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.

• Two of the masses are at the corner itself, so their distance from the axis (\( r \)) is \( 0 \).
• One mass is at a distance equal to the side length of the square (\( r = a \)) from the corner.
• The fourth mass is at a distance equal to the length of the diagonal of the square from the corner. The length of the diagonal \( d \) of a square with side \( a \) is \( d=\sqrt{a^2 + a^2}=\sqrt{2}a \).

Step3: Calculate the moment of inertia for each mass

• For the two masses at the corner (\( r = 0 \)): \( I_1 = m(0)^2=0 \) and \( I_2 = m(0)^2 = 0 \).
• For the mass at distance \( a \) from the corner: \( I_3=m a^2 \). Substituting \( m = 0.2\space kg \) and \( a = 0.22\space m \), we get \( I_3=0.2\times(0.22)^2 \).
• For the mass at distance \( \sqrt{2}a \) from the corner: \( I_4=m(\sqrt{2}a)^2=m\times2a^2 \). Substituting the values, \( I_4 = 0.2\times2\times(0.22)^2 \).

Step4: Calculate the total moment of inertia

The total moment of inertia \( I_{\text{total}}=I_1 + I_2+I_3 + I_4 \). Since \( I_1 = I_2 = 0 \), we have:

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First, calculate \( (0.22)^2=0.0484 \). Then, \( 0.2\times0.0484 = 0.00968 \). Then, \( 0.00968\times3=0.02904\space kg\cdot m^2 \).

Answer:

\( 0.02904 \) \( kg\cdot m^2 \)