Solutions to Exercises on Rotational Mechanics HC Verma's Concepts of Physics Part 1 (Q23-Q29)

 Q#23

A square plate of mass 120 g and edge 5.0 cm rotates about one of the edges. If it has a uniform angular acceleration of 0.20 rad/s², what torque acts on the plate?        

Answer:
Let the torque on the plate = T
Given, α = 0.20 rad/s²
Mass of the plate, M = 120 g = 0.12 kg
Edge of the plate, a = 5.0 cm = 0.05 m

M.I. of the plate about a line parallel to one of the edges and passing through the center I = Ma²/12
M.I. of the plate about the edge, I' = I + M(a/2)²

= Ma²/12 + Ma²/4
= Ma²{(1 + 3)/12}
= Ma²/3
= 0.12(0.05)(0.05/3) = 0.0001 kgm²

The torque T on the plate = I'α

= 0.0001(0.20 Nm)
= 2.0 x 10⁻⁵ Nm

Q#24
Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with the same angular acceleration.  

Answer:
As we have calculated the M.I. of the square plate about a diagonal in Problem number-15 = Ma²/12
The torque on the plate, T = Iα

= (Ma²/12)(0.20 Nm)
= 0.12(0.05)(0.05)(0.20/12 Nm)
= 0.5 x 10⁻⁵ Nm

Q#25

A flywheel of moment of inertia 5.0 kg-m²is rotated at a speed of 60 rad/s. Because of the friction at the axle, it comes to rest in 5.0 minutes. Find (a) the average torque of the friction. (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating.      

Answer:
(a) Let us calculate the angular retardation α of the wheel. Final angular speed ⍵' = 0, Initial angular speed ⍵ = 60 rad/s and time taken t = 5 min = 300 s.

⍵' = ⍵ - αt
0 = 60 – 300α
α = 0.20 rad/s²
M.I. = 5.0 kgm²

Hence the average torque of the friction = Iα
= 5.0(0.20) = 1.0 Nm

(b) Let the angle rotated
θ = ⍵t –  ½αt²
= 60(300) – 0.50(0.20)(300)300

= 9000 rad

Work done = Tθ

= 1.0(9000 J)
= 9.0 kJ

(c) Let the angular speed at t = 240 s (4 min) be ω''.
ω'' = ω – αt = 60 – 0.20(240)
= 60 – 48 =12 rad/s

Angular momentum of the wheel 1 min before it stops
= Iω''
= 5.0(12 kgm²/s)

= 60 kgm²/s

Q#26
Because of the friction between the water in oceans with the earth's surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by 0.0016 rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is 6.0 x 10²⁴ kg.      

Answer:
Time t = 100 years = 100(365) day = 36500 day.
Initial angular speed = ω = 2π rad/day
Final angular speed ω' = (2π – 0.0016) rad/day
Angular deceleration, α = ?

ω' = ω – αt
(2π – 0.0016) = 2π – α(36500)
α = 0.0016/36500 rad/day²
α = 0.0016/36500(24 x  3600)² rad/s²
= 5.87x10⁻¹⁸ rad/s²

M = 6.0 x 10²⁴ kg, R = 6400 x 1000 m
M.I. of the earth = (2/5)MR²

= 0.40(6.0 x 10²⁴)(6400000)² kg-m²
= 9.83 x 10³⁷ kgm²

Average torque of the friction force = Iα
= (9.83x10³⁷)(5.87 x 10⁻¹⁸)
= 5.77 x 10²⁰ Nm    
≈ 5.8 x 10²⁰ Nm

Q#27
A wheel rotating at a speed of 600rpm (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque.       

Answer:
Initial angular velocity = ω = 600 rpm = 10 rev/s
Final angular velocity =ω' = 0, Time = t = 10 s,
Angular deceleration α = ?

ω' = ω – αt
0 = 10 – α(10)
α = 1.0 rev/s²

 Let the angular velocity after t = 5 s be ω''
ω'' = ω – αt = 10 – 1.0(5) = 5.0 rev/s      

Q#28
A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 revs/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.      

Answer:
Initial angular speed = ω = 100 rev/min = 10/6 rev/s
Final angular speed = ω' = 0, Total revolutions covered Θ = 10 rev, Angular deceleration α = ?

We have, ω'² = ω² - 2αΘ
0 = 100/36 – 2(α)10
20α = 100/36
α = 5/36 rev/s² = 2π(5/36) rad/s² = 10π/36 rad/s²

M.I. of the wheel, I = mr²/2 = ½(10)(0.20)²
= 0.20 kgm²

If the force required to stop the wheel = F,
Torque of F = Fr = 0.20F

But the torque should be equal to Iα = 0.20(10π/36)
= 2π/36
Hence, 0.20F = 2π/36
F = 2π/7.2 = 0.87 N      

Q#29
A cylinder rotating at an angular speed of 50 revs/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed?        

Answer:
Let the time taken is t seconds. Initial angular speed ω = 50 rev/s, Final angular speed = ω' and the angular acceleration = α = 1 rev/s².
For the first cylinder,

 ω' = ω - αt = 50 - (1)t = 50-t          (i)
For the second cylinder, initial angular speed = 0,
Final angular speed = ω', α = 1 rev/s²

ω' = 0 + 1(t) = t
50 – t = t             {Putting the value of ω' from (i)}
2t = 50
t = 25 s    

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