Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

37 of 130

Question 37 of 130

The center of mass of a thin rectangular plate (fig - x ) with sides of length aa and bb, whose mass per unit area (σ)(\sigma) varies as σ=σ0xab\sigma=\frac{\sigma_0 x}{a b} (where σ0\sigma_0 is a constant), would be \qquad

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Answer: (A) (23a, b2)\left(\frac{2}{3} a, \frac{\mathrm{~b}}{2}\right)

dm=σdAd=σ(dx)(dy)=σ0xab(dx)(dy)xcom=∫xdm∫dm=∫x(σ0x)ab(dx)(dy)∫0σ0ab(dx)(dy)=∫0ax2dx∫0bdy∫0bxdx∫0bdy=2a3ycom=∫ydm∫dm=∫y(σ0xab)(dx)(dy)∫σ0xab(dx)(dy)\begin{aligned} & d m=\sigma d A \\ & \begin{aligned} d & =\sigma(d x)(d y)=\frac{\sigma_0 x}{a b}(d x)(d y) \\ x_{c o m} & =\frac{\int x d m}{\int d m}=\frac{\int x \frac{\left(\sigma_0 x\right)}{a b}(d x)(d y)}{\int_0 \frac{\sigma_0}{a b}(d x)(d y)} \\ & =\frac{\int_0^a x^2 d x \int_0^b d y}{\int_0^b x d x \int_0^b d y}=\frac{2 a}{3} \\ y_{c o m} & =\frac{\int y d m}{\int d m}=\frac{\int y\left(\frac{\sigma_0 x}{a b}\right)(d x)(d y)}{\int \frac{\sigma_0 x}{a b}(d x)(d y)}\end{aligned}\end{aligned}
=∫0axdx∫0bydy∫0axdx∫0bdy=b2=\frac{\int_0^a x d x \int_0^b y d y}{\int_0^a x d x \int_0^b d y}=\frac{b}{2}
i.e., r⃗com≡(2a3,b2)\vec{r}_{\mathrm{com}} \equiv\left(\frac{2 a}{3}, \frac{b}{2}\right)

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