Solutions to Exercises on Rotational Mechanics HC Verma's Concepts of Physics Part 1 (Q1-Q8)

 Q#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:
Given: 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)(4)²
= 200 revolutions
= 200(2π)
= 400π radians.

Q#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(5)²
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.

Q#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)²
= 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

Q#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

Q#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√5 revs/s  

Q#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        

Q#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  

Q#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.  

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