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

A spinning solid cylinder carries both rotational kinetic energy and angular momentum, and both come from the same quantity — its moment of inertia about the spin axis.For a uniform solid cylinder spinning about its own central (longitudinal) axis, the moment of inertia is \[I = \frac{1}{2}MR^2 \] where \(\displaystyle M\) is the mass of the cylinder and \(\displaystyle R\) is its radius. This is different from a hollow cylinder (\(\displaystyle I = MR^2\)) or a cylinder spinning about a diameter — always check which axis before picking the formula.Step $\displaystyle 1$: Moment of inertia about the axisGiven \(\displaystyle M = 20\ \text{kg}\), \(\displaystyle R = 0.25\ \text{m}\): \[I = \frac{1}{2}(20\ \text{kg})(0.25\ \text{m})^2 = \frac{1}{2}(20)(0.0625)\ \text{kg m}^2 = 0.625\ \text{kg m}^2 \]Step $\displaystyle 2$: Rotational kinetic energyThe rotational analogue of \(\displaystyle \tfrac12 mv^2\) is \[KE_{\text{rot}} = \frac{1}{2}I\omega^2 \] where \(\displaystyle \omega\) is the angular speed. Here \(\displaystyle \omega = 100\ \text{rad s}^{-1}\) is already in rad/s — the usual trap in these problems is forgetting to convert from rev/min or rev/s into rad/s before using this formula, but that conversion isn't needed here.\[KE_{\text{rot}} = \frac{1}{2}(0.625\ \text{kg m}^2)(100\ \text{rad s}^{-1})^2 = \frac{1}{2}(0.625)(10000)\ \text{J} = 3125\ \text{J} \]Step $\displaystyle 3$: Angular momentumThe rotational analogue of \(\displaystyle p = mv\) is \[L = I\omega \] \[L = (0.625\ \text{kg m}^2)(100\ \text{rad s}^{-1}) = 62.5\ \text{kg m}^2\text{s}^{-1} = 62.5\ \text{J s} \]\(\displaystyle L\) points along the axis of rotation (by the right-hand rule); the question asks only for its magnitude, which is the value computed above.Rounding: the given data — $\displaystyle 20$ kg, $\displaystyle 0.25$ m, $\displaystyle 100$ rad s⁻¹ — support three significant figures, so the final values are reported to three figures: \(\displaystyle KE_{\text{rot}} = 3.13\times10^{3}\ \text{J}\) (from the exact value $\displaystyle 3125$ J) and \(\displaystyle L = 62.5\ \text{J s}\), which already has three significant figures.Answer: Rotational kinetic energy ≈ \(\displaystyle 3.13\times10^{3}\ \text{J}\) ($\displaystyle 3125$ J); angular momentum about the axis = $\displaystyle 62.5$ J s.

No external torque acts on the child-turntable system, so angular momentum is conserved — but kinetic energy is not, because the child's own muscles do work.(a) The new angular speedThe turntable-plus-child system spins freely (no friction, no external forces trying to twist it), so there is no external torque about the rotation axis. By the law of conservation of angular momentum,\[L = I\omega = \text{constant} \]where \(\displaystyle I\) is the moment of inertia about the vertical axis and \(\displaystyle \omega\) is the angular speed. Pulling the arms in changes \(\displaystyle I\) but cannot change \(\displaystyle L\), since nothing outside the system supplies a torque.Let the outstretched-arms state be "$\displaystyle 1$" and the folded-arms state be "$\displaystyle 2$":\[I_1\omega_1 = I_2\omega_2 \]You are given \(\displaystyle \omega_1 = 40\ \text{rev/min}\) and \(\displaystyle I_2 = \dfrac{2}{5}I_1\). Substituting,\[I_1(40) = \left(\frac{2}{5}I_1\right)\omega_2 \]The \(\displaystyle I_1\) cancels (its actual value is never needed):\[\omega_2 = \frac{40}{2/5} = 40 \times \frac{5}{2} = 100\ \text{rev/min} \]A common slip here is to multiply by \(\displaystyle 2/5\) instead of dividing by it — moment of inertia decreases, so angular speed must increase, and \(\displaystyle 100 > 40\) checks out.\[\boxed{\omega_2 = 100\ \text{rev/min}} \](Kept to two significant figures, matching the two significant figures given in \(\displaystyle 40\ \text{rev/min}\).)(b) The new kinetic energy is largerRotational kinetic energy is \(\displaystyle KE = \dfrac{1}{2}I\omega^2\). It is useful to rewrite it in terms of the conserved quantity \(\displaystyle L = I\omega\):\[KE = \frac{1}{2}I\omega^2 = \frac{(I\omega)^2}{2I} = \frac{L^2}{2I} \]Since \(\displaystyle L\) is the same before and after (part a), \(\displaystyle KE\) varies only through \(\displaystyle I\), and it varies as \(\displaystyle 1/I\) — a smaller moment of inertia means a larger kinetic energy for the same angular momentum. Taking the ratio,\[\frac{KE_2}{KE_1} = \frac{L^2/(2I_2)}{L^2/(2I_1)} = \frac{I_1}{I_2} = \frac{I_1}{(2/5)I_1} = \frac{5}{2} = 2.5 \]So\[KE_2 = 2.5\,KE_1 \]which is indeed greater than the initial kinetic energy — the child's rotational kinetic energy increases by a factor of $\displaystyle 2.5$ when the arms are folded in.Where does this extra energy come from? Angular momentum is conserved here because the force each part of the child's body exerts on every other part is an internal force, and internal forces cannot change the total angular momentum of the system. But internal forces can do net work on the system — and that is exactly what happens. As the child pulls the arms inward, each arm is moving on a curved (roughly circular) path, so pulling it inward means pulling it against the outward (centrifugal, in the rotating frame) tendency of the spinning arm — the muscles must do positive work to drag the arm's mass closer to the axis while the arm is already moving tangentially. This muscular work is supplied by the chemical (biological) energy stored in the child's body, and it is this energy that is converted into the extra rotational kinetic energy. There is no violation of energy conservation: the rise in \(\displaystyle KE\) is paid for by the child's internal muscular effort, not created from nothing.Answer: (a) \(\displaystyle \omega_2 = 100\ \text{rev/min}\). (b) \(\displaystyle KE_2 = 2.5\,KE_1\) — the kinetic energy rises by a factor of $\displaystyle 2.5$; angular momentum stays fixed (no external torque), but the child's muscles do positive work pulling the arms in against their own rotational motion, and this internal muscular (biochemical) work is converted into the extra rotational kinetic energy.

A rope wound round a cylinder and pulled tangentially applies a torque equal to force times radius — and for a hollow cylinder, all the mass sits at the rim, which makes its moment of inertia the largest possible for its mass.Setting up the torque. The rope leaves the cylinder tangent to its surface, so the full force \(\displaystyle F \) acts at the rim, perpendicular to the radius. The torque about the axis is\[\tau = F R \]With \(\displaystyle F = 30 \text{ N} \) and \(\displaystyle R = 40 \text{ cm} = 0.40 \text{ m} \):\[\tau = 30 \text{ N} \times 0.40 \text{ m} = 12 \text{ N m} \]Moment of inertia of a hollow cylinder. Unlike a solid cylinder (where mass is spread from the axis out to the rim, giving \(\displaystyle I = \tfrac{1}{2}MR^2 \)), a hollow cylinder has essentially all its mass concentrated at radius \(\displaystyle R \), like a ring:\[I = MR^2 \]This is the step people get wrong — reaching for the solid-cylinder formula out of habit. Here \(\displaystyle M = 3 \text{ kg} \), so\[I = (3 \text{ kg})(0.40 \text{ m})^2 = 3 \times 0.16 \text{ kg m}^2 = 0.48 \text{ kg m}^2 \]Angular acceleration. Newton's second law for rotation, \(\displaystyle \tau = I\alpha \), gives\[\alpha = \frac{\tau}{I} = \frac{12 \text{ N m}}{0.48 \text{ kg m}^2} = 25 \text{ rad s}^{-2} \]Linear acceleration of the rope. The rope stays in contact with the rim (no slipping), so a point on the rim — and hence the rope — moves with linear acceleration\[a = \alpha R = (25 \text{ rad s}^{-2})(0.40 \text{ m}) = 10 \text{ m s}^{-2} \]As a check, notice this is just \(\displaystyle a = F/M = 30/3 = 10 \text{ m s}^{-2} \): for a hollow cylinder unwinding a rope from its own rim, the rim's linear acceleration always equals what the same force would give the whole mass moving freely — a special feature of \(\displaystyle I = MR^2 \), not true for a solid cylinder or sphere.All three input quantities ($\displaystyle 3$ kg, $\displaystyle 40$ cm, $\displaystyle 30$ N) carry two significant figures, so both results are kept to two significant figures.Answer: Angular acceleration \(\displaystyle \alpha = 25 \text{ rad s}^{-2} \); linear acceleration of the rope \(\displaystyle a = 10 \text{ m s}^{-2} \) (directed along the rope, tangent to the cylinder).

Power delivered by a torque spinning something at constant angular speed is \(\displaystyle P = \tau\omega \), not \(\displaystyle \tau\alpha \) — you multiply by the speed, not the acceleration.For rotational motion, the power input by an applied torque \(\displaystyle \tau \) that keeps a body turning at angular speed \(\displaystyle \omega \) is the rotational analogue of \(\displaystyle P = Fv \): \[P = \tau \omega \] where \(\displaystyle \tau \) is the torque the engine must supply (here, to overcome friction and keep the rotor from slowing down), and \(\displaystyle \omega \) is the (uniform) angular speed of the rotor.The note in the problem is the physics behind why any torque is needed at all: if the rotor really felt no friction, zero torque would keep it spinning at constant \(\displaystyle \omega \) forever (Newton's first law for rotation). Since a torque of $\displaystyle 180$ N m is needed just to hold \(\displaystyle \omega \) steady, that torque is entirely doing work against friction — and the engine, being $\displaystyle 100$% efficient, delivers exactly that much power as useful output.Substituting the values\[\tau = 180 \ \text{N m}, \qquad \omega = 200 \ \text{rad s}^{-1} \]\[P = \tau\omega = (180 \ \text{N m}) \times (200 \ \text{rad s}^{-1}) \]\[P = 36000 \ \text{W} \]Radians are dimensionless, so N m × rad s⁻¹ simplifies cleanly to watts — no extra angle unit survives in the answer.Significant figuresBoth given quantities ($\displaystyle 180$ N m and $\displaystyle 200$ rad s⁻¹) carry $\displaystyle 2$–$\displaystyle 3$ significant figures as printed; keeping $\displaystyle 2$ significant figures (the more conservative reading, since trailing zeros without a decimal point are ambiguous) gives\[P = 3.6 \times 10^{4} \ \text{W} = 36 \ \text{kW} \]This is power, not torque or energy — since the engine is $\displaystyle 100$% efficient, all of this power shows up as mechanical output; none is lost internally.Answer: \(\displaystyle P = 3.6 \times 10^{4} \ \text{W} = 36 \ \text{kW} \)

A hole cut out of a disc is equivalent to adding a disc of negative mass at the hole's location — superposition turns a missing region into an ordinary centre-of-mass calculation.Set up coordinates along the line joining the two centres: let \(\displaystyle O \), the centre of the original (uncut) disc, be the origin, and let the positive \(\displaystyle x \)-axis run through \(\displaystyle O' \), the centre of the circular hole, so \(\displaystyle O' \) sits at \(\displaystyle x = R/2 \).Because the disc is uniform, mass is proportional to area, with the same surface mass density \(\displaystyle \sigma \) (mass per unit area) throughout. For the original full disc of radius \(\displaystyle R \), \[M_1 = \sigma \pi R^2 . \] The disc-shaped piece removed to make the hole has radius \(\displaystyle R/2 \), so its mass is \[M_2 = \sigma \pi \left(\frac{R}{2}\right)^2 = \frac{\sigma \pi R^2}{4} = \frac{M_1}{4}. \]The trick is to see the full disc as two pieces added together: the remaining lamina (mass \(\displaystyle M_1 - M_2 \), centre of mass at the unknown position \(\displaystyle x_{cm} \) we want) plus the removed circular piece put back in place (mass \(\displaystyle M_2 \), at \(\displaystyle x = R/2 \)). Since the centre of mass of the whole disc is just \(\displaystyle O \) itself (\(\displaystyle x = 0 \)), the mass-weighted average of these two pieces must reproduce that, by the definition of centre of mass \(\displaystyle \sum m_i x_i = M x_{cm} \): \[(M_1 - M_2)\,x_{cm} + M_2 \cdot \frac{R}{2} = M_1 \cdot 0 . \]Solving for \(\displaystyle x_{cm} \): \[x_{cm} = -\frac{M_2 \left(R/2\right)}{M_1 - M_2}. \]Substituting \(\displaystyle M_2 = M_1/4 \): \[x_{cm} = -\frac{\left(\dfrac{M_1}{4}\right)\left(\dfrac{R}{2}\right)}{M_1 - \dfrac{M_1}{4}} = -\frac{\dfrac{M_1 R}{8}}{\dfrac{3M_1}{4}} = -\frac{R}{8}\times\frac{4}{3} = -\frac{R}{6}. \]Notice that \(\displaystyle M_1 \) — and with it \(\displaystyle \sigma \) and the actual mass of the disc — cancelled out completely. The result depends only on the geometry (the ratio of the two radii and the offset \(\displaystyle R/2 \)), so \(\displaystyle R/6 \) is an exact fraction, not a number to round.The minus sign is the part people get wrong. It is tempting to think the centre of gravity "moves toward" the missing material, but removing mass from one side pulls the balance point away from that side and toward the material that remains. Since the disc lies in a uniform gravitational field, its centre of gravity coincides with its centre of mass, so the same conclusion applies to both.The centre of gravity therefore lies on the line joining the centre of the original disc and the centre of the hole, at a distance \(\displaystyle R/6 \) from the centre of the original disc, on the side directly opposite the hole.Answer: The centre of gravity lies on the line joining the two centres, at a distance \(\displaystyle R/6 \) from the centre of the original disc, on the side opposite to the hole.

The stick's own weight can be treated as one force acting at its centre of mass — and the first balance point tells you exactly where that is.A metre stick is uniform, so its centre of mass sits at its geometric centre, the $\displaystyle 50.0$ cm mark. That is why, with nothing else on it, it balances on a knife edge placed at $\displaystyle 50.0$ cm: the stick's weight \(\displaystyle W\) acting downward at $\displaystyle 50.0$ cm is exactly balanced by the normal reaction of the knife edge acting upward at the same point, so there is no net torque about that point.Setting up the new balance conditionNow two $\displaystyle 5$ g coins are stacked at the $\displaystyle 12.0$ cm mark, and the knife edge has to move to $\displaystyle 45.0$ cm to restore balance. At the new pivot ($\displaystyle 45.0$ cm), the system is again in rotational equilibrium, so by the principle of moments (law of the lever):\[\text{sum of clockwise moments about the pivot} = \text{sum of anticlockwise moments about the pivot} \]Only two forces produce a torque about the $\displaystyle 45.0$ cm pivot: These two weights are on opposite sides of the pivot, so their torques oppose each other, which is exactly what balance requires.Distances of each force from the new pivot\[d_c = 45.0\ \text{cm} - 12.0\ \text{cm} = 33.0\ \text{cm} \qquad (\text{coins to pivot}) \]\[d_M = 50.0\ \text{cm} - 45.0\ \text{cm} = 5.0\ \text{cm} \qquad (\text{stick's centre of mass to pivot}) \]Applying the principle of moments\[m_c\, g\, d_c = M\, g\, d_M \]The acceleration due to gravity \(\displaystyle g\) appears on both sides and cancels — this is a balance of masses through their moment arms, not a comparison of a mass to a weight, so there is no need to know \(\displaystyle g\) at all here.\[m_c\, d_c = M\, d_M \]The mass of the coins is two $\displaystyle 5$ g coins stacked together:\[m_c = 2 \times 5\ \text{g} = 10\ \text{g} \]Substituting the values:\[(10\ \text{g})(33.0\ \text{cm}) = M(5.0\ \text{cm}) \]\[M = \frac{(10\ \text{g})(33.0\ \text{cm})}{5.0\ \text{cm}} = \frac{330\ \text{g·cm}}{5.0\ \text{cm}} \]\[M = 66\ \text{g} \]The centimetre units cancel cleanly, leaving mass in grams. Rounding to two significant figures (matching the two-significant-figure precision of the $\displaystyle 5$ g coin masses), the mass of the metre stick works out to a clean $\displaystyle 66$ g, so no further rounding is needed.Answer: The mass of the metre stick is $\displaystyle 66$ g.

The rotational kinetic energy formula \(\displaystyle K_{rot} = \tfrac{1}{2}I\omega^2 \) is the only place \(\displaystyle \omega\) appears, so the whole problem is about linking that to the translational kinetic energy you're told about.The oxygen molecule moves through the gas (translational motion of its centre of mass) and also spins about an axis through its centre, perpendicular to the line joining the two atoms (rotational motion). These are two separate kinds of kinetic energy, and the two formulas that describe them are:\[K_{trans} = \frac{1}{2}mv^2 , \qquad K_{rot} = \frac{1}{2}I\omega^2 \]where \(\displaystyle m\) is the mass of the molecule, \(\displaystyle v\) is its (mean) speed, \(\displaystyle I\) is its moment of inertia about the given axis, and \(\displaystyle \omega\) is the angular velocity you want.You are given:\[m = 5.30\times10^{-26}\ \text{kg}, \qquad I = 1.94\times10^{-46}\ \text{kg m}^2, \qquad v = 500\ \text{m/s} \]and the condition\[K_{rot} = \frac{2}{3}K_{trans} \]A step people miss here: \(\displaystyle v = 500\) m/s is used exactly as the "mean speed" of the molecule's centre of mass — it is not converted to anything else, and it belongs only in \(\displaystyle K_{trans}\), never in \(\displaystyle K_{rot}\).Substitute the two energy expressions into the given ratio:\[\frac{1}{2}I\omega^2 = \frac{2}{3}\left(\frac{1}{2}mv^2\right) \]The factors of \(\displaystyle \tfrac{1}{2}\) cancel, leaving\[I\omega^2 = \frac{2}{3}mv^2 \]Solve for \(\displaystyle \omega\):\[\omega = \sqrt{\frac{2mv^2}{3I}} \]Now substitute numbers, carrying units all the way through. First,\[v^2 = (500\ \text{m/s})^2 = 2.50\times10^{5}\ \text{m}^2/\text{s}^2 \]\[mv^2 = (5.30\times10^{-26}\ \text{kg})(2.50\times10^{5}\ \text{m}^2/\text{s}^2) = 1.325\times10^{-20}\ \text{kg m}^2/\text{s}^2 \]\[2mv^2 = 2.65\times10^{-20}\ \text{kg m}^2/\text{s}^2 \]\[3I = 3(1.94\times10^{-46}\ \text{kg m}^2) = 5.82\times10^{-46}\ \text{kg m}^2 \]\[\omega^2 = \frac{2mv^2}{3I} = \frac{2.65\times10^{-20}\ \text{kg m}^2/\text{s}^2}{5.82\times10^{-46}\ \text{kg m}^2} = 4.553\times10^{25}\ \text{s}^{-2} \]Notice the units: \(\displaystyle \text{kg m}^2/\text{s}^2\) divided by \(\displaystyle \text{kg m}^2\) leaves \(\displaystyle \text{s}^{-2}\), exactly what \(\displaystyle \omega^2\) should have (so \(\displaystyle \omega\) itself will come out in \(\displaystyle \text{s}^{-1}\), i.e. rad/s, since radians are dimensionless).Taking the square root:\[\omega = \sqrt{4.553\times10^{25}\ \text{s}^{-2}} = \sqrt{45.53\times10^{24}} = 6.748\times10^{12}\ \text{rad/s} \]All three given quantities (\(\displaystyle m\), \(\displaystyle I\), \(\displaystyle v\)) carry three significant figures, so the answer is rounded once, at the end, to three significant figures:\[\omega \approx 6.75\times10^{12}\ \text{rad/s} \]This is the average angular speed — a single number describing how fast the molecule tumbles about its centre, not a direction, since no particular spin axis orientation was specified in the problem beyond "perpendicular to the line joining the two atoms."Answer: \(\displaystyle \omega \approx 6.75\times10^{12}\ \text{rad/s}\)