Question

Moment of inertia of a rectangular plate about an axis passing through P and perpendicular to the plate is I. Then moment of PQR about an axis perpendicular to the plane of the plate:

• A. about P = I/2
• B. about R = I/2
• C. about P > I/2
• D. about R > I/2

Answer 1

Answer: (C)

about P > I/2

Answer 2

The problem asks us to find the moment of inertia of a triangular plate PQR, which is half of a rectangular plate PQRS, about an axis perpendicular to its plane. We are given the moment of inertia of the full rectangular plate about an axis passing through point P and perpendicular to the plate, denoted as I.

Let the rectangular plate PQRS have mass M, length l (side PQ or SR) and width b (side PS or QR). From the image, P is the top-right corner, Q is the bottom-right, R is the bottom-left, and S is the top-left. So, PQ has length l and QR has length b.

The moment of inertia of a uniform rectangular plate of mass M, length l, and width b about an axis passing through one of its corners (say P) and perpendicular to the plane is given by: $I = \frac{M(l^2 + b^2)}{3}$

The diagonal PR divides the rectangular plate into two congruent right-angled triangles: PQR and PSR. Since the plate is uniform, the mass of each triangle is $M_t = M/2$.

Let's set up a coordinate system with P at the origin for calculating $I_{PQR,P}$ (moment of inertia of triangle PQR about P). If P is at (0,0), then Q is at (l,0) and S is at (0,b). R would be at (l,b). So the coordinates are: P(0,0), Q(l,0), R(l,b), S(0,b). The side PQ has length l, and PS has length b. This matches the standard formula application. The given I is about P(0,0).

The moment of inertia of the rectangle about P(0,0) is $I = \int_0^l \int_0^b (x^2+y^2) \sigma dx dy$, where $\sigma = M/(lb)$ is the surface mass density. $I = \sigma \int_0^l [x^2y + y^3/3]_0^b dx = \sigma \int_0^l (x^2b + b^3/3) dx = \sigma [x^3b/3 + b^3x/3]_0^l$ $I = \sigma (\frac{l^3b}{3} + \frac{b^3l}{3}) = \frac{M}{lb} \frac{lb(l^2+b^2)}{3} = \frac{M(l^2+b^2)}{3}$. This confirms the formula.

Now, let's find the moment of inertia of triangle PQR about P. The vertices of triangle PQR are P(0,0), Q(l,0), R(l,b). The region PQR is bounded by the x-axis (PQ), the line x=l (QR), and the line PR. The equation of the line PR passing through (0,0) and (l,b) is $y = (b/l)x$. The integral for $I_{PQR,P}$ is: $I_{PQR,P} = \sigma \int_0^l \int_0^{(b/l)x} (x^2+y^2) dy dx$ $I_{PQR,P} = \sigma \int_0^l [x^2y + y^3/3]_0^{(b/l)x} dx$ $I_{PQR,P} = \sigma \int_0^l (x^2(b/l)x + \frac{1}{3}((b/l)x)^3) dx$ $I_{PQR,P} = \sigma \int_0^l (b/l x^3 + b^3/(3l^3) x^3) dx$ $I_{PQR,P} = \sigma (\frac{b}{l} + \frac{b^3}{3l^3}) \int_0^l x^3 dx$ $I_{PQR,P} = \sigma (\frac{b}{l} + \frac{b^3}{3l^3}) [\frac{x^4}{4}]_0^l$ $I_{PQR,P} = \sigma (\frac{b}{l} + \frac{b^3}{3l^3}) \frac{l^4}{4}$ $I_{PQR,P} = \sigma (\frac{bl^3}{4} + \frac{b^3l}{12})$ Substitute $\sigma = M/(lb)$: $I_{PQR,P} = \frac{M}{lb} (\frac{bl^3}{4} + \frac{b^3l}{12}) = M (\frac{l^2}{4} + \frac{b^2}{12}) = \frac{M(3l^2 + b^2)}{12}$

Now, compare $I_{PQR,P}$ with $I/2$. $I = \frac{M(l^2 + b^2)}{3} = \frac{4M(l^2 + b^2)}{12}$. So, $I/2 = \frac{2M(l^2 + b^2)}{12}$.

We need to compare $\frac{M(3l^2 + b^2)}{12}$ with $\frac{M(2l^2 + 2b^2)}{12}$. This simplifies to comparing $(3l^2 + b^2)$ with $(2l^2 + 2b^2)$. Subtracting the second expression from the first: $(3l^2 + b^2) - (2l^2 + 2b^2) = l^2 - b^2$.

The image shows that the length l (side PQ) is greater than the width b (side QR or PS). So, $l > b$. Therefore, $l^2 > b^2$, which implies $l^2 - b^2 > 0$. Thus, $(3l^2 + b^2) > (2l^2 + 2b^2)$. This means $I_{PQR,P} > I/2$.

So, the moment of inertia of PQR about P is greater than I/2. This matches option (C).

Let's quickly check $I_{PSR,P}$ to ensure the sum is I. The vertices of triangle PSR are P(0,0), S(0,b), R(l,b). The region PSR is bounded by the y-axis (PS), the line y=b (SR), and the line PR ($y=(b/l)x$). The integral for $I_{PSR,P}$ is: $I_{PSR,P} = \sigma \int_0^l \int_{(b/l)x}^b (x^2+y^2) dy dx$ $I_{PSR,P} = \sigma \int_0^l [x^2y + y^3/3]_{(b/l)x}^b dx$ $I_{PSR,P} = \sigma \int_0^l (x^2b + b^3/3 - (x^2(b/l)x + \frac{1}{3}((b/l)x)^3)) dx$ $I_{PSR,P} = \sigma \int_0^l (x^2b + b^3/3 - b/l x^3 - b^3/(3l^3) x^3) dx$ $I_{PSR,P} = \sigma [x^3b/3 + b^3x/3 - b/(l) x^4/4 - b^3/(3l^3) x^4/4]_0^l$ $I_{PSR,P} = \sigma (\frac{4l^3b + 4b^3l - 3bl^3 - b^3l}{12}) = \sigma (\frac{l^3b + 3b^3l}{12})$ Substitute $\sigma = M/(lb)$: $I_{PSR,P} = \frac{M}{lb} (\frac{l^3b + 3b^3l}{12}) = M (\frac{l^2}{12} + \frac{3b^2}{12}) = \frac{M(l^2 + 3b^2)}{12}$.

Summing $I_{PQR,P}$ and $I_{PSR,P}$: $I_{PQR,P} + I_{PSR,P} = \frac{M(3l^2 + b^2)}{12} + \frac{M(l^2 + 3b^2)}{12} = \frac{M(4l^2 + 4b^2)}{12} = \frac{M(l^2 + b^2)}{3} = I$. The calculations are consistent.

Since the problem states l > b (as inferred from the figure), $I_{PQR,P} > I/2$.