Centre of Mass

Answer: (d)

1. Take A as origin: $x_{cm} = \frac{m_A \cdot 0 + m_B \cdot 9}{m_A + m_B}$
2. $x_{cm} = \frac{4 \times 9}{2 + 4} = 6$ cm
3. So the answer is (d) $6$ cm from A.

Answer: (b)

1. Equal masses at A, B, C on a line; take A as origin, so the positions are $0$, $AB$ and $AC$.
2. $x_{cm} = \frac{m(0) + m(AB) + m(AC)}{3m} = \frac{AB + AC}{3}$
3. So the answer is (b) $\frac{AB + AC}{3}$.

Answer: (a)

1. Apex angle $120^\circ$, sides 10 m: half-base $= 10\sin 60^\circ = 5\sqrt{3}$ m, height $= 10\cos 60^\circ = 5$ m.
2. P is the mid-point of the 5 m bisector, so relative to P: 2 kg at $(-5\sqrt{3}, -2.5)$, 3 kg at $(5\sqrt{3}, -2.5)$, 15 kg at $(0, 2.5)$.
3. $x_{cm} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$
4. $y_{cm} = \frac{-5 - 7.5 + 37.5}{20} = 1.25$
5. So the answer is (a) $\left(\frac{\sqrt{3}}{4}, 1.25\right)$.

Answer: (b)

1. For two particles, $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$ is a weighted average of the two positions.
2. A weighted average of two points always lies on the line joining them, closer to the heavier mass (not at the mid-point unless $m_1 = m_2$).
3. So the answer is (b) on the line joining the two particles.

Answer: (c)

1. For the centre of mass to stay fixed, $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.
2. $m_1$ moves $d$ towards the CM, so $m_2$ must also move towards the CM by $d_2$ with $m_1 d = m_2 d_2$.
3. $d_2 = \frac{m_1}{m_2} d$
4. So the answer is (c) $\frac{m_1}{m_2} d$, towards the centre of mass.

Answer: 3

1. CM unchanged: $m_1 d_1 = m_2 d_2$
2. $3 \times 2 = 2 \times d_2$
3. $d_2 = 3$ cm

Answer: (b)

1. Each uniform ball acts as a point mass $m$ at its centre.
2. Three equal masses at the vertices of a triangle have their CM at the centroid.
3. The centroid is the point of intersection of the medians.
4. So the answer is (b) point of intersection of medians.

Answer: 3

1. Masses $2M$ at $(0,0)$, $(4,0)$ and $(0,4)$ m.
2. $x_{cm} = \frac{4}{3}$ m, $y_{cm} = \frac{4}{3}$ m
3. $|\vec{r}_{cm}| = \frac{4}{3}\sqrt{2} = \frac{4\sqrt{2}}{3}$
4. So $x = 3$.

Answer: (d)

1. Weight $W$ acts at the CM, distance $x$ from A and $(r - x)$ from B.
2. Torque about B is zero: $N_A \cdot r = W(r - x)$
3. $N_A = \frac{W(r - x)}{r}$
4. So the answer is (d) $\frac{W \cdot (r - x)}{r}$.

Answer: (b)

1. Take the 10 kg mass as origin: $x_{cm} = \frac{20 \times 10}{10 + 20}$
2. $x_{cm} = \frac{200}{30} = \frac{20}{3}$ m
3. So the answer is (b) $\frac{20}{3}$ m.

Answer: (a)

1. $\vec{r}_{cm} = \frac{(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{4} = -2\hat{i} - \hat{j} + \hat{k}$
2. $|\vec{r}_{cm}| = \sqrt{4 + 1 + 1} = \sqrt{6}$
3. $|\hat{i} + 2\hat{j} + \hat{k}| = \sqrt{1 + 4 + 1} = \sqrt{6}$
4. So the answer is (a) $\hat{i} + 2\hat{j} + \hat{k}$.

Answer: (c)

1. CM unchanged: $m_1 d_1 = m_2 d_2$, with both blocks moving towards the CM.
2. $10 \times 6 = 30 \times d_2 \Rightarrow d_2 = 2$ cm
3. Moving towards the CM means the 30 kg block moves towards the 10 kg block.
4. So the answer is (c) 2 cm towards the 10 kg block.

Answer: 2

1. Masses $M$ at $(0,0)$, $(3,0)$ and $(0,3)$ m.
2. $x_{cm} = \frac{3}{3} = 1$ m, $y_{cm} = 1$ m
3. $|\vec{r}_{cm}| = \sqrt{1^2 + 1^2} = \sqrt{2}$ m
4. So $x = 2$.

Answer: (b)

1. Take the 5 kg particle as origin: $x_{cm} = \frac{10 \times 1}{5 + 10}$ m
2. $x_{cm} = \frac{2}{3}$ m $\approx 0.67$ m $= 67$ cm
3. So the answer is (b) 67 cm.

Answer: (c)

1. For the centre of mass to stay fixed, $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.
2. $m_1$ moves $d$ towards the CM, so $m_2$ must move towards the CM by $d_2$ with $m_1 d = m_2 d_2$.
3. $d_2 = \frac{m_1}{m_2} d$
4. So the answer is (c) $\frac{m_1}{m_2} d$, towards the centre of mass.

Answer: (d)

1. Masses $M$ at $(0,0)$, $(2,0)$ and $(0,2)$ m.
2. $x_{cm} = \frac{M(0 + 2 + 0)}{3M} = \frac{2}{3}$ m, $y_{cm} = \frac{M(0 + 0 + 2)}{3M} = \frac{2}{3}$ m
3. So the answer is (d) $\frac{2}{3}(\hat{i} + \hat{j})$.

Answer: (b)

1. Origin at the 1 kg mass: 1.5 kg at $(3, 0)$ cm and 2.5 kg at $(0, 4)$ cm.
2. $x_{cm} = \frac{1.5 \times 3}{5} = 0.9$ cm
3. $y_{cm} = \frac{2.5 \times 4}{5} = 2.0$ cm
4. So the answer is (b) 0.9 cm right and 2.0 cm above the 1 kg mass.

Answer: (c)

1. Positions: 50 g at $(0, 0)$, 100 g at $(1, 0)$, 150 g at $\left(0.5, \frac{\sqrt{3}}{2}\right)$ m.
2. $x_{cm} = \frac{100(1) + 150(0.5)}{300} = \frac{175}{300} = \frac{7}{12}$ m
3. $y_{cm} = \frac{150 \cdot \frac{\sqrt{3}}{2}}{300} = \frac{\sqrt{3}}{4}$ m
4. So the answer is (c) $\left(\frac{7}{12}m, \frac{\sqrt{3}}{4}m\right)$.

Answer: (d)

1. (1) False: CM and centre of gravity coincide only when $g$ is uniform over the body.
2. (2) Taken as correct: in uniform $g$ the gravitational torque about the CM is zero.
3. (3) False: a couple gives zero net force, so it produces only rotation.
4. (4) True: $MA = \frac{\text{load}}{\text{effort}} > 1$ means a small effort lifts a large load.
5. So the answer is (d) (2) and (4).

Answer: (a)

1. $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400}$
2. $x_{cm} = \frac{20000 + 28000}{1200} = 40$ cm
3. So the answer is (a) 40 cm.

Answer: (a)

1. $\vec{r}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$
2. $= \frac{(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{4} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$
3. So the answer is (a) $-2\hat{i} - \hat{j} + \hat{k}$.

Answer: (c)

1. CM unchanged: $m_1 \Delta x_1 + m_2 \Delta x_2 = 0 \Rightarrow m_1 d = m_2 d_2$
2. $d_2 = \frac{m_1}{m_2} d$ (towards the CM)
3. So the answer is (c) $\frac{m_1}{m_2} d$.

Answer: (b)

1. CM unchanged: $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$, so the displacements are opposite and $m_1 d = m_2 d_2$.
2. $d_2 = \frac{m_1}{m_2} d$ (towards the CM)
3. So the answer is (b) $\frac{m_1}{m_2} d$.

Answer: (d)

1. Each identical ball acts as an equal point mass at its centre.
2. Three equal masses at the vertices of a triangle have their CM at the centroid, where the medians meet.
3. So the answer is (d) point of intersection of the medians.

Answer: (d)

1. $\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$ depends only on the masses and their positions (hence on relative distances).
2. Forces change the motion of the centre of mass, not its location for a given arrangement.
3. So the answer is (d) forces acting on the particles.

Answer: (b)

1. $\sigma \propto x$ only, so along $y$ the plate is uniform: $y_{cm} = \frac{b}{2}$.
2. $x_{cm} = \frac{\int_0^a x \cdot x\,dx}{\int_0^a x\,dx} = \frac{a^3/3}{a^2/2} = \frac{2a}{3}$
3. So the answer is (b) $\left(\frac{2}{3}a, \frac{b}{2}\right)$.

Answer: 8

1. $M = \int_0^L \rho_0\left(1 - \frac{x^2}{L^2}\right)dx = \frac{2\rho_0 L}{3}$
2. $\int_0^L x\rho\,dx = \rho_0\left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4}$
3. $x_{cm} = \frac{L^2/4}{2L/3} = \frac{3L}{8}$
4. So $\alpha = 8$.

Answer: 2

1. For a uniform semicircular wire, $y_{cm} = \frac{2R}{\pi}$ from the centre.
2. Comparing with $\frac{xR}{\pi}$: $|x| = 2$

Answer: 4

1. In the figure $a$ is the radius of the disc.
2. For a uniform quarter disc of radius $a$, $x_{cm} = y_{cm} = \frac{4a}{3\pi}$.
3. Comparing with $\frac{x}{3}\frac{a}{\pi}$: $x = 4$

Answer: 3

1. For a solid hemisphere, CM is at $\frac{3R}{8}$ from the centre of the flat face.
2. $x = \frac{3 \times 8}{8} = 3$ cm

Answer: (c)

1. $M = \int_0^L \left(a + \frac{bx^2}{L^2}\right)dx = aL + \frac{bL}{3}$
2. $\int_0^L x\rho\,dx = \frac{aL^2}{2} + \frac{bL^2}{4}$
3. $x_{cm} = \frac{L(2a + b)/4}{(3a + b)/3} = \frac{3}{4}\left(\frac{2a + b}{3a + b}\right)L$
4. So the answer is (c).

Answer: (d)

1. For a solid cone the CM lies on the axis at $\frac{h}{4}$ from the base.
2. Distance from vertex $= h - \frac{h}{4} = \frac{3h}{4}$
3. So the answer is (d) $\frac{3h}{4}$.

Answer: (c)

1. $M = aL + \frac{bL}{2}$, $\int_0^L x\mu\,dx = \frac{aL^2}{2} + \frac{bL^2}{3}$
2. $x_{cm} = \frac{L(3a + 2b)}{3(2a + b)} = \frac{7L}{12}$
3. $12(3a + 2b) = 21(2a + b) \Rightarrow 3b = 6a \Rightarrow b = 2a$
4. So the answer is (c) $2a = b$.

Answer: (d)

1. $\lambda = \lambda_o + kx$; $M = \lambda_o L + \frac{kL^2}{2} \Rightarrow kL^2 = 2(M - \lambda_o L)$
2. $\int_0^L x\lambda\,dx = \frac{\lambda_o L^2}{2} + \frac{kL^3}{3} = \frac{\lambda_o L^2}{2} + \frac{2(M - \lambda_o L)L}{3}$
3. $x_{cm} = \frac{1}{M}\left(\frac{2ML}{3} - \frac{\lambda_o L^2}{6}\right) = \frac{2L}{3} - \frac{\lambda_o L^2}{6M}$
4. So the answer is (d).

Answer: (a)

1. $X_{CM} = \frac{\int_0^L x^{n+1}dx}{\int_0^L x^n dx} = \frac{n + 1}{n + 2}L$
2. At $n = 0$: $X_{CM} = \frac{L}{2}$ (uniform rod).
3. As $n$ increases, $X_{CM}$ rises steadily and approaches $L$ as $n \to \infty$, rising fast at first then levelling off.
4. So the answer is (a).

Answer: (b)

1. $\lambda = kx$, so $x_{cm} = \frac{\int_0^L x \cdot kx\,dx}{\int_0^L kx\,dx} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}$
2. $x_{cm} = \frac{2 \times 3}{3} = 2$ m
3. So the answer is (b) 2 m.

Answer: (c)

1. Uniform rods: mass $\propto$ length. Vertical rod $2m$ with CM at $(0, L)$; horizontal rod $m$ with CM at $\left(\frac{L}{2}, 0\right)$.
2. $x_{cm} = \frac{m \cdot \frac{L}{2}}{3m} = \frac{L}{6}$
3. $y_{cm} = \frac{2m \cdot L}{3m} = \frac{2L}{3}$
4. So the answer is (c) $\left(\frac{L}{6}, \frac{2L}{3}\right)$.

Answer: (d)

1. Each rod has mass $m$ with CM at its mid-point: $(a, 0)$, $\left(0, \frac{a}{2}\right)$ and $\left(a, \frac{a}{2}\right)$.
2. $x_{cm} = \frac{a + 0 + a}{3} = \frac{2a}{3}$
3. $y_{cm} = \frac{0 + \frac{a}{2} + \frac{a}{2}}{3} = \frac{a}{3}$
4. So the answer is (d) $\left(\frac{2a}{3}, \frac{a}{3}\right)$.

Answer: (a)

1. Mass $\propto$ length: vertical part $3L$ with CM at $(0, 1.5L)$; horizontal part $2L$ with CM at $(L, 0)$.
2. $x_{cm} = \frac{2L \cdot L}{5L} = 0.4L = 4$ cm
3. $y_{cm} = \frac{3L \cdot 1.5L}{5L} = 0.9L = 9$ cm
4. So the answer is (a) $4\hat{i} + 9\hat{j}$.

Answer: (d)

1. Hole touches the edge, so its centre is at $20 - 5 = 15$ cm from O. Mass $\propto$ area $\propto r^2$.
2. Disc: $20^2 = 400$ units at $x=0$; hole: $5^2 = 25$ units at $x = 15$ cm.
3. $x_{cm} = \frac{400(0) - 25(15)}{400-25} = -\frac{375}{375} = -1$ cm
4. So the answer is (d) $1.0$ cm (on the side opposite the hole).

Answer: 15

1. Plate = full rectangle $3\times 2$ (area $6$, centre $(1.5,\ 1)$) minus square $1\le x\le 2,\ 1\le y\le 2$ (area $1$, centre $(1.5,\ 1.5)$).
2. The shape is symmetric about $x = 1.5$, so $x_{cm} = 1.5$.
3. $y_{cm} = \frac{6(1) - 1(1.5)}{6-1} = \frac{4.5}{5} = 0.9$
4. $\frac{x_{cm}}{y_{cm}} = \frac{1.5}{0.9} = \frac{5}{3} = \frac{15}{9}$
5. So $n = 15$.

Answer: (c)

1. Full disc: area $\pi a^2$, centre at $x = a$. Hole: area $\frac{\pi a^2}{4}$, centre at $x = \frac{3a}{2}$.
2. $x_{cm} = \frac{\pi a^2 (a) - \frac{\pi a^2}{4}\cdot\frac{3a}{2}}{\pi a^2 - \frac{\pi a^2}{4}} = \frac{a - \frac{3a}{8}}{\frac{3}{4}}$
3. $x_{cm} = \frac{5a}{8}\cdot\frac{4}{3} = \frac{5a}{6}$
4. So the answer is (c) $\frac{5}{6}a$.

Answer: (a)

1. Split the flag into two rectangles of uniform density, so mass $\propto$ area.
2. Rectangle 1: $0\le x\le 1,\ 0\le y\le 3$, area $3$, centre $(0.5,\ 1.5)$.
3. Rectangle 2: $1\le x\le 2,\ 2\le y\le 3$, area $1$, centre $(1.5,\ 2.5)$.
4. $x_{cm} = \frac{3(0.5)+1(1.5)}{4} = 0.75$ m, $\ y_{cm} = \frac{3(1.5)+1(2.5)}{4} = 1.75$ m
5. So the answer is (a) $(0.75\text{ m},\ 1.75\text{ m})$.

Answer: (d)

1. Take C as origin. The cavity touches the sphere surface, so $CO = R - 1$; mass $\propto$ volume $\propto r^3$.
2. $x_G = \frac{-1^3(R-1)}{R^3 - 1} = -\frac{R-1}{(R-1)(R^2+R+1)} = -\frac{1}{R^2+R+1}$
3. G lies on the cavity surface on the far side: $x_G = (R-1) - 1 = -(2-R)$.
4. $\frac{1}{R^2+R+1} = 2 - R \Rightarrow (R^2+R+1)(2-R) = 1$
5. So the answer is (d).

Answer: 23

1. Mass $\propto$ area. Disc: $\pi a^2$ at O. Square hole: $\frac{a^2}{4}$ at $x = \frac{a}{2}$.
2. $x_{cm} = \frac{-\frac{a^2}{4}\cdot\frac{a}{2}}{\pi a^2 - \frac{a^2}{4}} = -\frac{a}{8\pi - 2}$
3. $8\pi - 2 \approx 23.1$
4. So $X \approx 23$.

Answer: (c)

1. The bar has three straight parts (the left vertical line is the y-axis, not the bar); mass $\propto$ length.
2. Top part: length $2L$, centre $(L,\ L)$. Vertical part: length $L$, centre $(2L,\ L/2)$. Bottom part: length $L$, centre $(2.5L,\ 0)$.
3. Total length $4L$: $x_{cm} = \frac{2L(L)+L(2L)+L(2.5L)}{4L} = \frac{13}{8}L$
4. $y_{cm} = \frac{2L(L)+L(L/2)+L(0)}{4L} = \frac{5}{8}L$
5. So the answer is (c) $\vec{r}_{cm} = \frac{13}{8}L\,\hat{x} + \frac{5}{8}L\,\hat{y}$.