Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 7 - System Of Particles And Rotational Motion

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Page No 50:

Question 7.1:

For which of the following does the centre of mass lie outside the body ?
(a) A pencil
(b) A shotput
(c) A dice
(d) A bangle

Answer:

In a bangle its centre of mass lies at its geometrical centre.
Hence, the correct answer is option (d).

Page No 50:

Question 7.2:

Which of the following points is the likely position of the centre of mass of the system shown in Fig. 7.1?

(a) A
(b) B
(c) C
(d) D

Answer:

Because of presence of sand (more dense than air) in the lower hemisphere , centre of mass will get shifted by some distance below the geometrical centre of the sphere.

Hence, the correct answer is option (c).

Page No 50:

Question 7.3:

A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:

(a) mva ê_x
(b) 2mva ê_x
(c) ymv ê_x
(d) 2ymv ê_x

Answer:

Angular momentum,
$\vec{L} = \vec{r} \times \vec{p} So, ∆ \vec{L} = \vec{r} \times ∆ \vec{p} = \vec{r} \times m (∆ \vec{v}) ∆ \vec{v} = {\vec{v}}_{f} - {\vec{v}}_{i} = - v {\vec{e}}_{y} - v {\vec{e}}_{y} = - 2 v {\vec{e}}_{y} and \overset{⇀}{r} = y {\vec{e}}_{y} + a {\vec{e}}_{z} So, \vec{L} = (y {\vec{e}}_{y} + a {\vec{e}}_{z}) \times (- 2 v {\vec{e}}_{y}) = - a {\vec{e}}_{z} m v (- 2) = + 2 a m v {\vec{e}}_{z}$

Hence, the correct answer is option (b).

Page No 51:

Question 7.4:

When a disc rotates with uniform angular velocity, which of the following is not true?
(a) The sense of rotation remains same.
(b) The orientation of the axis of rotation remains same.
(c) The speed of rotation is non-zero and remains same.
(d) The angular acceleration is non-zero and remains same.

Answer:

If angular velocity ω is constant , then angular acceleration,

$α = \frac{d ω}{d t} = 0$
Hence, the correct answer is option (d)

Page No 51:

Question 7.5:

A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind (Fig. 7.3). The moment of inertia about the
z-axis is then

(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner.

Answer:

Moment of Inertia, $I = m a s s \times {(r a d i u s o f g y r a t i o n)}^{2}$

So, due to the shift of some mass from Q to centre decreases the value of effective radius of gyration of the body, which will cause a decrease in value of I .
Hence, the correct answer is option (a).

Page No 51:

Question 7.6:

In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane,
(a) I
(b) II
(c) III
(d) IV

Answer:

As the mass at Q is shifted to centre of plate , hence the new centre of mass of the plate will shift in the opposite quadrant of mass Q , that is in III quadrant.
Hence, the correct answer is option (c).

Page No 51:

Question 7.7:

The density of a non-uniform rod of length 1 m is given by ρ(x) = a(1 + bx²) where a and b are constants and 0 ≤ x ≤ 1.
The centre of mass of the rod will be at
(a) $\frac{3 (2 + b)}{4 (3 + b)}$

(b) $\frac{4 (2 + b)}{3 (3 + b)}$

(c) $\frac{3 (3 + b)}{4 (2 + b)}$

(d) $\frac{4 (3 + b)}{3 (2 + b)}$

Answer:

Since, ρ(x) = a(1 + bx²)
So, if b = 0, ρ(x) = a = constant, which tells us that centre of mass of the rod should lie at 0.5 m in this case.
We see that only option (a) satisfies the condition at b = 0.
Hence , the correct answer is option (a).

Page No 52:

Question 7.8:

A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round).
The speed of the round afterwards is
(a) 2ω
(b) ω
(c) $\frac{ω}{2}$
(d) 0

Answer:

Since there is no external torque acting on the Merry-go-round , so its angular momentum remains constant before and after the jumping off of the person of mass M.

I₁ω₁=I₂ω₂

I₁ = 2M R², ω_{1 =}ω .

I₂=M R²,
ω₂= ?
So, on putting the above values we get ω₂= 2ω.
Hence, the correct answer is option (a).

Page No 52:

Question 7.9:

Choose the correct alternatives:
(a) For a general rotational motion, angular momentum L and angular velocity ω need not be parallel.
(b) For a rotational motion about a fixed axis, angular momentum L and angular velocity ω are always parallel.
(c) For a general translational motion, momentum p and velocity v are always parallel.
(d) For a general translational motion, acceleration a and velocity v are always parallel.

Answer:

In a general rotational motion along a non-symmetrical axis, angular momentum L and angular velocity ω need not be parallel.
Also , for a general translational motion along any path, momentum p and velocity v are always parallel.
Hence, the correct options are (a) and (c).

Page No 52:

Question 7.10:

Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines.

Choose the correct options:
(a) Angular momentum l₁ of particle 1 about A is l₁ = mvd₁
(b) Angular momentum l₂ of particle 2 about A is l₂ = mvr₂
(c) Total angular momentum of the system about A is l = mv(r₁ + r₂)
(d) Total angular momentum of the system about A is l = mv(d₂ – d₁) ⊗ represents a unit vector coming out of the page.⊗ represents a unit vector going into the page.

Answer:

From the diagram given in the question, it is clear that d₁< d_2^.
Angular momentum of particle 1 is,
l₁= mvd₁ outwards the page
Angular momentum of particle 2 is,
l₂= mvd₂ into the page
As, d₁< d₂
so, mvd₁< mvd₂
Total angular momentum of the system about A, l = l₂ – l₁= mv(d₂ – d₁) ⊗ into the page.
Hence, the correct options are (a) and (d).

Page No 52:

Question 7.11:

The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?
(a) The forces may be acting radially from a point on the axis.
(b) The forces may be acting on the axis of rotation.
(c) The forces may be acting parallel to the axis of rotation.
(d) The torque caused by some forces may be equal and opposite to that caused by other forces.

Answer:

Torque = $\overset{⇀}{τ} = \overset{⇀}{r} \times \overset{⇀}{F} = r F \sin θ \hat{n}$

(a) If the forces are acting radially from a point on the axis, then θ = 0^o , which will give torque = rFsinθ = 0 .
(b) If the forces may be acting on the axis of rotation, then r = 0 , so torque = 0.
(c) If the forces may be acting parallel to the axis of rotation, then component of force perpendicular to radius is zero ,hence torque = 0.
(d) If the torque caused by some forces may be equal and opposite to that caused by other forces, then net torque should be 0.
Hence, the correct options are (a), (b), (c) and (d).

Page No 53:

Question 7.12:

Figure 7.5 shows a lamina in x-y plane. Two axes z and z' pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z'-axis than the z-axis.)

(a) Torque τ caused by F about z axis is along $- \hat{k}$ .
(b) Torque τ' caused by F about z' axis is along $- \hat{k}$ .
(c) Torque τ caused by F about z axis is greater in magnitude than that about z' axis.
(d) Total torque is given be τ = τ + τ'.

Answer:

As shown in the diagram, point P is closer to z'-axis than z-axis.

Torque
$= \vec{r} \times \vec{F} Direction of torque can be determined by right hand rule .$
(a) False,
Torque is along $\hat{k}$ as force F is in the right of z axis pointing as shown in the diagram given in the question.
(b) True,
Torque is along $- \hat{k}$ as force F is in the left of z' axis pointing as shown in the diagram given in the question.
(c) True, as perpendicular distance between force and the z axis is more than the force and the z' axis.(The point P is closer to z′-axis than the z-axis)
(d) False, there is no sense in adding torques about 2 different axes.

Page No 53:

Question 7.13:

With reference to Fig. 7.6 of a cube of edge a and mass m, state whether the following are true or false.
(O is the centre of the cube.)

(a) The moment of inertia of cube about z-axis is I_z = I_x+ I_y
(b) The moment of inertia of cube about z' is $I'_{z} = I_{z} + \frac{m a^{2}}{2}$
(c) The moment of inertia of cube about z'' is $I_{z} + \frac{m a^{2}}{2}$
(d) I_x= I_y

Answer:

(a) False, as perpendicular axis theorem is valid for lamina in 2D only but the cube is a three dimensional object.
(b) True,
As moment of inertia of an object about an axis passing through its centre of mass O is,
$= I_{z} and Z' - axis is parallel to the Z - axis also, mass of the object is m, perpendicular distance between Z' - axis and Z - axis is \frac{a}{\sqrt{2}} So, I'_{z} = I_{z} + m \times {(\frac{a}{\sqrt{2}})}^{2} \Rightarrow I'_{z} = I_{z} + \frac{m a^{2}}{2}$
(c) False,
As Z-axis is not parallel to the Z''-axis, so we can not apply parallel axis theorem in this case as above.
(d) True,
By symmetry of the cube about x-axis and y-axis, moment of inertia I_x= I_y

Page No 53:

Question 7.14:

The centre of gravity of a body on the earth coincides with its centre of mass for a ‘small’ object whereas for an ‘extended’ object it may not. What is the qualitative meaning of ‘small’ and ‘extended’ in this regard?
For which of the following the two coincides? A building, a pond, a lake, a mountain?

Answer:

In general, we compare an object with the dimension of the earth. An object is said to be small if its vertical height is very small compared to the radius of the earth, otherwise it is extended.
(a) The buildings and ponds are small objects, so we may say that their centre of gravity may coincide with the centre of mass.
(b) But, a deep lake and a mountain are considered as extended objects, so their centre of gravity may not coincide with the centre of mass.

Page No 53:

Question 7.15:

Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

Answer:

As, moment of inertia,
$I = \sum_{} m_{i} {r_{i}}^{2}$
It means, moment of inertia will be more if the mass of an object is distributed at a larger distance from the axis of rotation.
For a hollow cylinder all the mass particles are at a distance of R from the axis (i.e., at their optimum distance), but for a solid sphere the distribution of the mass particles is uniform from centre to radius R.
So, the moment of inertia of the sphere is smaller than that of a hollow cylinder about about an axis passing through their axes of symmetry.

Page No 53:

Question 7.16:

The variation of angular position θ, of a point on a rotating rigid body, with time t is shown in Fig. 7.7. Is the body rotating clock-wise or anti-clockwise?

Answer:

Here slope of
$θ vs t curve is positive, So, \frac{d θ}{d t} = + ve \Rightarrow angular velocity, ω = + ve$
By sign convention, if a rigid body is rotating anticlockwise direction, its angular velocity is assumed as positive.
So, the body is rotating anticlockwise by convention.

Page No 54:

Question 7.17:

A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in Fig. 7.8. Match the following (most appropriate choice):

(a) mg/4 < F < mg /2	(i) Cube will move up.
(b) F > mg/2	(ii) Cube will not exhibit motion.
(c) F > mg	(iii) Cube will begin to rotate and slip at A.
(d) F = mg/4	(iv) Normal reaction effectively at a/3 from A, no motion.

Answer:

The correct match is:
(a) - (ii)
(b) - (iii)
(c) - (i)
(d) - (iv)

The cube will move up only if :
F − mg > 0
F > mg
Length of each side of the cube is a.
Torque of mg about point A,
$T_{1} = m g \times \frac{a}{2} anticlockwise Totque of force F about point A, T_{2} = F \times a clockwise Totque of normal reaction about point A is zero as it is just in touch with point A . So, the cube will rotate if, T_{2} > T_{1}, F \times a > m g \times \frac{a}{2} \Rightarrow F > \frac{m g}{2}$

Now if the the effective normal reaction (N = mg − F) is at point a/3 from point A, then torque of normal reaction about A is,
$T_{3} = N \times \frac{a}{3} = (m g - F) \times \frac{a}{3} anticlockwise So, the cube will not be in rotational motion if, T_{1} - T_{2} + T_{3} = 0 F \times a - m g \times \frac{a}{2} + (m g - F) \times \frac{a}{3} = 0 \Rightarrow \frac{2 F}{3} = \frac{m g}{6} \Rightarrow F = \frac{m g}{4}$
Thus, the cube will not exhibit motion if,
   $\frac{m g}{4} < F < \frac{m g}{2}$

Page No 54:

Question 7.18:

A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Fig. 7.9). The sphere is struck horizontally at a height h from the floor. Match the following:

(a) h = R/2	(i) Sphere rolls without slipping with a constant velocity and no loss of energy.
(b) h = R	(ii) Sphere spins clockwise, loses energy by friction.
(c) h = 3R/2	(iii) Sphere spins anti-clockwise, loses energy by friction.
(d) h = 7R/5	(iv) Sphere has only a translational motion, looses energy by friction.

Answer:

The correct match is:
(a) - (iii)
(b) - (iv)
(c) - (ii)
(d) - (i)

Given mass of the sphere = m, radius = R, height of the floor = h
The sphere will roll without slipping when, $v = ω R$
â€‹ Where, v = linear velocity of the sphere and $ω$ = angular velocity of the sphere.

Angular momentum of sphere, about centre of mass,
$m v (h - R) = I ω \Rightarrow m v (h - R) = \frac{2}{5} m R^{2} ω = \frac{2}{5} m v R (As, I = \frac{2}{5} m R^{2}) \Rightarrow h = \frac{7}{5} R$
Hence, the sphere will roll without slipping with a constant velocity and there is no loss of energy when $h = \frac{7}{5} R$
Now, torque due to applied force (F)about centre of mass, τ =F(h−R)
If τ = 0 then h = R
So, the sphere will have only translational motion and It would lose energy by friction for h = R
Now if h > R, torque about centre of mass must be positive and sphere will spin clockwise.
And if h < R, torque about centre of mass must be negative and sphere will spin anticlockwise.

The correct match is:
(a) - (iii)
(b) - (iv)
(c) - (ii)
(d) - (i),Ï„â†’>

Page No 54:

Question 7.19:

The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

Answer:

No, not necessarily.
Given that sum of forces acting on a rigid body is,
$\sum_{} {\vec{F}}_{i} \neq 0 Also, vector sum of torque due to force s about a point X, \sum_{} {\vec{r}}_{i} \times {\vec{F}}_{i} = 0 Now, vector sum of torque due to forces about a point X', which is some distance apart fom the point X . \sum_{} ({\vec{r}}_{i} - \vec{a}) \times {\vec{F}}_{i} = \sum_{} {\vec{r}}_{i} \times {\vec{F}}_{i} - \sum \vec{a} \times {\vec{F}}_{i} = 0 - \sum \vec{a} \times \vec{F} \sum \vec{a} \times \vec{F} may or may not be zero,$
This means that torque about the new arbitrary point may not necessarily be zero.

Page No 54:

Question 7.20:

A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

Answer:

A wheel is symmetric about its center i.e. distribution of mass particle is symmetric about center of mass. Internal elastic forces give rise to the centripetal acceleration of its particles in a wheel. These forces acts in pairs and cancel each other due to symmetry. Hence, wheel is in equilibrium and no external force of torque is required to maintain the motion.
When the wheel is half, the mass distribution is not symmetric about center of mass. Therefore, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to maintain the motion of the half wheel.

Page No 54:

Question 7.21:

A door is hinged at one end and is free to rotate about a vertical axis (Fig. 7.10). Does its weight cause any torque about this axis?
Give reason for your answer.

Answer:

No.
As door will rotate about y-axis while closing or opening it. Weight acts in downward direction, so torque due to the weight do not contribute in closing or opening the door.
(Torque $= \vec{r} \times \vec{F}$ )

Page No 55:

Question 7.22:

(n – 1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.

Answer:

Let position vector of center of mass of (n – 1) equal point masses be $\vec{x}$ .
Also, the position vector of n^thcorner or the polygon is $\vec{a}$ . Let if an object of mass m be placed at the n^th corner, then position vector of the system be at origin.
So, position vector of center of mass of the system be,
$\frac{(n - 1) m \vec{x} + m \vec{a}}{(n - 1) m + m} = 0 \Rightarrow \vec{x} = - \frac{1}{n - 1} \vec{a}$

Page No 55:

Question 7.23:

Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

Answer:

(a). Centre of mass of a uniform half-disc:
Let mass of the half disc be M and radius R.
Its, density be,
$= \frac{m a s s}{a r e a} = \frac{M}{(\frac{π R^{2}}{2})} = \frac{2 M}{π R^{2}} Let, choose a small mass element whose COM be (0, \frac{2 r}{π}) d m = ρ π r d r So, X_{c m} = 0 (by symmetry) and Y_{c m} = \frac{\int_{0}^{R} (\frac{2 r}{π}) ρ π r d r}{M} = \frac{2 ρ (R^{3} - 0^{3})}{3 M} = \frac{2 (\frac{2 M}{π R^{2}}) R^{3}}{3 M} = \frac{4 R}{3 π} Centre of mass of the half disc, (0, \frac{4 R}{3 π}) .$
(b) In similar manner we can get the COM of the quarter disc be:
$(\frac{4 R}{3 π}, \frac{4 R}{3 π}) .$

Page No 55:

Question 7.24:

Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω₁ and ω₂are brought into contact face to face with their axes of rotation coincident.
(a) Does the law of conservation of angular momentum apply to the situation? why?
(b) Find the angular speed of the two-disc system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.

Answer:

(a) Yes, law of conservation of angular momentum is applicable to this situation because no net external torque is acting on the system during the process.
(b) According to law of conservation of angular momentum:
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω \Rightarrow ω = \frac{I_{1} ω_{1} + I_{2} ω_{2}}{(I_{1} + I_{2})}$
(c) Initial and final kinetic energy,
$K E_{i} = \frac{1}{2} I_{1} {ω_{1}}^{2} + \frac{1}{2} I_{2} {ω_{2}}^{2} K E_{f} = \frac{1}{2} I ω^{2} = \frac{1}{2} (I_{1} + I_{2}) {(\frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}})}^{2} = \frac{1}{2} {\frac{(I_{1} ω_{1} + I_{2} ω_{2})}{(I_{1} + I_{2})}}^{2}$
So, loss in kinetic energy,
$= K E_{i} - K E_{f} = \frac{I_{1} I_{2}}{2 (I_{1} - I_{2})} {(ω_{1} - ω_{2})}^{2}$
(d) This loss in kinetic energy of the system is due to the work against the friction between the two discs.

Page No 55:

Question 7.25:

A disc of radius R is rotating with an angular speed ω₀ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is μ_k.
(a) What was the velocity of its centre of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c).
(e) What condition should be satisfied for rolling to begin?
(f ) Calculate the time taken for the rolling to begin.

Answer:

(a) Zero, as, before being brought in contact with the table, the disc was in pure rotational motion (only) about its axis passing through the center. So, V_cm= 0.
(b) Linear velocity of a point on the rim decreases when the rotating disc is placed in contact with the table’s surface due to friction.
(c) The linear velocity of center of mass of the disc will increases.
(d) Frictional force is responsible for the increase in linear velocity of COM of disc and decrease in linear velocity of a point on the rim decreases when the rotating disc is placed in contact with the table’s surface.
(e) The disc will roll only when, the velocity of center of mass of the disc,
$V_{c m} = ω r (where, ω = angular velocity and r = radius of the disc)$
(f) Friction force acting on the disc,
$F = μ_{k} m g, So, acceleration of COM, a_{c m} = \frac{μ_{k} m g}{m} = μ_{k} g \Rightarrow angular acceleration, α = \frac{F R}{I} = \frac{μ_{k} m g R}{I} Velocity of COM, v_{cm} = u_{c m} + a_{c m} t = 0 + μ_{k} g t a l s o, ω = ω_{o} + α t \Rightarrow t = \frac{ω - ω_{o}}{α} If disc is rolling without slipping, then : \Rightarrow t = \frac{R ω_{o}}{μ_{k} g (1 + \frac{m R^{2}}{I})}$

Page No 55:

Question 7.26:

Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R + δ) . They are now brought in contact (δ → 0).
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?

Answer:

(a) Frictional force on smaller drum of radius (R) is in upwards direction ( which is rotating anticlockwise), while frictional force on larger drum of radius (2R) is in downward direction ( which is rotating clockwise) as shown in figure.

Velocity of larger drum at the point of contact, $v' = ω \times 2 R$
Velocity of smaller drum at the point of contact, $v = ω \times R$

(b) F = F′ = F′′ where F' and F′′ are external forces through support.
So, net external force on the system must be zero.
But the net external torque on the system $= 3 R \times F$ in anticlockwise direction.

(c) Let ω₁ (anticlockwise) and ω₂ (clockwise) be the final angular velocities of the smaller and the larger drum.
When, there will be no friction, then velocity at the point of contact must be same:
Rω₁ = 2 R ω₂
$\Rightarrow \frac{ω_{1}}{ω_{2}} = \frac{2}{1}$

Page No 56:

Question 7.27:

A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses (Fig. 7.11).

Show that $(i) \frac{I_{x R}}{I_{x S}} < 1; (ii) \frac{I_{y R}}{I_{y S}} > 1; (iii) \frac{I_{z R}}{I_{z S}} > 1$ .

Answer:

M = mass of the rectangular plate = mass of the square plate
Area of plate S = Area of plate R
And $c^{2} = a b$
Moment of inertia of plate R about x-axis and y -axis is
$I_{x R} = \frac{M b^{2}}{12}$ .
$I_{y R} = \frac{M a^{2}}{12}$ .
Similarly, moment of inertia of plate S about x-axis and y -axis is,
$I_{x S} = \frac{M c^{2}}{12}$
$I_{y S} = \frac{M c^{2}}{12}$
Now,
$(i) \frac{I_{x R}}{I_{x S}} = \frac{\frac{M b^{2}}{12}}{\frac{M c^{2}}{12}} = \frac{b^{2}}{c^{2}} < 1 (a s, c^{2} = a b a n d a > b) (ii) \frac{I_{y R}}{I_{y S}} = \frac{\frac{M a^{2}}{12}}{\frac{M c^{2}}{12}} = \frac{a^{2}}{c^{2}} > 1 (iii) \frac{I_{z R}}{I_{z S}} = \frac{I_{x R} + I_{y R}}{I_{x S} + I_{y S}} = \frac{a^{2} + b^{2}}{2 c^{2}} > 1$

Page No 56:

Question 7.28:

A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is μ (Fig 7.12). Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Answer:

Let f be the friction force acting on the disc, and a be the acceleration.
$F - f = m a . . . (1) As disc is rolling without slipping, Torque = 0 + f r = I α = (\frac{1}{2} m r^{2}) \frac{a}{r} \Rightarrow m a = 2 f . . . (2) S olving equation (1) and (2), we get : F = 3 f \Rightarrow f = \frac{F}{3} As, the disc rolls without slipping, then friction force at the point of contact : f \leq μ m g \Rightarrow \frac{F}{3} \leq μ m g \Rightarrow F \leq 3 μ m g$

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