NCERT Solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion - Param Himalaya

NCERT Solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion - Param Himalaya 

6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? 

Solution : 

(i) The centre of mass of sphere is located at its centre. 

(ii) The centre of mass of cylinder is at the centre of its axis of symmetry. 

(iii) The centre of mass of a ring is at centre of the ring. 

(iv) The centre of mass of a cube is at its geometrical centre. 

The centre of mass may lie outside the body also, e.g., in case a ring, a hollow sphere, a hollow cylinder, a hollow cube etc.

6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = $10^{-10}$ m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 

Solution : 

In the HCl molecule


Distance between H and Cl atoms = 1.27 A° 

Mass of H atoms = 35.5 m 

Let the centre of mass of the system lie at a distance x from the Cl atoms.

Distance of the centre of mass from the H atom = (1.27 - X)

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore , we can have : 

$\frac{m(1.27-x)+35.5mx}{m+35.5mx} = 0$

$m(1.27-X) = -35.5X$

$\therefore X= - \frac{1.27}{35.5-1} = - 0.037 A^{o}$

Here , the negative sign indicates that the centre of mass lies at the left of the molecule. Hence , the centre of mass of HCl molecule lies $0.037 A^{o}$ from the Cl atoms. 

6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ? 

Solution : 

It is necessary that an external force is applied on a system to change its motion. In this case, the boy is running arbitrarily on trolley with velocity v, which produces only internal forces in the system. The internal force cannot affect the motion of the system. As there is no external force on the system, so the speed of the centre of mass of the trolley and child system will remain same.

6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. 

Solution : 

The following figure shows two vectors a and b at an angle θ extended to form a parallelogram.

From the above figure in ΔOMN, 

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

sinθ= MN/OM = MN /| b | 

MN=| b |sinθ  ---- (1)

The magnitude for cross product of a and b is, | a×b |=| a || b |sinθ 

From the equation (1), 

| a×b |=( OK )( MN ) 

Multiply and divide by 2. 

| a×b |=( OK )( MN )× 2/ 2 

=2×Area of ΔOMK 

Area of ΔOMK= 1/2 | a×b | 

Thus, the area of the triangle contained between the vectors a and b is one half the magnitude of the a×b.

6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c. 

Solution: 

$\vec{OJ} = \vec{a}$ , $\vec{OL} = \vec{b}$ and $\vec{OK} = \vec{c}$ , 

$\hat{n}$ is a unit vector along OJ perpendicular to the plane containing 

$\vec{b}$ and $\vec{c}$ 

Now , $\vec{b} \times \vec{c} = bcsin90°\hat{n} =bc\hat{n}$

$\vec{a}.(\vec{b} \times \vec{c}) = a.(bc\vec{n})$

= $abccos\theta \hat{n}$ = abc cos0°=abc

The volume of the parallelepiped.

6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components $p_{x} , p_{y} and p_{z}$ . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. 

Solution: 

Linear momentum 

$\vec{p} = p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k}$

Position vector of the body 

$\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$

Angular momentum, 

$\vec{I} = \vec{r} \times \vec{p}$

= $(p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k}) \times (x\hat{i}+y\hat{j}+z\hat{k})$

=$\begin{array}{l}\begin{bmatrix} \hat{i} & \hat{j} & \hat{k}\\ x & y & z\\ p_{x}& p_{y} & p_{z} \end{bmatrix}\end{array}$

=$\hat{i}(yp_{z}-zp_{y})+\hat{j}(zp_{x}-xp_{z})+\hat{k}(xp_{y}-yp_{x})$

6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken. 

6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end


6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. 

6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? 

6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 

6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.

 6.14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

 6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. 

6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? 

6.17 The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Previous Post Next Post