KKMishra's Tutorials: Solutions to Problems on "ROTATIONAL MECHANICS" - H C Verma's Concepts of Physics, Part-I, Chapter-10, EXERCISES Q16 to Q30

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ROTATIONAL MECHANICS:--

EXERCISES-(Q16 to Q30)

16. The surface density (mass/area) of a circular disc of radius a depends on the distance from the center as ρ(r) = A+Br. Find its moment of inertia about the line perpendicular to the plane of the disc through its center.

ANSWER: Consider a ring of radius r from the center with a width dr. Area of the ring = 2πr.dr
Mass of the ring = (A+Br)*2πr.dr
M.I. of this ring about the axis perpendicular to the plane of the disc through its center =(A+Br)*2πr.dr* r² = (2πr³A+2πr⁴B).dr
M.I of the disc = ∫(2πr³A+2πr⁴B).dr
{Limit of integration from r =0 to r = a}
= [2πr⁴A/4+2πr⁵B/5], putting the limits we get,
=2π{Aa⁴/4+Ba⁵/5}

17. A particle of mass m is projected with a speed u at an angle θ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

ANSWER: Let t be time taken to reach the highest point. At the highest point velocity in the vertical direction is =0. Hence,
0 = u.sinθ -gt
→ t = u.sinθ/g
Horizontal distance traveled in this time = d = u.cosθu.sinθ/g
= u²sinθ.cosθ/g
The weight of the particle W= mg
Position vector of the particle at the highest point taking the point of projection as origin = r

Figure for Q - 17

The torque of the weight about the point of projection = r x W
= Weight x perpendicular distance of the line of weight from the origin.
= mg*d
= mg*u²sinθ.cosθ/g
= mu²sinθ.cosθ
The direction of the torque will be perpendicular to the plane of motion.

18. A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is torque zero?

ANSWER: Taking the point of suspension as origin, position vector of the bob = r
The weight of the bob = w
The torque of the weight of the bob about the point of suspension
= r x w
= wl.sinθ {Direction of the torque will be perpendicular to the plane of motion}

In the above expression, w and l are constant, θ is variable. The torque will be zero when the value of sinθ is zero. sinθ will be zero for θ = 0. Which will be when the bob is at the lowest point.

19. When a force of 6.0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?

Figure for Q-19

ANSWER: Let us resolve the force of 6.0 N along the wrench and perpendicular to it.

Figure for Q-19

The component of the force along the wrench, P = 6.0*cos30°
Since it passes through the nut, its torque about the nut = 0
The component of the force perpendicular to the wrench, P' =6.0*sin30° =3.0 N
The torque of this force about the nut = 3.0*8/100 =0.24 N-m
This much torque is required to just move the nut.
The torque of the force F about the nut = F*16/100 N-m
Equating the two we get,
F*16/100 =0.24
→F = 24/16 =3/2 =1.5 N

20. Calculate the total torque acting on the body shown in figure (10-E2) about the point O.

Figure for QQ-20

ANSWER: We take the torque positive which produces an anticlockwise effect.
Total torque = 20*0.04*sin30° + 5*0 + 15*0.06*sin37° - 10*0.04*sin90° N-m
= 20*0.02+15*0.036-0.40 N-m
= 0.40+0.54-0.40 N-m
= 0.54 N-m, Anticlockwise

21. A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center.

ANSWER: Three forces act on the block. Weight mg, Friction F and Normal force N. The weight acts at the center of the block. Its components along the plane are mg.sinθ and perpendicular to the plane mg.cosθ. Since the block slides with uniform speed, the friction F = mg.sinθ. But these two equal forces are not in line. While mg.sinθ acts at the center C, the friction F acts on the surface. These two antiparallel forces are a/2 distance apart. So, the torque produced by them = mg.sinθ*a/2 =½mga.sinθ. (Clockwise in the figure below).

Figure for Q - 21

Since the block does not rotate, it means there is an equal and opposite torque on the block. Consider the other force, the component of weight perpendicular to the plane mg.cosθ which acts at the center and the normal force N which does not pass through the center C but at some point down the plane. (In previous chapters we considered N to act through the center because we did not consider the torque then). Since in the perpendicular direction there is no movement, hence N = mg.cosθ and they act some distance apart. The torque produced by the normal force N about the center = Torque produced by the friction F about the center =½mga.sinθ.

22. A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the center. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.

ANSWER: The torque on the rod, T =F*L/4
M.I. of the rod, I= mL²/12
Angular acceleration, α = T/I = FL*12/4mL²
= 3FL/mL²
The angle rotated by the rod during time t = θ = 0*t+½αt²
= ½*(3FL/mL²)*t²
= 3FLt²/2mL²
= 3Ft²/2mL

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
=0.04*0.0025 = 0.0001 kg-m²
The torque T on the plate = I'α
=0.0001*0.20 N-m
=2.0x10⁻⁵ N-m

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 N-m
= 0.12*0.05*0.05*0.20/12 N-m
= 0.01*0.05*0.01 N-m
= 0.000005 N-m
= 0.5x10⁻⁵ N-m

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=60/300 =0.20 rad/s²
M.I. = 5.0 kg-m²
Hence the average torque of the friction = Iα
=5.0*0.20 =1.0 N-m

(b) Let the angle rotated θ = ⍵t-½αt²
=60*300-0.50*0.20*300*300
=18000-0.10*90000
=18000-9000
=9000 rad
Work done = Tθ
=1.0*9000 J
=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 kg-m²/s
= 60 kg-m²/s

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 6.0x10²⁴ 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*3600)² rad/s²
= 5.87x10⁻¹⁸ rad/s²
M = 6.0x10²⁴ kg, R = 6400x1000 m
M.I. of the earth = (2/5)MR²
= 0.40*6.0x10²⁴*6400000*6400000 kg-m²
= 9.83x10³⁷
Average torque of the friction force = Iα
= 9.83x10³⁷*5.87x10⁻¹⁸
= 5.77x10²⁰ N-m
≈ 5.8x10²⁰ N-m

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

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*0.20
= 0.20 kg-m²
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

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

30. A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s². At what time will the body have kinetic energy same as the initial value if the torque continues to act?

ANSWER: Angular speed ω = 20 rad/s, Deceleration α = 2 rad/s². Since the torque decelerates rotating body, the body will come to rest but the torque continues, hence the body will again accelerate in opposite direction. When after acceleration it gets an angular speed of -ω, the kinetic energy of the body will again be equal to the initial value because of KE = ½Iω² and it depends on the numerical value of ω.
Now we have Final angular speed, ω' = -ω = -20 rad/s
Initial angular speed = ω = 20 rad/s
Deceleration α = 2 rad/s²
If the time taken be t, from the formula ω'=ω-αt
-ω = ω -αt
→αt = 2ω
→t = 2ω/α = 2*20/2 = 20 s

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