Open Question
The three 200 g masses in FIGURE EX12.11 are connected by massless, rigid rods. (b) What is the triangle's kinetic energy if it rotates about the axis at 5.0 rev/s?
13. Rotational Inertia & Energy
Intro to Moment of Inertia
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A system is made of two small masses (MLEFT = 3 kg, MRIGHT = 4 kg) attached to the ends of a 5 kg, 2-m long thin rod, as shown. Calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left.
A
8.67 kg•m2
B
15.7 kg•m2
C
17.7 kg•m2
D
22.7 kg•m2
Verified step by step guidance1
Identify the components of the system: two point masses and a rod. The left mass (MLEFT) is 3 kg, the right mass (MRIGHT) is 4 kg, and the rod is 5 kg with a length of 2 m.
Calculate the moment of inertia for the left mass (MLEFT) about the axis. Since the axis is through this mass, its distance from the axis is 0, so its moment of inertia is I_LEFT = 0.
Calculate the moment of inertia for the right mass (MRIGHT) about the axis. Use the formula for point masses: I = m * r^2, where m is the mass and r is the distance from the axis. Here, r is the length of the rod (2 m), so I_RIGHT = 4 kg * (2 m)^2.
Calculate the moment of inertia for the rod about the axis. Since the axis is at the end of the rod, use the formula I = (1/3) * m * L^2, where m is the mass of the rod and L is its length. Substitute the given values to find I_ROD.
Sum the moments of inertia of all components to find the total moment of inertia of the system: I_TOTAL = I_LEFT + I_RIGHT + I_ROD.
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