Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:
A12
B24
C48
D36
Answer & Solution
Correct answer: B. 24
Each mass at distance 2 m from center.
We have I = sum m × r².
That is I = 2 × (3 × 4).
So I = 24 kg m².
Wait, recompute: 3 × 2² + 3 × 2² = 12 + 12 = 24. So I = 24. Hmm option B says 24 — I should select B.
Keep practising Unit 2A: Newton's Laws and Translational Equilibrium
Browse all Unit 2A: Newton's Laws and Translational Equilibrium questions for AP Physics 1, or explore more Physics topics.
Related questions
At what temperature does kT equal 1 eV?A car travels 100 km at 50 km/h, then 100 km at 100 km/h. Average speed:Photon of wavelength 200 nm strikes metal with work function 2 eV. KE of fastest photoelecWork done in stretching wire by 1 mm is W. Work to stretch it by 2 mm (same wire, within eMagnetic flux through a coil changes from 5 mWb to 1 mWb in 0.1 s. Induced EMF:Bohr radius for hydrogen is 0.53 Å. Radius of 3rd orbit:Two coherent sources of intensity I and 4I interfere. Maximum intensity in resulting patteFor a body in SHM with amplitude A, at displacement x, ratio of KE to total energy: