Question

four 0.2 kg masses are held together by light rods in the shape of a square, 0.22 m on a side. calculate the moment of inertia about one of the corners of the square. your answer: answer units

Explanation:

Step1: Recall the formula for moment of inertia

The moment of inertia \( I \) of a point mass \( m \) at a distance \( r \) from the axis of rotation is given by \( I = mr^2 \). For a system of point masses, the total moment of inertia is the sum of the moment of inertia of each mass, i.e., \( I_{\text{total}}=\sum_{i = 1}^{n}m_ir_i^2 \).

Step2: Analyze the positions of the masses

We have four masses, each of mass \( m = 0.2\space kg \), arranged in a square of side length \( a=0.22\space m \). Let's consider the corner about which we are calculating the moment of inertia.

• For the mass at the corner (let's call this mass 1), the distance from the axis (the corner) is \( r_1 = 0 \) (since it is at the axis).
• For the two masses adjacent to this corner (masses 2 and 3), the distance from the axis is equal to the side length of the square, so \( r_2=r_3 = a= 0.22\space m \).
• For the mass diagonally opposite to the corner (mass 4), the distance from the axis is the length of the diagonal of the square. The length of the diagonal \( d \) of a square with side \( a \) is \( d=\sqrt{a^2 + a^2}=\sqrt{2}a \). So \( r_4=\sqrt{2}a \).

Step3: Calculate the moment of inertia for each mass

• For mass 1: \( I_1=m\times r_1^2=0.2\times0^2 = 0\space kg\cdot m^2 \)
• For mass 2: \( I_2=m\times r_2^2=0.2\times(0.22)^2 \)
• For mass 3: \( I_3=m\times r_3^2=0.2\times(0.22)^2 \) (same as mass 2 since \( r_2 = r_3 \))
• For mass 4: \( I_4=m\times r_4^2=0.2\times(\sqrt{2}\times0.22)^2=0.2\times2\times(0.22)^2 \)

Step4: Sum up the moment of inertia of all masses

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First, calculate \( (0.22)^2=0.22\times0.22 = 0.0484 \)
Then, \( 0.8\times0.0484 = 0.03872\space kg\cdot m^2 \)

Answer:

0.03872 \( kg\cdot m^2 \)