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An isosceles triangular plate has mass M and base length l. The apex is at the origin, the angle at the apex is 90 degrees, and the base is parallel to the x-axis. The moment of inertia of the plate about the x-axis is (1/k) * M * l². Find the value of k.
- 6
- 12
- 24
- 3
Correct answer: 12
Solution
Apex at origin, apex angle 90 deg, base parallel to x-axis. The two equal sides slope outward at 45 deg from the y-axis (the axis of symmetry). At height y from apex, the width of the strip = 2y*tan(45 deg) = 2y. The base is at y = h. Since base length l = 2h*tan(45 deg) = 2h, so h = l/2. Area = (1/2)*base*height = (1/2)*l*(l/2) = l²/4. Surface mass density sigma = M / (l²/4) = 4M/l². For a horizontal strip at height y (from apex), width = 2y, thickness dy: dm = sigma * 2y * dy = (4M/l²) * 2y dy = 8M*y dy / l². Moment of inertia about x-axis (at origin, horizontal): Ix = integral of y² dm from 0 to l/2 = integral₀^(l/2) y² * (8M/l²) * y dy = (8M/l²) * [y⁴/4]₀^(l/2) = (8M/l²) * (l/2)⁴ / 4 = (8M/l²) * l⁴/(16*4) = (8M/l²) * l⁴/64 = Ml²/8. So 1/k = 1/8, k = 8. Hmm let me recheck with the base at y=h=l/2 above origin: Ix = (8M/l²)*(l/2)⁴/4 = 8M/l² * l⁴/64 = Ml²/8. So k = 8.
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