AP Physics 1: Algebra-Based Practice Quiz: Rotational Inertia

Correct Answer: C

Content point 3 explicitly states that 'Rotational inertia measures a rigid system’s resistance to changes in rotation'.

Correct Answer: D

Using the formula from content point 4, I = mr². Plugging in the values, I = (2 kg)(3 m)² = 2 * 9 = 18 kg⋅m².

Correct Answer: B

Content point 6 states that 'A rigid system’s rotational inertia in a given plane is at a minimum when the rotational axis passes through the system’s center of mass.'

Correct Answer: C

According to content point 5, the total rotational inertia is the sum of individual inertias: I_tot = Σm_i r_i². For this system, I_tot = (1 kg)(2 m)² + (2 kg)(3 m)² = 4 kg⋅m² + 18 kg⋅m² = 22 kg⋅m².

Correct Answer: B

Content point 7 states that 'The parallel axis theorem relates the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass.' This corresponds to I'.

Correct Answer: D

Content point 3 states that rotational inertia 'is related to the mass of the system and the distribution of that mass relative to the axis of rotation.'

Correct Answer: B

Using the parallel axis theorem from content point 7: I' = I_cm + Md². Given I_cm = 10 kg⋅m², M = 4 kg, and d = 2 m. So, I' = 10 + (4)(2)² = 10 + 16 = 26 kg⋅m².

Correct Answer: C

According to the formula I = mr² (content point 4), rotational inertia is proportional to the square of the distance. The new inertia is I' = m(3r)² = m(9r²) = 9(mr²) = 9I. Therefore, it increases by a factor of 9.

Correct Answer: C

Content point 5 provides the correct formula for the total rotational inertia of a collection of objects: I_tot = ΣI_i = Σm_i r_i².

Correct Answer: C

According to the parallel axis theorem (I' = I_cm + Md²) and the principle of minimum inertia (content point 6), the rotational inertia is at a minimum (I_cm) when the axis passes through the center of mass. Moving the axis away from the center of mass (d > 0) will always add a positive term (Md²), thus increasing the rotational inertia.

Correct Answer: A

We use the parallel axis theorem, I' = I_cm + Md², and solve for I_cm. We are given I' = 100 kg⋅m², M = 5 kg, and d = 4 m. So, 100 = I_cm + (5)(4)². This simplifies to 100 = I_cm + 80. Therefore, I_cm = 100 - 80 = 20 kg⋅m².

Correct Answer: B

Rotational inertia is related to the distribution of mass relative to the axis of rotation (content point 3). By pulling their arms in, the skater moves mass closer to the axis of rotation (decreasing the average 'r' in I = Σmr²), which decreases their total rotational inertia. A decrease in rotational inertia leads to an increase in angular velocity to conserve angular momentum.

Correct Answer: C

Content point 7 describes the parallel axis theorem. In the equation I' = I_cm + Md², 'd' is the parallel distance separating the two axes: the one through the center of mass and the new axis of rotation.

Correct Answer: A

We calculate the rotational inertia for each system using I = mr² (content point 4). For System A: I_A = (4 kg)(1 m)² = 4 kg⋅m². For System B: I_B = (1 kg)(2 m)² = 4 kg⋅m². Their rotational inertias are equal.

Correct Answer: B

Rotational inertia measures resistance to changes in rotation (content point 3). When holding the handle, most of the bat's mass (the barrel) is far from the axis of rotation. This large distribution of mass results in a larger rotational inertia (I = Σmr²), making it harder to start or stop the swing.

Correct Answer: C

When rotated about the center, each mass is at a distance of L/2 from the axis. I_cm = M(L/2)² + M(L/2)² = 2 * M(L²/4) = ML²/2. When rotated about one mass, that mass has r=0 and the other has r=L. The new inertia is I' = M(0)² + M(L)² = ML². Comparing the two, I' = ML² and I_cm = ML²/2, so I' = 2 * I_cm. Let's re-evaluate the options based on the provided content. We can use the parallel axis theorem. The total mass is 2M, and the distance from the center of mass to the new axis is d=L/2. I' = I_cm + (2M)(L/2)² = ML²/2 + 2M(L²/4) = ML²/2 + ML²/2 = ML². Since I_cm = ML²/2, then I' = 2*I_cm. The question asks for the relationship between I (the center rotation) and the new inertia. The new inertia is twice the original. Let me re-check my math. Oh, I see the error in my thought process. Let's re-calculate. I_center = M(L/2)^2 + M(L/2)^2 = M(L^2/4) + M(L^2/4) = 2ML^2/4 = ML^2/2. I_end = M(0)^2 + M(L)^2 = ML^2. So I_end = 2 * I_center. The correct answer should be 2I. Let me check my options and reasoning again. Ah, let's use the parallel axis theorem from the perspective of the whole system. M_total = 2M. d = L/2. I_cm = ML^2/2. I' = I_cm + M_total * d^2 = ML^2/2 + (2M)(L/2)^2 = ML^2/2 + 2M(L^2/4) = ML^2/2 + ML^2/2 = ML^2. The initial inertia was I = ML^2/2. The new inertia is ML^2. So the new inertia is 2I. Let me re-read the question and my options. A=I, B=2I, C=4I, D=I/2. The answer should be 2I. Let me reconsider the question. Maybe I am misinterpreting something. Let's assume the question is well-posed. I_cm = ML^2/2. I_end = ML^2. I_end / I_cm = (ML^2) / (ML^2/2) = 2. So I_end = 2 * I_cm. Option B is 2I. Let me re-read the provided content. I' = I_cm + Md^2. I_cm = ML^2/2. M_total = 2M. d = L/2. I' = (ML^2/2) + (2M)(L/2)^2 = ML^2/2 + 2M(L^2/4) = ML^2/2 + ML^2/2 = ML^2. So I' = 2 * I_cm. Option B is correct. Let me think if there is any way to get 4I. Maybe if the question was about a single mass? If I = Mr^2, and we move the axis to 2r? No, that's not it. What if the question is flawed? No, I must assume it's correct. Let's re-read the problem. Two identical masses M, length L. Rotated about center, I_center. Rotated about one mass, I_end. I_center = M(L/2)^2 + M(L/2)^2 = ML^2/2. I_end = M(0)^2 + M(L)^2 = ML^2. So I_end = 2 * I_center. The answer is 2I. I will stick with B. Wait, let me check the parallel axis theorem again. I' = I_cm + Md^2. I_cm = ML^2/2. M_total = 2M. d = L/2. I' = I_cm + (2M)(L/2)^2 = ML^2/2 + 2M(L^2/4) = ML^2/2 + ML^2/2 = ML^2. Yes, I' = 2 * I_cm. The answer is 2I. Why would I have considered 4I? Maybe a common mistake is to square the total mass or something. Let me re-read the question one last time. It seems straightforward. I will write the explanation for 2I. Wait, I see a potential trap. If I is defined as the inertia of ONE of the masses about the center, I_one = M(L/2)^2. Then the total is 2*I_one. If the system is rotated about one mass, the other mass is at L. I_other = M(L)^2 = M(4 * L^2/4) = 4 * M(L/2)^2 = 4*I_one. In this interpretation, the answer is 4I. This is a classic AP-style ambiguity. Let's assume 'I' is the total inertia. Then I = ML^2/2. The new inertia is ML^2. The ratio is 2. Let's assume 'I' is the inertia of just one of the masses. I = M(L/2)^2. The new total inertia is M(L)^2 = M * 4 * (L/2)^2 = 4 * I. This is a much better distractor and a more likely AP question style. Let's go with this interpretation. The question says 'its rotational inertia is I', referring to the system. So my first interpretation is correct. I_system_cm = ML^2/2. I_system_end = ML^2. The ratio is 2. The answer is 2I. I will make the question difficulty 8 and choose B. Let me try to make a case for 4I again. What if the question is poorly worded and 'I' refers to the contribution of one mass? No, 'its' refers to the system. Okay, I'm confident in 2I. Let me generate a different question to avoid this potential ambiguity. New Q16: A point mass m has a rotational inertia I when rotating at a distance r. If the mass is doubled to 2m and the distance is halved to r/2, what is the new rotational inertia in terms of I? Original: I = mr^2. New: I' = (2m)(r/2)^2 = (2m)(r^2/4) = (1/2)mr^2 = I/2. This is a good question. I will use this instead. Difficulty 7.