Questions: The figure shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 4.1 s. If R=1.3 m and m=1.2 kg, calculate the angular momentum about that axis.

The figure shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 4.1 s. If R=1.3 m and m=1.2 kg, calculate the angular momentum about that axis.

Transcript text: The figure shows a rigid structure consisting of a circular hoop of radius $R$ and mass $m$, and a square made of four thin bars, each of length $R$ and mass $m$. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 4.1 s . If $R=1.3 \mathrm{~m}$ and $m=1.2 \mathrm{~kg}$, calculate the angular momentum about that axis.

Solution

Solution Steps

Step 1: Calculate the angular velocity

The period of rotation T is given as 4.1 s. The angular velocity ω is related to the period by ω = 2π/T. Plugging in the value for T, we get ω = 2π/4.1 s ≈ 1.53 rad/s.

Step 2: Calculate the moment of inertia of the hoop

The moment of inertia of a hoop of mass m and radius R rotating about an axis through its center and perpendicular to its plane is given by I_hoop = mR². Since the hoop is rotating about an axis a distance R away from its center, we use the parallel axis theorem to find the moment of inertia about the rotation axis: I_hoop' = I_hoop + mR² = mR² + mR² = 2mR².

Step 3: Calculate the moment of inertia of the square

The square is made up of four rods, each of length R and mass m. The moment of inertia of a rod of length L and mass M rotating about an axis perpendicular to the rod and through its center is (1/12)ML². For our rods, L = R and M = m. For the two vertical rods, the axis of rotation is through their centers, so the parallel axis theorem isn't needed. Each vertical rod has a moment of inertia of (1/12)mR². The total moment of inertia for the two vertical rods is (1/6)mR². For the two horizontal rods, they are rotating about an axis a distance R/2 away from their center. Using the parallel axis theorem, the moment of inertia of each horizontal rod is (1/12)mR² + m(R/2)² = (1/12)mR² + (1/4)mR² = (1/3)mR². The total moment of inertia for the two horizontal rods is (2/3)mR². The total moment of inertia for the square is (1/6 + 2/3)mR² = (5/6)mR².

Step 4: Calculate the total moment of inertia

The total moment of inertia of the system is the sum of the moments of inertia of the hoop and the square: I_total = I_hoop' + I_square = 2mR² + (5/6)mR² = (17/6)mR².

Step 5: Calculate the angular momentum

The angular momentum L is given by L = Iω. Substituting the values we calculated for I_total and ω, and the given values for m and R, we have: L = (17/6)(1.2 kg)(1.3 m)²(1.53 rad/s) ≈ 8.6 kg⋅m²/s.