KKMishra's Tutorials: Solutions to Problems on "ROTATIONAL MECHANICS" - H C Verma's Concepts of Physics, Part-I, Chapter-10, EXERCISES Q1 to Q15

My Channel on YouTube → SimplePhysics with KK

For links to

other chapters - See bottom of the page

ROTATIONAL MECHANICS:--

EXERCISES-(Q1 to Q15)

1. A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches 100 revs/second in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds.

ANSWER: α = angular acceleration, ω = Initial angular velocity = 0, ω' = Final angular velocity =100 revs/s and t = time = 4 s.

We know that ω' = ω + αt

→100 = 0 + α*4

→α =100/4 = 25 rev/s².

The angle rotated is analogous to the distance in the three equations of kinematics, hence

θ = ωt+½αt²

=0 + ½*25*4²

=200 revolutions

=200*2π

=400π radians.

2. A wheel rotating with uniform angular acceleration covers 50 revolutions in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds.

ANSWER: θ = ωt + ½αt²

50 = 0 + ½α*5²

→25α = 100

→α = 4 revs/s²

Let the angular velocity after 5 seconds = ω', we have the equation

ω' = ω + αt

= 0 + 4*5

=20 revs/s.

3. A wheel starting from rest is uniformly accelerated at 4 rads/s² for 10 seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in the next 10 seconds. Find the total angle rotated by the wheel.

ANSWER: We have three parts of rotations each having duration 10 seconds. First uniform acceleration then uniform rotation without acceleration and finally uniform deceleration. Since the final angular velocity of the first part is same as the initial velocity of the third part, the angle rotated in the first and third part will be the same.

So total angle rotated = θ +θ' +θ"

=θ+θ'+θ {Since θ=θ}

=2θ+θ' ----------------------- (i)

Now θ =ωt+½αt²

=0*10+½*4*10*10

=200 radians

And θ' = ω'*t ={ω+αt}t

={0+4*10}10

=400 radians

Finally, total angle rotated from (i)

=2θ+θ'

=2*200+400

=800 rad

4. A body rotates about a fixed axis with an angular acceleration of one radian/second/second. Through what angle does it rotate during the time in which its angular velocity increases from 5 rad/s to 15 rad/s.

ANSWER: We have α = 1 rad/s²
ω = 5 rad/s, ω' = 15 rad/s, θ = ?
We have ω'² = ω² + 2αθ
→15² = 5²+2*1*θ
→2θ =225-25 =200
→θ = 100 rad

5. Find the angular velocity of a body rotating with an acceleration of 2 revs/s² as it completes the 5th revolution after the start.

ANSWER: We have θ = 5 revolutions
ω = 0, α = 2 revs/s², ω' = ?
Since ω'² = ω² + 2αθ
→ω'² = 0² + 2*2*5
→ω'² = 2²*5
→ω' = 2√5 revs/s

6. A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rads/s. Find the linear speed of

(a) a point on the rim,
(b) the middle point of a radius.

ANSWER: We have ω = 20 rad/s, r = 10 cm = 0.10 m
(a) The linear speed of a point on the rim = ωr
= 20 rad/s * 0.10 m
= 2 m/s
(b) For middle point on the radius r' = 0.10/2 m = 0.05 m
The linear speed of the middle point of a radius = ωr'
= 20 rad/s * 0.05 m
= 1 m/s

7. A disc rotates about its axis with a constant angular acceleration of 4 rad/s². Find the radial and tangential accelerations of a particle at a distance of 1 cm from the axis at the end of the first second after the disc starts rotating.

ANSWER: We have α = 4 rad/s², r = 1 cm = 0.01 m
Tangential acceleration = a = rα = 0.01 m*4 rad/s² =0.04 m/s²
= 4 cm/s²
For the radial acceleration we need to know ω at the end of first second. ω=0+αt =4*1 =4 rad/s
Radial acceleation = ω²r = 4²*0.01 =0.16 m/s² = 16 cm/s²

8. A block hangs from a string wrapped on a disc of radius 20 cm free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is 10 rad/s at some instant, with what speed is the block going down at that instant?

ANSWER: We have r = 20 cm = 0.20 m
ω = 10 rad/s, v =?
v = ωr = 10*0.20 =2 m/s
v is the linear speed of the rim of the disc and hence of the string. So the block is going down with a speed of 2 m/s.

9. Three particles each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertia of the system about an axis

(a) joining two of the particles and
(b) passing through one of the particles and perpendicular to the plane of the particles.

ANSWER: (a) Let us draw a diagram as below,

Diagram for Q-9

Mass of each of the particle m = 200 g = 0.20 kg.
The side of the equilateral triangle r = 10 cm = 0.10 m
The axis OO' is joining two of the particles.
The distance of the two particles on the axis from the axis = 0
The distance of the third particle from the axis = r.cos30°
= 0.10*√3/2 = 0.10*1.732/2 =0.10*0.866 = 0.087 m
So, the moment of inertia about the axis OO'
= mass*distance²
= 0.20*(0.087)² = 0.20*0.0075 = 0.0015 =1.5x10-³ kg-m²

(b) Let us draw a diagram in this case as below,

Diagram for Q-9, Case (b)

The distance of the particle on the axis from the axis OO' =0
The distance of other two particles from the axis OO' = r = 0.10 m
The moment of inertia of the system = mr²+mr²+m*0²
=2mr²
=2*0.20*0.10²
=0.40*0.01
=0.004 kg-m²
=4.0x10-³ kg-m²

10. Particles of masses 1 g, 2 g, 3 g, ..... 100 g are kept at the marks 1 cm, 2 cm, 3 cm, ...... 100 cm respectively on a meter scale. Find the moment of Inertia of the system of particles about a perpendicular bisector of a meter scale.

ANSWER: Clearly the axis passes through the 50 cm mark, hence the weight of 50 g is at zero distance from the axis. Towards the decreasing side, there are 49 weights while on the increasing side there are 50 weights. See the diagram below:-

Diagram for Q-10

The 1 g mass is at 49 cm away from the axis and the 49 g mass is at 1 cm away. On the other side, 51 g mass is at 1 cm away from the axis while the 100 g mass is at 50 cm.
Moment of inertia of the system about the given axis
={1*49²+2*48²+3*47²+...+49*1²+51*1²+52*2²+ ....+98*48²+99*49²+100*50²}/10000000
[Division by 10000000 is to convert g into kg and cm to m]
={100*50²+(1+99)*49²+(2+98)*48² + .....+(49+51)*1²}/10000000
=100{50²+49²+48²+ ...+2²+1²}/10000000
={1²+2²+3²+ .....+48²+49²+50²}/100000
={50³/3+50²/2+50/6}/100000
[Since sum Sn = n³/3+n²/2+n/6]
={125000*2+2500*3+50}/600000
={250000+7500+50}/600000
=257550/600000
=0.43 kg-m²

11. Find the moment of inertia of a pair of spheres, each having a mass m and radius r, kept in contact about the tangent passing through the point of contact.

ANSWER: The Moment of inertia of one sphere about the axis passing through its center I = (2/5)mr²
The Moment of inertia of one sphere about an axis parallel to this axis at a distance r = I + mr² = (2/5)mr²+mr² = (7/5)mr²

Diagram for Q-11

In the given problem each sphere's moment of inertia is required about an axis parallel to the axis through its center. So, the required M.I. of the system = 2*(7/5)mr² = (14/5)mr²

12. The moment of inertia of a uniform rod of mass 0.50 kg and length 1 m is 0.10 kg-m² about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod.

ANSWER: m =0.50 kg, l =1 m
Moment of inertia about its perpendicular bisector (Through CoM and middle point)
I = ml²/12
=0.50*1²/12
=1/24 kg-m²
Let the moment of inertia given in the problem be at a distance r from the middle point of the rod, I' = 0.10 kg-m² (Given)
So, I' = I + mr²
→0.10 = 1/24 +0.50r²
→1+0.50*24r² =0.10*24
→12r² =2.4-1=1.4
→r² = 1.4/12 = 0.467/4
→r= 0.68/2
→0.34 m

13. Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles.

ANSWER: The Moment of inertia of the ring about an axis perpendicular to its plane and passing through its center = I =mr² (Where m is its assumed mass).
A line perpendicular to the plane of the ring and passing through one of its particles will be parallel to the previous axis and at a distance = r from that. From the parallel axis theorem, the moment of inertia of the ring about this new line I' = I+mr² =mr²+mr² =2mr²
If the radius of gyration of the ring about this new line is a, then I'=ma². Equating the two expressions of I' we get,
ma² = 2mr²
→a² = 2r²
→a=√2r.

14. The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the center.

ANSWER: The Moment of inertia of the disc about a line perpendicular to the disc and passing through its center I = mr²/2. Let a line parallel to this axis and at a distance d about which the radius of gyration is = r. M.I. about this axis =I+md² =mr²/2+md²
But this M.I. = mr² (radius of gyration =r)
Hence mr²/2+md² =mr²
→d² = r²/2
→d = r/√2.

15. Find the moment of inertia of a uniform square plate of mass m and edge a, about one of its diagonals.

ANSWER: Let the moment of inertia of the uniform square plate about one of its diagonal be I, then M.I about other diagonal will also be I. Since it is a square plate both these diagonal axes are perpendicular. Let I' be the M.I. of the square plate about the axis perpendicular to the plate and passing through the center. Now, these three axes are mutually perpendicular and passing through the CoM. Hence from the perpendicular axis theorem of M.I.,
I+I =I'
→2I = ma²/6
→I = ma²/12

Alternative Method
It can also be derived from the basic principle. Let us draw a diagram as below,

Diagram for Q-15

Mass per unit area of the square plate = m/a²
Consider a strip (FG) of thickness dx at a distance x from the diagonal (axis). DE = a/√2, DH = DE - EH = a/√2 - x.
FG = 2*GH = 2*DH = 2(a/√2-x) =√2a-2x
Area of the strip FG = (√2a-2x).dx
Mass of strip FG = (m/a²).(√2a-2x).dx
M.I. of strip about OO' = (m/a²).(√2a-2x).dx*x²
M.I. of the square plate about the axis OO'
=2∫(m/a²)x²(√2a-2x).dx
{Limits of integration from x=0 to x=a/√2}
=(2m/a²)∫(√2ax²-2x³)dx
=(2m/a²)[√2ax³/3-2x⁴/4]
Apply the limits, and we get
=(2m/a²)[√2a.a³/(3*2√2)-½*a⁴/4]
=(2m/a²)[a⁴/6-a⁴/8]
=m[a²/3-a²/4]
=ma²/12

===<<<O>>>===

Links to the chapters -