Question

19 mark for review three small objects with the masses indicated in the figure are arranged in a circle as shown. the object of mass 2m0 is at the top of the circle, and the two objects of mass m0 can be located anywhere in the regions on the circle defined by 0 < θ < 90°. point p is located at the center of the circle. relative to point p, where is the center of mass of the three - object system? a at point p for 0 < θ < 90°. b directly above point p for 0 < θ < 90°. c directly below point p for 0 < θ < 90°. d directly below point p for 0 < θ < 60°, at point p for θ = 60°, and directly above point p for 60° < θ < 90°.

Explanation:

Step1: Set up coordinate system

Let the center of the circle (point P) be the origin \((0,0)\) of a two - dimensional coordinate system, with the positive y - axis pointing upwards. The mass \(m_1 = 2m_0\) is at \((0,R)\) (where \(R\) is the radius of the circle), and the two masses \(m_2=m_3 = m_0\) are at \((R\cos\theta,-R\sin\theta)\) and \((-R\cos\theta,-R\sin\theta)\) respectively.

Step2: Calculate the y - coordinate of the center of mass \(y_{cm}\)

The formula for the y - coordinate of the center of mass of a system of particles is \(y_{cm}=\frac{\sum_{i = 1}^{n}m_iy_i}{M}\), where \(M=\sum_{i = 1}^{n}m_i\). Here, \(M=2m_0 + m_0+m_0=4m_0\), \(y_1 = R\), \(y_2=-R\sin\theta\), \(y_3=-R\sin\theta\). Then \(y_{cm}=\frac{2m_0\times R+m_0\times(-R\sin\theta)+m_0\times(-R\sin\theta)}{4m_0}=\frac{2m_0R - 2m_0R\sin\theta}{4m_0}=\frac{R(1 - \sin\theta)}{2}\). Since \(0<\theta<90^{\circ}\), \(\sin\theta\in(0,1)\), and \(y_{cm}>0\).

Step3: Calculate the x - coordinate of the center of mass \(x_{cm}\)

The formula for the x - coordinate of the center of mass is \(x_{cm}=\frac{\sum_{i = 1}^{n}m_ix_i}{M}\). Here, \(x_1 = 0\), \(x_2 = R\cos\theta\), \(x_3=-R\cos\theta\). Then \(x_{cm}=\frac{2m_0\times0+m_0\times R\cos\theta+m_0\times(-R\cos\theta)}{4m_0}=0\).

Answer:

B. Directly above Point P for \(0 < \theta < 90^{\circ}\)