{s}=I_{s}\alpha_{s}\). So \(fr=\frac{2}{5}m_{s}r^{2}\alpha_{s}\), which gives \(\alpha_{s}=\frac{5f}{2m_{s}r}\).
Since \(r\alpha_{d}=r\alpha_{s}\) (no - slipping condition), \(\frac{2(T - f)}{m_{d}}=\frac{5f}{2m_{s}}\).

Step2: Substitute the given values

Given \(m_{d}=1.4kg\), \(m_{s}=3.5kg\), and \(T = 6.6N\).
Substitute into \(\frac{2(T - f)}{m_{d}}=\frac{5f}{2m_{s}}\):

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Answer:

\(4.40\)