Question

Point particle A has a mass of 192 g and is located at (x, y, z) = (−3.00 cm, 5.00 cm, 0), point particle B has a mass of 300 g and is at (6.00 cm, 0, 0), and point particle C has a mass of 501 g and is at (−5.00 cm, −4.00 cm, 0). The z-axis is perpendicular to the xy-plane and points out of the page. Part A: Find the rotational inertia of the system of particles shown in the figure assuming the system rotates around the x-axis. Part B: Find the rotational inertia of the system of point particles shown in the figure assuming the system rotates about the y-axis. Part C: Find the rotational inertia of the system of point particles shown in the figure assuming the system rotates about the z-axis. Part D: What is the x-coordinate of the center of mass of the system? Part E: What is the y-coordinate of the center of mass of the system?

Answer 1

Part A: Rotational inertia of the system when rotating around the x-axis can be found by calculating the moment of inertia for each particle and summing them up. Part B: Same method for calculating rotational inertia, but with new perpendicular distances. Part C: Same method for calculating rotational inertia as Part A, since particles lie in the same plane.

Step-by-step explanation:

Part A:

To find the rotational inertia of the system when rotating around the x-axis, we need to calculate the moment of inertia for each point particle and then sum them up. The moment of inertia of a point particle rotating about an axis is given by the formula I = m*r^2, where m is the mass of the particle and r is the perpendicular distance from the axis of rotation to the particle. The rotational inertia of the system is the sum of the individual moment of inertia for each particle. Let's calculate:

Particle A: IA = mA*rA2 = 0.192 kg*(5 cm)2

Particle B: IB = mB*rB2 = 0.3 kg*(6 cm)2

Particle C: IC = mC*rC2 = 0.501 kg*((-4 cm)^2)

Rotational inertia of the system: Isystem = IA + IB + IC

Part B:

To find the rotational inertia of the system when rotating around the y-axis, we need to calculate the new perpendicular distances for each particle from the y-axis and use the same formula. Let's calculate:

Particle A: IA = mA*r'A2 = 0.192 kg*((-3 cm)^2)

Particle B: IB = mB*rB2 = 0.3 kg*(6 cm)2

Particle C: IC = mC*r'C2 = 0.501 kg*((-4 cm)^2)

Rotational inertia of the system: Isystem = IA + IB + IC

Part C:

To find the rotational inertia of the system when rotating around the z-axis, we need to calculate the new perpendicular distances for each particle from the z-axis and use the same formula. Since all particles lie in the same plane, the new perpendicular distances will be the same as the original distances. Let's calculate:

Particle A: IA = mA*rA2 = 0.192 kg*(5 cm)2

Particle B: IB = mB*rB2 = 0.3 kg*(6 cm)2

Particle C: IC = mC*rC2 = 0.501 kg*((-4 cm)^2)

Rotational inertia of the system: Isystem = IA + IB + IC

Part D:

The x-coordinate of the center of mass can be calculated using the formula xcm = (mA*xA + mB*xB + mC*xC) / (mA + mB + mC), where m is the mass of each particle and x is the x-coordinate of each particle. Let's calculate:

xcm = (0.192 kg*(-3 cm) + 0.3 kg*(6 cm) + 0.501 kg*(-5 cm)) / (0.192 kg + 0.3 kg + 0.501 kg)

Part E:

The y-coordinate of the center of mass can be calculated using the same formula as the x-coordinate, but using the y-coordinate of each particle. Let's calculate:

ycm = (mA*yA + mB*yB + mC*yC) / (mA + mB + mC)