NCERT Solutions · Class 11 Physics Systems of Particles and Rotational Motion

The centre of mass of a uniform, symmetric body sits at its geometric centre — this follows purely from symmetry, with no need to do an integral.For a continuous body the centre of mass is defined by \[\vec{R} = \frac{\int \vec{r}\,dm}{\int dm}, \] where \(\displaystyle \vec{r}\) is the position vector of a mass element \(\displaystyle dm\), measured from some chosen origin.Now choose the origin at the geometric centre of a body that has uniform mass density and possesses point symmetry about that centre (the sphere, the cylinder's axis-midpoint, the ring's centre, and the cube's centre all have this property). For every volume element \(\displaystyle dm\) sitting at position \(\displaystyle \vec{r}\), symmetry guarantees an equal-mass element \(\displaystyle dm\) sitting at the mirror-image position \(\displaystyle -\vec{r}\), because the density is the same everywhere and the shape is symmetric about that point. Each such pair contributes \[\vec{r}\,dm + (-\vec{r})\,dm = 0 \] to the numerator. Summing over the whole body, every element is matched by its mirror partner, so \[\int \vec{r}\,dm = 0 \implies \vec{R} = 0, \] i.e., the centre of mass coincides with the point about which the symmetry holds. This argument needs no calculus of the actual shape — only the existence of the mirror-pairing — which is why it works identically for all four bodies: In each case the location is the geometric centre of symmetry — no working masses or coordinates are needed, only the mirror-symmetry argument above.Does the centre of mass necessarily lie inside the body? No.The derivation above locates the centre of mass at a geometric point of symmetry — it says nothing about whether that point is occupied by the material of the body. The ring is the counter-example already sitting in the list: its centre of mass is at the centre of the circle, but there is no mass there at all — the material of the ring lies only on the circumference, and the centre is empty space. The same is true of any hollow, curved, or hoop-like uniform body (a bangle, a spherical shell, a hollow cube shell): the mass distribution can be symmetric about a point that the body's material never actually reaches.So a solid, filled, convex body like the sphere, the solid cylinder, or the solid cube does have its centre of mass on its own material (or at least within its interior volume), but this is not guaranteed for every uniform body — a body need only be shaped so that its material surrounds empty space at the point of symmetry for the centre of mass to fall outside the body itself.Answer: (i) sphere — at its centre; (ii) cylinder — at the midpoint of its axis; (iii) ring — at the centre of the ring; (iv) cube — at the point of intersection of its body diagonals (its centroid). No, the centre of mass need not lie inside the body — for a ring (or any hollow/hoop-shaped uniform body) it falls in the empty space at the centre, outside the material of the body.

The centre of mass of two point masses lies on the line joining them, much closer to the heavier one.For two particles of masses \(\displaystyle m_1 \) and \(\displaystyle m_2 \) sitting at positions \(\displaystyle x_1 \) and \(\displaystyle x_2 \) on a line, the centre of mass is at\[x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]Since nearly all the mass of an atom sits in its nucleus, treat each atom as a point mass at its nucleus. Put the origin at the hydrogen nucleus, with the x-axis running along the H–Cl bond. Then the chlorine nucleus sits at\[x_{Cl} = 1.27\text{ Å} \]the given internuclear separation, and the hydrogen nucleus is at \(\displaystyle x_H = 0 \).Let the mass of a hydrogen atom be \(\displaystyle m \). Since a chlorine atom is $\displaystyle 35.5$ times as massive as a hydrogen atom,\[m_H = m, \qquad m_{Cl} = 35.5\,m \]Substituting into the centre-of-mass formula:\[x_{cm} = \frac{m(0) + 35.5m(1.27\text{ Å})}{m + 35.5m} = \frac{35.5 \times 1.27}{36.5}\text{ Å} \]Notice the mass \(\displaystyle m \) cancels out — the position of the CM depends only on the mass ratio $\displaystyle 35.5$ : $\displaystyle 1$, not on the actual mass of a hydrogen atom.\[x_{cm} = \frac{45.085}{36.5}\text{ Å} = 1.2352\text{ Å} \]Aside on significant figures: both given quantities — the separation \(\displaystyle 1.27\text{ Å} \) and the mass ratio \(\displaystyle 35.5 \) — carry three significant figures, so this raw quotient carries one digit more precision than the data justifies. Round only now, at the end, to three significant figures:\[x_{cm} \approx 1.24\text{ Å, measured from the hydrogen nucleus} \]Check how far this is from the chlorine nucleus: \(\displaystyle 1.27\text{ Å} - 1.24\text{ Å} \approx 0.03\text{ Å} \). So the centre of mass sits only about \(\displaystyle 0.03\text{ Å} \) from the chlorine nucleus — essentially on top of the chlorine atom. This makes physical sense: chlorine is $\displaystyle 35.5$ times heavier than hydrogen, so it dominates the mass-weighted average position, pulling the CM almost all the way to itself and leaving hydrogen's much smaller mass to barely shift it back toward the H end.Answer: The centre of mass lies on the line joining the two nuclei, about \(\displaystyle 1.24 \times 10^{-10}\text{ m} = 1.24\text{ Å}\) from the hydrogen nucleus (equivalently, only about \(\displaystyle 0.03\text{ Å}\) from the chlorine nucleus) — that is, very close to the chlorine atom.

The speed of the centre of mass does not change — internal forces cannot move it. The child running on the trolley only pushes on the trolley and is pushed back by it; these are forces within the system, and by Newton's third law they occur in equal-and-opposite pairs. Internal forces always cancel out when you sum the total force on a system, so they cannot alter how the centre of mass moves.The floor is smooth, so there is no friction between the trolley's wheels and the ground — meaning no external horizontal force acts on the (trolley + child) system at all. (Gravity and the normal reaction from the floor are external forces too, but they act vertically and simply balance each other; they have no horizontal component to change the motion along the floor.)Newton's second law applied to a system of particles says \[F_{\text{ext}} = M\,\frac{dv_{\text{cm}}}{dt} \] where \(\displaystyle F_{\text{ext}}\) is the net external force on the system, \(\displaystyle M\) is its total mass, and \(\displaystyle v_{\text{cm}}\) is the velocity of its centre of mass. Since \(\displaystyle F_{\text{ext}} = 0\) here (no horizontal external force), \[\frac{dv_{\text{cm}}}{dt} = 0 \] so \(\displaystyle v_{\text{cm}}\) stays constant. Before the child starts running, the whole system — trolley and child together — moves uniformly with speed \(\displaystyle V\), so the centre of mass already has speed \(\displaystyle V\) at that instant. Because the centre-of-mass velocity cannot change without an external force, it remains \(\displaystyle V\) no matter how the child moves about on the trolley afterward.This is the same reasoning used for an internal explosion or a person walking inside a stationary boat: whatever redistribution of mass happens inside the system, the centre of mass keeps moving (or stays at rest) exactly as it would have if nothing inside had stirred, because internal forces alone can never change the total momentum of the system.Answer: The centre of mass continues to move with the same speed V — running about inside the trolley does not change it, since no external horizontal force acts on the system.

The area of a triangle is half its base times its height, and the cross product's magnitude is built out of exactly that height.Draw the two vectors from a common point \(\displaystyle O\), so that \(\displaystyle \vec{OA}=\vec a\) and \(\displaystyle \vec{OB}=\vec b\). The third side of the triangle is then \(\displaystyle \vec{AB}=\vec b-\vec a\), and triangle \(\displaystyle OAB\) is "the triangle contained between \(\displaystyle \vec a\) and \(\displaystyle \vec b\)." Let \(\displaystyle \theta\) be the angle between \(\displaystyle \vec a\) and \(\displaystyle \vec b\) at \(\displaystyle O\).Step $\displaystyle 1$ — write the triangle's area as base times height. Take \(\displaystyle OA\) as the base, so the base length is \(\displaystyle |\vec a|\). Drop a perpendicular from \(\displaystyle B\) onto the line \(\displaystyle OA\), meeting it at \(\displaystyle N\); \(\displaystyle BN\) is the height of the triangle. In the right triangle \(\displaystyle OBN\), the angle at \(\displaystyle O\) is \(\displaystyle \theta\) and the hypotenuse is \(\displaystyle OB=|\vec b|\), so \[BN=|\vec b|\sin\theta . \] Hence \[\text{Area of }\triangle OAB=\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\,|\vec a|\,|\vec b|\sin\theta . \]Step $\displaystyle 2$ — recognize this as half the cross product's magnitude. By definition, the magnitude of the vector (cross) product of \(\displaystyle \vec a\) and \(\displaystyle \vec b\) is \[|\vec a\times\vec b|=|\vec a|\,|\vec b|\sin\theta , \] where \(\displaystyle \theta\;(0\le\theta\le\pi)\) is the angle between them, so \(\displaystyle \sin\theta\ge 0\) and this is automatically non‑negative, matching an area. Comparing the two boxed expressions, \[\text{Area of }\triangle OAB=\frac{1}{2}\,|\vec a\times\vec b| . \]This is the step people rush past: the height of the triangle, \(\displaystyle |\vec b|\sin\theta\), is precisely the factor that makes \(\displaystyle |\vec a\times\vec b|\) different from a simple product of lengths — the cross product already "contains" a base-times-height combination, which is why it connects to area at all, not just to length.Check by coordinates (independent confirmation). Put \(\displaystyle O\) at the origin with everything in the \(\displaystyle xy\)-plane: \(\displaystyle \vec a=(a_1,a_2,0)\), \(\displaystyle \vec b=(b_1,b_2,0)\), so \(\displaystyle A=(a_1,a_2)\), \(\displaystyle B=(b_1,b_2)\). The standard coordinate (shoelace) formula for the area of a triangle with one vertex at the origin is \[\text{Area}=\frac{1}{2}\,|a_1b_2-a_2b_1| . \] Meanwhile \[\vec a\times\vec b=(0,\,0,\,a_1b_2-a_2b_1)\ \Rightarrow\ |\vec a\times\vec b|=|a_1b_2-a_2b_1| . \] So again \(\displaystyle \text{Area}=\frac{1}{2}|\vec a\times\vec b|\) — the same result by an entirely different route, which is a useful cross-check whenever you are unsure about a geometric argument.Why this had to be true geometrically, too. The parallelogram with adjacent sides \(\displaystyle \vec a\) and \(\displaystyle \vec b\) (vertices \(\displaystyle O\), \(\displaystyle A\), \(\displaystyle C=A+B\), \(\displaystyle B\)) has area \(\displaystyle |\vec a\times\vec b|\) (shown in the previous exercise). Its diagonal \(\displaystyle AB\) cuts that parallelogram into two congruent triangles, one of which is exactly triangle \(\displaystyle OAB\) — the triangle "contained between \(\displaystyle \vec a\) and \(\displaystyle \vec b\)." A diagonal always bisects a parallelogram's area, so each triangle gets exactly half: \(\displaystyle \frac{1}{2}|\vec a\times\vec b|\), the same answer as above.Answer: Area of the triangle contained between \(\displaystyle \vec a\) and \(\displaystyle \vec b\) \(\displaystyle = \dfrac{1}{2}\,|\vec a\times\vec b|\), since \(\displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta\) equals twice the triangle's base-times-height.

The scalar triple product is just "base area times height" written with vectors — the cross product builds the base, the dot product measures the height.Set up the parallelepiped with edges \(\displaystyle \vec{a} \), \(\displaystyle \vec{b} \), \(\displaystyle \vec{c} \) starting from a common corner. Take \(\displaystyle \vec{b} \) and \(\displaystyle \vec{c} \) as the two edges of the base, so the base is the parallelogram they span, and \(\displaystyle \vec{a} \) is the edge that tilts up out of that base.Step $\displaystyle 1$ — The cross product gives the base area. By the definition of the vector (cross) product, if \(\displaystyle \theta \) is the angle between \(\displaystyle \vec{b} \) and \(\displaystyle \vec{c} \), \[\vec{b}\times\vec{c} = (bc\sin\theta)\,\hat{n} \] where \(\displaystyle b = |\vec b| \), \(\displaystyle c = |\vec c| \), and \(\displaystyle \hat{n} \) is the unit vector normal to the plane of \(\displaystyle \vec{b} \) and \(\displaystyle \vec{c} \) (direction fixed by the right-hand rule). The factor \(\displaystyle bc\sin\theta \) is exactly the area of the parallelogram with sides \(\displaystyle \vec b \) and \(\displaystyle \vec c \) — that parallelogram is the base of the parallelepiped. So\[|\vec{b}\times\vec{c}| = \text{base area}. \]Step $\displaystyle 2$ — Dotting with \(\displaystyle \vec a \) picks out the height. By the definition of the scalar (dot) product, for any two vectors the dot product is (magnitude)×(magnitude)×(cosine of the angle between them). Let \(\displaystyle \alpha \) be the angle between \(\displaystyle \vec a \) and \(\displaystyle \hat n \). Then\[\vec{a}\cdot(\vec{b}\times\vec{c}) = \vec a \cdot (bc\sin\theta)\hat n = (bc\sin\theta)\,(a\cos\alpha), \]using \(\displaystyle \vec a \cdot \hat n = a\cos\alpha \) since \(\displaystyle \hat n \) is a unit vector.Step $\displaystyle 3$ — \(\displaystyle a\cos\alpha \) is the perpendicular height of the box. This is the step most people skip over: \(\displaystyle \hat n \) is not just "some direction" — it is normal to the base plane. So \(\displaystyle a\cos\alpha \), the projection of \(\displaystyle \vec a \) onto \(\displaystyle \hat n \), is precisely the perpendicular distance from the base plane up to the tip of \(\displaystyle \vec a \). That perpendicular distance is the height \(\displaystyle h \) of the parallelepiped (the height is measured perpendicular to the base, not along the slanted edge \(\displaystyle \vec a \) itself).Step $\displaystyle 4$ — Base area times height is volume. Putting the two pieces together, \[\vec{a}\cdot(\vec{b}\times\vec{c}) = \underbrace{(bc\sin\theta)}_{\text{base area}}\;\underbrace{(a\cos\alpha)}_{\text{height }h} = (\text{base area})\times h = V, \] which is exactly the volume of the parallelepiped built on \(\displaystyle \vec a, \vec b, \vec c \).Why "in magnitude" and not simply "equal": if \(\displaystyle \vec a \) happens to point to the opposite side of the base plane from \(\displaystyle \hat n \), the angle \(\displaystyle \alpha \) is obtuse and \(\displaystyle \cos\alpha \) is negative, so \(\displaystyle \vec a\cdot(\vec b\times\vec c) \) comes out negative — but a volume is never negative. The geometry (the box itself) hasn't changed; only the sign flips with the choice of which way \(\displaystyle \hat n \) points relative to \(\displaystyle \vec a \). Taking the absolute value removes this sign ambiguity:\[|\vec{a}\cdot(\vec{b}\times\vec{c})| = V. \]Answer: \(\displaystyle \vec a\cdot(\vec b\times\vec c) = (bc\sin\theta)(a\cos\alpha) = (\text{base area})\times(\text{height}) \), so its magnitude \(\displaystyle |\vec a\cdot(\vec b\times\vec c)| \) equals the volume of the parallelepiped formed on \(\displaystyle \vec a, \vec b, \vec c \).

Angular momentum is a cross product, and a cross product's components come from a determinant — write that out and the "only z survives" result falls out of z = 0.The angular momentum of a particle about the origin is defined by \[\vec{l} = \vec{r} \times \vec{p} \] where \(\displaystyle \vec{r} = (x, y, z) \) is the position vector of the particle and \(\displaystyle \vec{p} = (p_x, p_y, p_z) \) is its linear momentum.Setting up the cross productWriting the cross product as a determinant, \[\vec{l} = \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix} \]Expanding along the top row gives the three Cartesian components of \(\displaystyle \vec{l} \): \[l_x = y\,p_z - z\,p_y \] \[l_y = z\,p_x - x\,p_z \] \[l_z = x\,p_y - y\,p_x \]These three expressions hold for a particle in general, three‑dimensional motion. Each component pairs the two coordinates (or momenta) that are not along that axis — this is the pattern to remember rather than memorising the determinant: \(\displaystyle l_x \) uses \(\displaystyle y,z \) and \(\displaystyle p_y,p_z \); \(\displaystyle l_y \) uses \(\displaystyle z,x \) and \(\displaystyle p_z,p_x \); \(\displaystyle l_z \) uses \(\displaystyle x,y \) and \(\displaystyle p_x,p_y \).Restricting the motion to the x-y planeIf the particle moves only in the x-y plane, then at every instant its z-coordinate is zero, \[z = 0 \] and since z never changes (it stays zero for all time), the z-component of momentum is also zero, \[p_z = \frac{dz}{dt} \cdot m = 0 \]This is the point that is easy to miss: it is not enough that \(\displaystyle z=0 \) at one instant — the particle must have \(\displaystyle z=0 \) throughout its motion for \(\displaystyle p_z \) to vanish too. Substituting both \(\displaystyle z = 0 \) and \(\displaystyle p_z = 0 \) into the three components above: \[l_x = y\,p_z - z\,p_y = y(0) - (0)p_y = 0 \] \[l_y = z\,p_x - x\,p_z = (0)p_x - x(0) = 0 \] \[l_z = x\,p_y - y\,p_x \quad \text{(unchanged — no } z \text{ or } p_z \text{ appears in it)} \]So \(\displaystyle l_x = 0 \) and \(\displaystyle l_y = 0 \), while \(\displaystyle l_z = x\,p_y - y\,p_x \) survives and is, in general, non-zero.Why this makes sense physically\(\displaystyle \vec{r} \) and \(\displaystyle \vec{p} \) both lie in the x-y plane when the motion is confined to that plane. The cross product of any two vectors lying in a plane is a vector perpendicular to that plane — here, along the z-axis. That is exactly what the algebra shows: the angular momentum has no component within the plane of motion, only a component perpendicular to it.**Answer: For general motion, \(\displaystyle l_x = y p_z - z p_y \), \(\displaystyle l_y = z p_x - x p_z \), \(\displaystyle l_z = x p_y - y p_x \). When the particle is confined to the x-y plane (\(\displaystyle z = 0 \), \(\displaystyle p_z = 0 \)), \(\displaystyle l_x = l_y = 0 \) and only \(\displaystyle l_z = x p_y - y p_x \) survives — the angular momentum has only a z-component, perpendicular to the plane of motion.

Angular momentum does not depend on where you put the reference point, as long as the system's total linear momentum is zero — and that is exactly the situation here.Set up axes so the common direction of the two lines is \(\displaystyle \hat x \) and the perpendicular direction joining them is \(\displaystyle \hat y \). Particle $\displaystyle 1$ moves with velocity \(\displaystyle \vec v_1 = v\hat x \) along one line; particle $\displaystyle 2$, on the parallel line a perpendicular distance \(\displaystyle d \) away, moves with velocity \(\displaystyle \vec v_2 = -v\hat x \) — same speed \(\displaystyle v \), opposite direction. Both particles have mass \(\displaystyle m \).Step $\displaystyle 1$: the shift-of-origin rule for angular momentum.For a system of particles, the angular momentum about a point O is \(\displaystyle \vec L_O = \sum_i \vec r_i \times \vec p_i \), where \(\displaystyle \vec r_i \) is the position vector of particle \(\displaystyle i \) measured from O and \(\displaystyle \vec p_i = m_i\vec v_i \) is its momentum. If a second point \(\displaystyle O' \) is displaced from O by a fixed vector \(\displaystyle \vec a \), every particle's position vector relative to \(\displaystyle O' \) is \(\displaystyle \vec r_i' = \vec r_i - \vec a \), so\[\vec L_{O'} = \sum_i \vec r_i' \times \vec p_i = \sum_i (\vec r_i - \vec a)\times \vec p_i = \sum_i \vec r_i \times \vec p_i \;-\; \vec a \times \sum_i \vec p_i \]\[\vec L_{O'} = \vec L_O - \vec a \times \vec P, \qquad \vec P = \sum_i \vec p_i \ \text{(total linear momentum of the system)} \]This is the general rule: angular momentum about two different points can differ only by \(\displaystyle -\vec a \times \vec P \). If the system's total momentum \(\displaystyle \vec P \) is zero, that extra term vanishes for every choice of \(\displaystyle \vec a \) — so \(\displaystyle \vec L \) is the same about every point in space, not just the one point you happened to pick. This is the identity the whole proof rests on, so get it down before touching the two-particle setup.Step $\displaystyle 2$: check that \(\displaystyle \vec P = 0 \) here.\[\vec P = m\vec v_1 + m\vec v_2 = mv\hat x + m(-v\hat x) = \vec 0 \]The two particles have equal mass and equal speed but opposite velocity, so their momenta cancel exactly. By Step $\displaystyle 1$, this alone already guarantees \(\displaystyle \vec L_{O'} = \vec L_O \) for any two points \(\displaystyle O \) and \(\displaystyle O' \) — which is precisely the result the question asks you to show. What remains is to pin down what that common value of \(\displaystyle \vec L \) actually is.Step $\displaystyle 3$: compute \(\displaystyle \vec L \) about one convenient point, and watch it survive unchanged.Take the reference point O at particle $\displaystyle 2$'s own instantaneous position, so \(\displaystyle \vec r_2 = \vec 0 \). Let \(\displaystyle s \) be how far ahead particle $\displaystyle 1$ is along the \(\displaystyle \hat x \) direction at this instant; since the two lines stay a fixed perpendicular distance \(\displaystyle d \) apart,\[\vec r_1 = s\hat x + d\hat y \]Using \(\displaystyle \vec L = \vec r_1 \times \vec p_1 + \vec r_2\times \vec p_2 \), with \(\displaystyle \vec p_1 = mv\hat x \) and \(\displaystyle \vec r_2 = \vec 0 \):\[\vec L = (s\hat x + d\hat y)\times (mv\hat x) = mvs\,(\hat x\times\hat x) + mvd\,(\hat y \times \hat x) \]Since \(\displaystyle \hat x\times\hat x = \vec 0 \) and \(\displaystyle \hat y\times\hat x = -\hat z \) (the standard right-hand-rule identity, the reverse of \(\displaystyle \hat x\times\hat y=\hat z \)),\[\vec L = -mvd\,\hat z \]The along-the-line coordinate \(\displaystyle s \) has cancelled completely — it never appears in the final expression. That is the mark of a genuinely position-independent result: it means \(\displaystyle \vec L \) does not change as the particles slide along their lines with time, and (by Step $\displaystyle 1$, since \(\displaystyle \vec P=\vec 0 \)) it also does not change if the reference point O is moved anywhere else in space, on either line or off both of them.The direction \(\displaystyle -\hat z \) is perpendicular to the plane containing the two parallel lines — along the axis you get by curling the right-hand fingers from \(\displaystyle \vec r_1 \) toward \(\displaystyle \vec p_1 \), i.e. into the page if \(\displaystyle \hat x \) points along particle $\displaystyle 1$'s velocity and \(\displaystyle \hat y \) points from particle $\displaystyle 2$'s line across to particle $\displaystyle 1$'s line. The magnitude is \(\displaystyle mvd \), built only from the fixed perpendicular separation \(\displaystyle d \), never from how far along the lines the particles happen to be.This is the answer the question is after: the two-particle system's angular momentum has magnitude \(\displaystyle mvd \), directed perpendicular to the plane of the two lines, and this value is exactly the same no matter which point in space is chosen as the reference — because the system's total linear momentum is zero, so shifting the reference point by any vector \(\displaystyle \vec a \) changes \(\displaystyle \vec L \) only by the term \(\displaystyle -\vec a\times\vec P \), which vanishes identically.Answer: \(\displaystyle \vec L = mvd \), directed perpendicular to the plane containing the two lines, and this value is the same about every point in space, since the particles' equal-and-opposite momenta make the system's total linear momentum \(\displaystyle \vec P=\vec 0 \), so \(\displaystyle \vec L_{O'}-\vec L_O=-\vec a\times\vec P=\vec 0 \) for any shift \(\displaystyle \vec a \) of the reference point.

A rigid body at rest needs both zero net force and zero net torque — those give three independent equations, enough to pin down the two string tensions and the unknown position of the centre of gravity together.Call the bar's ends \(\displaystyle A\) (left) and \(\displaystyle B\) (right), with length \(\displaystyle L = 2\ \text{m}\). The string at \(\displaystyle A\) makes angle \(\displaystyle \theta_1 = 36.9^\circ\) with the vertical and carries tension \(\displaystyle T_1\); the string at \(\displaystyle B\) makes angle \(\displaystyle \theta_2 = 53.1^\circ\) with the vertical and carries tension \(\displaystyle T_2\). The bar's weight \(\displaystyle W\) acts straight down at its centre of gravity, a distance \(\displaystyle d\) from \(\displaystyle A\) — the quantity to find.Step $\displaystyle 1$ — Resolve forces along the horizontal and vertical. The bar hangs in equilibrium, so both components of the net force are zero:\[\text{Horizontal:}\quad T_1\sin\theta_1 = T_2\sin\theta_2 \qquad (i) \] \[\text{Vertical:}\quad T_1\cos\theta_1 + T_2\cos\theta_2 = W \qquad (ii) \]Step $\displaystyle 2$ — Balance torques about the left end \(\displaystyle A\). About a point on the bar, only the vertical component of a force produces a torque (a horizontal force applied anywhere on a horizontal bar has its line of action running along the bar itself, so its moment arm about any other point on that same line is zero — this is the step people skip). Taking torques about \(\displaystyle A\) — where \(\displaystyle T_1\) itself has zero moment arm —\[T_2\cos\theta_2 \cdot L = W \cdot d \qquad (iii) \]Step $\displaystyle 3$ — Eliminate the tensions. From (i), \(\displaystyle T_1 = T_2\,\dfrac{\sin\theta_2}{\sin\theta_1}\). Substituting into (ii):\[T_2\,\frac{\sin\theta_2\cos\theta_1}{\sin\theta_1} + T_2\cos\theta_2 = W \;\Longrightarrow\; T_2\cdot\frac{\sin\theta_2\cos\theta_1+\cos\theta_2\sin\theta_1}{\sin\theta_1}=W \]The numerator is the sine addition formula, \(\displaystyle \sin(\theta_1+\theta_2)\). Here \(\displaystyle \theta_1+\theta_2 = 36.9^\circ+53.1^\circ = 90^\circ\), and \(\displaystyle \sin 90^\circ = 1\), so this simplifies neatly to\[T_2 = W\sin\theta_1 \]Step $\displaystyle 4$ — Substitute back into the torque equation. From (iii):\[d = \frac{T_2\cos\theta_2\cdot L}{W} = \frac{W\sin\theta_1\cos\theta_2\cdot L}{W} = L\sin\theta_1\cos\theta_2 \]The weight \(\displaystyle W\) cancels — \(\displaystyle d\) depends only on the two angles and the bar's length, never on how heavy the bar actually is (a bar of any weight, hung at these two angles, has its centre of gravity at the same fraction of its length).Step $\displaystyle 5$ — Put in the numbers. \(\displaystyle 36.9^\circ\) and \(\displaystyle 53.1^\circ\) are the two acute angles of a $\displaystyle 3$–$\displaystyle 4$–$\displaystyle 5$ right triangle, so \(\displaystyle \sin 36.9^\circ = \cos 53.1^\circ = 0.600\) and \(\displaystyle \cos 36.9^\circ = \sin 53.1^\circ = 0.800\) (as a check, these satisfy \(\displaystyle \theta_1+\theta_2=90^\circ\) exactly, which is what made Step $\displaystyle 3$ collapse so cleanly).\[d = (2\ \text{m})(0.600)(0.600) = 0.72\ \text{m} \]Step $\displaystyle 6$ — Sanity check. Working back, \(\displaystyle T_2 = 0.600\,W\) and \(\displaystyle T_1 = T_2\sin\theta_2/\sin\theta_1 = 0.600\,W \times 0.800/0.600 = 0.800\,W\); checking (ii): \(\displaystyle 0.800W(0.800)+0.600W(0.600) = 0.64W+0.36W = 1.00\,W\) — consistent. Since \(\displaystyle T_1>T_2\), the end \(\displaystyle A\) carries more of the weight, so the centre of gravity should sit closer to \(\displaystyle A\) than to the midpoint — and indeed \(\displaystyle 0.72\ \text{m}\) is less than half the $\displaystyle 2$ m length, as it must be.The angles are given to three significant figures and the length is exact, but the trig values used $\displaystyle (0.600, 0.800)$ carry three significant figures, so the product is reported to the same precision.Answer: d = $\displaystyle 0.72$ m from the left end of the bar (the end whose string makes the $\displaystyle 36.9$° angle with the vertical).

A car in equilibrium on level ground has zero net force and zero net torque — that single fact, applied twice, pins down both wheel reactions.The car's mass is \(\displaystyle M = 1800 \) kg. "Weighs $\displaystyle 1800$ kg" is loose speech for mass; the actual downward weight force uses \(\displaystyle g \), not the mass number itself: \[W = Mg = (1800\ \text{kg})(9.8\ \text{m/s}^2) = 17640\ \text{N} \]Set up the geometry along the length of the car. Let the front axle be at point \(\displaystyle F\) and the back axle at point \(\displaystyle B\), a distance \[d = FB = 1.8\ \text{m} \] apart. The centre of gravity \(\displaystyle G\) lies \(\displaystyle 1.05\) m behind the front axle and therefore \[d_2 = d - 1.05\ \text{m} = 1.8 - 1.05 = 0.75\ \text{m} \] in front of the back axle.Three vertical forces act on the car: the weight \(\displaystyle W\) downward at \(\displaystyle G\), the total normal reaction \(\displaystyle R_F\) from the ground pushing up through the front axle (shared by the two front wheels), and the total normal reaction \(\displaystyle R_B\) pushing up through the back axle (shared by the two back wheels). The ground is level, so all three forces are vertical and parallel — no horizontal equation is needed.Translational equilibrium (net force zero): \[R_F + R_B = W \qquad \text{...(i)} \]Rotational equilibrium. Take torques about the front axle \(\displaystyle F\). Choosing this axis is the useful trick here: since \(\displaystyle R_F\) acts at \(\displaystyle F\), its moment arm about \(\displaystyle F\) is zero, so \(\displaystyle R_F\) drops out of the equation entirely and only \(\displaystyle R_B\) and \(\displaystyle W\) remain. The weight, acting at distance \(\displaystyle 1.05\) m from \(\displaystyle F\), and the reaction \(\displaystyle R_B\), acting at distance \(\displaystyle d = 1.8\) m from \(\displaystyle F\), must produce equal and opposite torques: \[R_B \times d = W \times 1.05\ \text{m} \] \[R_B = \frac{W \times 1.05}{d} = \frac{(17640\ \text{N})(1.05\ \text{m})}{1.8\ \text{m}} = 10290\ \text{N} \]From (i): \[R_F = W - R_B = 17640\ \text{N} - 10290\ \text{N} = 7350\ \text{N} \]As a check, take torques about the back axle \(\displaystyle B\) instead — this time \(\displaystyle R_B\) drops out, and the weight acts at distance \(\displaystyle d_2 = 0.75\) m from \(\displaystyle B\): \[R_F \times d = W \times d_2 \quad\Rightarrow\quad R_F = \frac{(17640\ \text{N})(0.75\ \text{m})}{1.8\ \text{m}} = 7350\ \text{N} \] This agrees with the value found above, confirming the arithmetic.\(\displaystyle R_F\) and \(\displaystyle R_B\) are each the combined reaction on a pair of wheels, not on one wheel. The car's chassis is symmetric left-to-right about its long axis, and the weight acts in that same plane of symmetry, so nothing distinguishes the left wheel of a pair from the right one — each wheel of a pair therefore carries exactly half of that axle's total reaction: \[\text{Force on each front wheel} = \frac{R_F}{2} = \frac{7350\ \text{N}}{2} = 3675\ \text{N} \] \[\text{Force on each back wheel} = \frac{R_B}{2} = \frac{10290\ \text{N}}{2} = 5145\ \text{N} \]Both forces act vertically upward on the car (Newton's third law pair to the wheel pressing down on the ground). The inputs are given to three or four significant figures (\(\displaystyle 1.05\) m to three, \(\displaystyle 1800\) kg and \(\displaystyle 1.8\) m taken as exact to the digits shown), so the results are kept at that same precision rather than rounded further — the numbers $\displaystyle 3675$ N and $\displaystyle 5145$ N already reflect the sensible number of figures for this data.As a sanity check, note the back wheels — closer to the centre of gravity — carry more of the car's weight than the front wheels, and the two together add back up to the full weight: \(\displaystyle 3675 \times 2 + 5145 \times 2 = 7350 + 10290 = 17640\ \text{N} = W\).Answer: Each front wheel bears a normal force of $\displaystyle 3675$ N, and each back wheel bears a normal force of $\displaystyle 5145$ N, both directed vertically upward from the level ground.

Same torque, same time — the one with the smaller moment of inertia speeds up faster.Newton's second law for rotation says \[\tau = I\alpha, \] where \(\displaystyle \tau\) is the applied torque, \(\displaystyle I\) is the moment of inertia about the rotation axis, and \(\displaystyle \alpha\) is the angular acceleration produced. Starting from rest, if the torque acts for a time \(\displaystyle t\) the angular speed gained is \[\omega = \alpha t = \frac{\tau}{I}\,t. \] For a given torque \(\displaystyle \tau\) and a given time \(\displaystyle t\), \(\displaystyle \omega\) is inversely proportional to \(\displaystyle I\): the smaller the moment of inertia, the larger the angular acceleration, and so the larger the angular speed built up in that time.Now compare the two moments of inertia. Both bodies have the same mass \(\displaystyle M\) and the same radius \(\displaystyle R\).For a hollow cylinder rotating about its own (symmetry) axis, all the mass lies at the rim, at distance \(\displaystyle R\) from the axis: \[I_{\text{cylinder}} = MR^2. \]For a solid sphere rotating about an axis through its centre, the mass is spread through the volume, most of it closer to the axis than \(\displaystyle R\): \[I_{\text{sphere}} = \frac{2}{5}MR^2. \]The trap here is to think that "same mass and same radius" means the two bodies must respond identically to the same torque. Moment of inertia depends on how that mass is arranged relative to the axis, not just on \(\displaystyle M\) and \(\displaystyle R\) alone. A hollow shell keeps every bit of its mass out at the maximum distance \(\displaystyle R\), while a solid sphere has mass filling the interior at distances smaller than \(\displaystyle R\), giving it a smaller \(\displaystyle I\) for the same \(\displaystyle M\) and \(\displaystyle R\).Since \[I_{\text{sphere}} = \frac{2}{5}MR^2 \;<\; MR^2 = I_{\text{cylinder}}, \] and \(\displaystyle \omega = \tau t/I\) with \(\displaystyle \tau\) and \(\displaystyle t\) the same for both, the sphere's smaller moment of inertia gives it the larger angular speed after the same time: \[\omega_{\text{sphere}} = \frac{\tau t}{\tfrac{2}{5}MR^2} = \frac{5}{2}\cdot\frac{\tau t}{MR^2} \;>\; \omega_{\text{cylinder}} = \frac{\tau t}{MR^2}. \] In fact the sphere ends up spinning \(\displaystyle 5/2 = 2.5\) times as fast as the cylinder, for the same applied torque and the same elapsed time.Answer: The solid sphere acquires the greater angular speed. Because \(\displaystyle I_{\text{sphere}} = \tfrac{2}{5}MR^2\) is smaller than \(\displaystyle I_{\text{cylinder}} = MR^2\) for the same mass and radius, the same torque produces a larger angular acceleration in the sphere, giving \(\displaystyle \omega_{\text{sphere}} = \tfrac{5}{2}\,\omega_{\text{cylinder}}\) after the same time.