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ROTATIONAL MECHANICS:--
EXERCISES-(Q46 to Q60)
46. A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end, Calculate (a) the angular momentum of the rod about the axis of rotation. (b) the speed of the center of the rod and (c) it's kinetic energy.
ANSWER: (a) Mass M = 300 g = 0.300 kg, Length L = 50 cm =0.50 m
The moment of inertia about the center I = ML²/12
The moment of inertia about an end I' = I + Mr²
{Where r = L/2 = 0.25 m}
= ML²/12 + Mr²
= 0.30*0.5*0.5/12 + 0.30*0.25*0.25
= 0.025 kg-m²
Angular speed = ⍵ = 2 rad/s
Hence the angular momentum of the rod about the axis of rotation = I'⍵
= 0.025*2 = 0.05 kg-m²/s
(b) The speed of the center of the rod = v = ⍵r = 2*0.25 =0.50 m/s = 50 cm/s
(c) The kinetic energy of the rod = ½I'⍵² = 0.50*0.025*2² J
= 0.50*4*0.025 J
= 2.0*0.025 J
= 0.050 J
47. A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.
ANSWER: Mass M = 2.0 kg, Edge a = 10 cm = 0.10 m.
The moment of inertia about the diagonal = Ma²/12
= 2.0*(0.10²)/12
= 0.01/6 =1/600 kg-m²
Torque = 0.10 N-m
Angular acceleration α = Torque/M.I.
= 0.10/(1/600) = 60 rad/s²
Angular speed at t = 5 s, ⍵ = 0 + 60*5 =300 rad/s
Hence the angular momentum at this instant = I⍵ = (1/600)*300
=1/2 = 0.50 kg-m²/s
Kinetic Energy = E = ½I⍵²
= ½*(1/600)*300²
=90000/1200 =900/12 =75 J
48. Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1.5x10⁸ km.
ANSWER: Let the mass of the earth = M.
Its radius r = 6400 km = 64x10⁵ m, Radius of the orbit R = 1.5x10⁸ km =1.5x10¹¹ m.
The angular speed of the spin of the earth = ⍵
= 2π/24*3600 rad/s
The angular speed of the revolution of the earth ⍵' = 2π/365*24*3600 rad/s
If the moment of inertia of the earth about its axis = I =(2/5)Mr²
Then angular momentum = Γ =I⍵
The moment of inertia of the earth about the sun I' = I+MR²
= (2/5)Mr² +MR² = M(2r²/5 + R²)
The angular momentum about the sun = I'⍵'
Hence the ratio of the angular momentum
= I⍵/I'⍵'
= (2/5)Mr²⍵/M(2r²/5+R²)⍵'
= ⍵/(1+5R²/2r²)⍵'
= (2π/24*3600)/(1+5*2.25x10²²/2x64x64x10¹⁰)*(2π/365*24*3600)
= 365/0.00137x10¹²
= 265784.9x10⁻¹²
= 2.66x10⁵x10⁻¹²
= 2.66x10⁻⁷
49. Two particles of mass m₁ and m₂ are joined by a light rigid rod of length r. The system rotates at an angular speed ⍵ about an axis through the center of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is L = µr²⍵ where µ is the reduced mass of the system defined as µ = m₁m₂/(m₁+m₂).
ANSWER: Let the center of mass of the system from the mass m₁ be x, then x = m₂r/(m₁+m₂)
The moment of inertia of the system about the CoM
I = m₁{m₂r/(m₁+m₂)}² + m₂{m₁r/(m₁+m₂)}²
= [m₁m₂²+m₁²m₂]*r²/(m₁+m₂)²
= [m₁m₂(m₁+m₂)]*r²/(m₁+m₂)²
= m₁m₂r²/(m₁+m₂)
= [m₁m₂/(m₁+m₂)]r²
= µr²
Since the angular speed = ⍵
Hence the angular momentum Γ = I⍵ =µr²⍵.
Hence proved.
50. A dumbbell consists of two identical small balls of mass ½ kg each connected to the two ends of a 50 cm long light rod. The dumbbell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5.0 N acts on one of the masses in the direction of its velocity for 0.10 s. Find the new angular velocity of the system.
ANSWER: The moment of inertia of the system I = 2mr²
= 2*½*0.25*0.25
= 0.0625 kg-m²
⍵ = 10 rad/s
Time of impulse = t = 0.10 s
Torque of the impulsive force = T = 5*0.25 =1.25 N-m
Impulsive torque =T*t
=1.25*0.10
=0.125 Nm-s
change in angular momentum = Impulsive torque
If ⍵' is the new angular velocity of the system, then
I⍵'-I⍵ = Tt
⍵' = ⍵+Tt/I
= 10 + 0.125/0.0625
= 10 + 2
= 12 rad/s
51. A wheel of moment of inertia 0.500 kg - m² and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.
ANSWER: The moment of inertia I = 0.500 kg-m²
r = 0.20 m, ⍵ = 20 rad/s.
The mass of the stationary particle m = 200 g = 0.20 kg
Hence the moment of inertia of the system with the particle
= I' =I+mr²
= 0.50 + 0.20*0.20*0.20 kg-m²
= 0.50+0.008 kg-m²
= 0.508 kg-m²
Since there is no external torque on the system, the angular momentum will be conserved. Hence
I'⍵' =I⍵
→0.508⍵' = 0.50*20
→⍵' = 10/0.508 rad/s = 19.7 rad/s
52. A diver, having a moment of inertia of 6.0 kg-m² about an axis through its center of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5 kg-m², what will be the new angular speed?
ANSWER: The initial moment of inertia I = 6.0 kg-m², ⍵ = 2 rad/s. The angular momentum = Г = I⍵ = 6.0*2 =12 kg-m²/s.
The final moment of inertia I' = 5.0 kg-m². Let the final angular speed = ⍵'.
Final angular momentum = Г' = I'⍵' = 5⍵' kg-m²/s
Since there is no external torque on the diver, his angular momentum will be conserved. i.e.
Г' = Г
→5⍵' = 12
→⍵' = 12/5 =2.4 rad/s
53. A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 kg-m² to 2 kg-m², what will b the new angular speed?
ANSWER: The initial moment of inertia I = 6 kg-m²,
⍵ = 120 rev/minute
Final M.I. = I' = 2 kg-m²
⍵' =?
Since the angular momentum will be conserved in the absence of a torque, So I'⍵' = I⍵
→⍵' = I⍵/I' =6*120/2 =360 rev/minute.
54. A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of "the platform plus the boy system" is 3.0x10⁻³ kg-m² and that of the umbrella is 2.0x10⁻³ kg-m². The boy starts spinning the umbrella about the axis at an angular speed of 2.0 rev/s with respect to himself. find the angular velocity imparted to the platform.
ANSWER: Let the angular velocity of the platform and boy = ⍵'
The angular velocity of the umbrella with respect to the ground = ⍵ = ⍵'+2 rev/minute
The angular momentum given to the umbrella = I⍵
=2.0x10⁻³*(⍵'+2) units
=(⍵'+2)*2.0x10⁻³ units
The M.I. of the platform+the boy = I' = 3.0x10⁻³ kg-m²
The angular momentum of the platform =I'⍵'
Since on the system of "The boy+platform+umbrella" there is no external force or torque, hence its angular momentum will be conserved. Initial angular momentum = 0, hence the final angular momentum I⍵+I'⍵' = 0
→I'⍵' = -I⍵
→⍵' = -(⍵'+2)*2.0x10⁻³/ 3.0x10⁻³
→3⍵' = -2⍵'-4
→5⍵' = -4
→⍵' = -4/5 =-0.80 rev/s
The negative sign denotes the direction of angular velocity which is opposite to the umbrella.
55. A wheel of moment of inertia 0.10 kg-m² is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.
ANSWER: The moment of inertia of the first wheel = I =0.10 kg-m²
Its angular speed ⍵ = 160 rev/minute.
The angular speed of the second wheel ⍵' = 300 rev/minute
Let the moment of inertia of the second wheel =I'
The total angular momentum of both the wheels = I⍵+I'⍵'
The angular speed of the system after the coupling = ⍵" = 200 rev/minute
The final angular momentum of the system =(I+I')⍵"
The angular momentum will be conserved, hence
I⍵+I'⍵' = (I+I')⍵"
→0.10*160+I'*300 = (0.10+I')*200
→16+300I' = 20+200I'
→100I' = 20-16 = 4
→I' = 4/100 = 0.04 kg-m²
56. A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is v horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.
ANSWER: The linear momentum of the ball =mv
The angular momentum of the ball about the center of the platform which is transferred to the system = moment of the linear momentum about the center = mvR
The moment of inertia of the system after the ball is caught = I+MR²+mR² = I+(M+m)R²
If the final angular speed of the system =⍵
its angular momentum = {I+(M+m)R²}⍵
Since the angular momentum will be conserved, hence
{I+(M+m)R²}⍵ = mvR
→⍵ = mvR/{I+(M+m)R²}
57. Suppose the platform of the previous problem is brought to rest with the ball in hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed v as seen by his friend. Find the angular velocity with which the platform will start rotating.
ANSWER: The moment of inertia of the platform and the kid
=I+MR²
It is initially at rest, so its angular momentum with the ball is zero.
If the angular velocity of the platform after throwing the ball is ⍵, then angular momentum =(I+MR²)⍵
The angular momentum of the ball after it is thrown =mvR
Since the angular momentum will be conserved, the total angular momentum of the ball and the platform with the boy will also be zero.
(I+MR²)⍵+mvR = 0
→⍵ = -mvR/(I+MR²)
The negative sign denotes the instantaneous speed of the boy will be opposite to the ball.
58. Suppose the platform with the kid in the previous problem is rotating in an anticlockwise direction at an angular speed of ⍵. The kid starts walking along the rim with a speed v relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.
ANSWER: The initial angular momentum of the system = (I+MR²)⍵
Let the new angular speed of the platform = ⍵'
The speed of the rim = ⍵'R
Hence the speed of the boy with respect to the ground = ⍵'R+v
Angular momentum of the boy = M(⍵'R+v)R
The angular momentum of the platform = I⍵'
Since the total angular momentum will be conserved,
I⍵'+M(⍵'R+v)R = (I+MR²)⍵
→I⍵'+M⍵'R²+MvR = (I+MR²)⍵
→(I+MR²)⍵' = (I+MR²)⍵-MvR
→⍵' = {(I+MR²)⍵-MvR}/(I+MR²)
→⍵' = ⍵ - MvR/(I+MR²)
59. A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate
(a) the speed of the center of mass, (b) the angular speed of the rod about the center the center of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the center of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.
ANSWER: (a) The impulse of the force = Ft
The initial speed of the center of mass = u =0
If the final speed of the center of mass is v, then
Change of linear momentum = Impulse
→mv-m*0 = Ft
→mv = Ft
→v = Ft/m
(b) Let the angular speed of the rod about the center of mass =⍵
The moment of inertia of the rod about the CoM and perpendicular to the rod = I = ml²/12
Final angular momentum = I⍵
Initial angular momentum = zero
The torque of the force about the center of mass T = Fl/2
The angular impulse = Tt = Flt/2
The change in angular momentum = angular impulse
→I⍵-0 = Flt/2
→⍵ = Flt/2I = Flt/2(ml²/12)
=6Flt/ml²
=6Ft/ml
(c) The kinetic energy of the rod
Linear kinetic energy + Rotational kinetic energy
= ½mv²+½I⍵²
= ½m(Ft/m)²+½(ml²/12)(6Ft/ml)²
=F²t²/2m + 36F²t²*/24m
=(F²t²/m)*(½+3/2)
=2F²t²/m
(d) The angular momentum = I⍵ =(ml²/12)*(6Ft/ml)
=6Ftml²/12ml
=Ftl/2
60. A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance traveled by the center of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.
ANSWER: Let the mass of the particle and the rod be m and M respectively. If the velocity of the particle is v initially and the angular speed of the rod after the strike = ⍵.
Taking the rod and the particle as a system, there is no external force or torque on the system.
The linear and angular momentum will be conserved. If the velocity of the center of mass after the strike is v', then
Mv' = mv
→v' = mv/M ...........................(i)
The angular momentum of the particle about the center = mvL/2
The angular momentum of the rod =I⍵ = ML²⍵/12
Equating the two,
ML²⍵/12 = mvL/2
→ML⍵/6 = mv
→⍵ = 6mv/ML
Time taken in turning through a right angle
t = π/2⍵ = πML/12mv
The distance traveled by the center of mass of the rod in this time
=v't
=(mv/M)*(πML/12mv)
=πL/12
When M = 4m
v' = mv/4m = v/4
And, ⍵ = 6mv/ML = 6mv/4mL = 3v/2L
Initial kinetic energy of the system = ½mv²
Final kinetic energy of the system = Translational K.E +Rotational K.E
=½Mv'²+½I⍵²
=½*4m*v²/16 + ½*ML²/12*(3v/2L)²
=mv²/8 +(4mL²/24)*(9v²/4L²)
=mv²/8 + 9mv²/24
=(1/8+9/24)mv²
={(3+9)/24}mv²
=12mv²/24
=½mv²
= Initial Kinetic energy of the system
Since the initial and final kinetic energy of the system is equal hence the collision is elastic.
ANSWER: (a) Mass M = 300 g = 0.300 kg, Length L = 50 cm =0.50 m
The moment of inertia about the center I = ML²/12
The moment of inertia about an end I' = I + Mr²
{Where r = L/2 = 0.25 m}
= ML²/12 + Mr²
= 0.30*0.5*0.5/12 + 0.30*0.25*0.25
= 0.025 kg-m²
Angular speed = ⍵ = 2 rad/s
Hence the angular momentum of the rod about the axis of rotation = I'⍵
= 0.025*2 = 0.05 kg-m²/s
(b) The speed of the center of the rod = v = ⍵r = 2*0.25 =0.50 m/s = 50 cm/s
(c) The kinetic energy of the rod = ½I'⍵² = 0.50*0.025*2² J
= 0.50*4*0.025 J
= 2.0*0.025 J
= 0.050 J
47. A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.
ANSWER: Mass M = 2.0 kg, Edge a = 10 cm = 0.10 m.
The moment of inertia about the diagonal = Ma²/12
= 2.0*(0.10²)/12
= 0.01/6 =1/600 kg-m²
Torque = 0.10 N-m
Angular acceleration α = Torque/M.I.
= 0.10/(1/600) = 60 rad/s²
Angular speed at t = 5 s, ⍵ = 0 + 60*5 =300 rad/s
Hence the angular momentum at this instant = I⍵ = (1/600)*300
=1/2 = 0.50 kg-m²/s
Kinetic Energy = E = ½I⍵²
= ½*(1/600)*300²
=90000/1200 =900/12 =75 J
48. Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1.5x10⁸ km.
ANSWER: Let the mass of the earth = M.
Its radius r = 6400 km = 64x10⁵ m, Radius of the orbit R = 1.5x10⁸ km =1.5x10¹¹ m.
The angular speed of the spin of the earth = ⍵
= 2π/24*3600 rad/s
The angular speed of the revolution of the earth ⍵' = 2π/365*24*3600 rad/s
If the moment of inertia of the earth about its axis = I =(2/5)Mr²
Then angular momentum = Γ =I⍵
The moment of inertia of the earth about the sun I' = I+MR²
= (2/5)Mr² +MR² = M(2r²/5 + R²)
The angular momentum about the sun = I'⍵'
Hence the ratio of the angular momentum
= I⍵/I'⍵'
= (2/5)Mr²⍵/M(2r²/5+R²)⍵'
= ⍵/(1+5R²/2r²)⍵'
= (2π/24*3600)/(1+5*2.25x10²²/2x64x64x10¹⁰)*(2π/365*24*3600)
= 365/0.00137x10¹²
= 265784.9x10⁻¹²
= 2.66x10⁵x10⁻¹²
= 2.66x10⁻⁷
49. Two particles of mass m₁ and m₂ are joined by a light rigid rod of length r. The system rotates at an angular speed ⍵ about an axis through the center of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is L = µr²⍵ where µ is the reduced mass of the system defined as µ = m₁m₂/(m₁+m₂).
ANSWER: Let the center of mass of the system from the mass m₁ be x, then x = m₂r/(m₁+m₂)
The moment of inertia of the system about the CoM
I = m₁{m₂r/(m₁+m₂)}² + m₂{m₁r/(m₁+m₂)}²
= [m₁m₂²+m₁²m₂]*r²/(m₁+m₂)²
= [m₁m₂(m₁+m₂)]*r²/(m₁+m₂)²
= m₁m₂r²/(m₁+m₂)
= [m₁m₂/(m₁+m₂)]r²
= µr²
Since the angular speed = ⍵
Hence the angular momentum Γ = I⍵ =µr²⍵.
Hence proved.
50. A dumbbell consists of two identical small balls of mass ½ kg each connected to the two ends of a 50 cm long light rod. The dumbbell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5.0 N acts on one of the masses in the direction of its velocity for 0.10 s. Find the new angular velocity of the system.
ANSWER: The moment of inertia of the system I = 2mr²
= 2*½*0.25*0.25
= 0.0625 kg-m²
⍵ = 10 rad/s
Time of impulse = t = 0.10 s
Torque of the impulsive force = T = 5*0.25 =1.25 N-m
Impulsive torque =T*t
=1.25*0.10
=0.125 Nm-s
change in angular momentum = Impulsive torque
If ⍵' is the new angular velocity of the system, then
I⍵'-I⍵ = Tt
⍵' = ⍵+Tt/I
= 10 + 0.125/0.0625
= 10 + 2
= 12 rad/s
51. A wheel of moment of inertia 0.500 kg - m² and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.
ANSWER: The moment of inertia I = 0.500 kg-m²
r = 0.20 m, ⍵ = 20 rad/s.
The mass of the stationary particle m = 200 g = 0.20 kg
Hence the moment of inertia of the system with the particle
= I' =I+mr²
= 0.50 + 0.20*0.20*0.20 kg-m²
= 0.50+0.008 kg-m²
= 0.508 kg-m²
Since there is no external torque on the system, the angular momentum will be conserved. Hence
I'⍵' =I⍵
→0.508⍵' = 0.50*20
→⍵' = 10/0.508 rad/s = 19.7 rad/s
52. A diver, having a moment of inertia of 6.0 kg-m² about an axis through its center of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5 kg-m², what will be the new angular speed?
ANSWER: The initial moment of inertia I = 6.0 kg-m², ⍵ = 2 rad/s. The angular momentum = Г = I⍵ = 6.0*2 =12 kg-m²/s.
The final moment of inertia I' = 5.0 kg-m². Let the final angular speed = ⍵'.
Final angular momentum = Г' = I'⍵' = 5⍵' kg-m²/s
Since there is no external torque on the diver, his angular momentum will be conserved. i.e.
Г' = Г
→5⍵' = 12
→⍵' = 12/5 =2.4 rad/s
53. A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 kg-m² to 2 kg-m², what will b the new angular speed?
ANSWER: The initial moment of inertia I = 6 kg-m²,
⍵ = 120 rev/minute
Final M.I. = I' = 2 kg-m²
⍵' =?
Since the angular momentum will be conserved in the absence of a torque, So I'⍵' = I⍵
→⍵' = I⍵/I' =6*120/2 =360 rev/minute.
54. A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of "the platform plus the boy system" is 3.0x10⁻³ kg-m² and that of the umbrella is 2.0x10⁻³ kg-m². The boy starts spinning the umbrella about the axis at an angular speed of 2.0 rev/s with respect to himself. find the angular velocity imparted to the platform.
ANSWER: Let the angular velocity of the platform and boy = ⍵'
The angular velocity of the umbrella with respect to the ground = ⍵ = ⍵'+2 rev/minute
The angular momentum given to the umbrella = I⍵
=2.0x10⁻³*(⍵'+2) units
=(⍵'+2)*2.0x10⁻³ units
The M.I. of the platform+the boy = I' = 3.0x10⁻³ kg-m²
The angular momentum of the platform =I'⍵'
Since on the system of "The boy+platform+umbrella" there is no external force or torque, hence its angular momentum will be conserved. Initial angular momentum = 0, hence the final angular momentum I⍵+I'⍵' = 0
→I'⍵' = -I⍵
→⍵' = -(⍵'+2)*2.0x
→3⍵' = -2⍵'-4
→5⍵' = -4
→⍵' = -4/5 =-0.80 rev/s
The negative sign denotes the direction of angular velocity which is opposite to the umbrella.
55. A wheel of moment of inertia 0.10 kg-m² is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.
ANSWER: The moment of inertia of the first wheel = I =0.10 kg-m²
Its angular speed ⍵ = 160 rev/minute.
The angular speed of the second wheel ⍵' = 300 rev/minute
Let the moment of inertia of the second wheel =I'
The total angular momentum of both the wheels = I⍵+I'⍵'
The angular speed of the system after the coupling = ⍵" = 200 rev/minute
The final angular momentum of the system =(I+I')⍵"
The angular momentum will be conserved, hence
I⍵+I'⍵' = (I+I')⍵"
→0.10*160+I'*300 = (0.10+I')*200
→16+300I' = 20+200I'
→100I' = 20-16 = 4
→I' = 4/100 = 0.04 kg-m²
56. A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is v horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.
ANSWER: The linear momentum of the ball =mv
The angular momentum of the ball about the center of the platform which is transferred to the system = moment of the linear momentum about the center = mvR
The moment of inertia of the system after the ball is caught = I+MR²+mR² = I+(M+m)R²
If the final angular speed of the system =⍵
its angular momentum = {I+(M+m)R²}⍵
Since the angular momentum will be conserved, hence
{I+(M+m)R²}⍵ = mvR
→⍵ = mvR/{I+(M+m)R²}
57. Suppose the platform of the previous problem is brought to rest with the ball in hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed v as seen by his friend. Find the angular velocity with which the platform will start rotating.
ANSWER: The moment of inertia of the platform and the kid
=I+MR²
It is initially at rest, so its angular momentum with the ball is zero.
If the angular velocity of the platform after throwing the ball is ⍵, then angular momentum =(I+MR²)⍵
The angular momentum of the ball after it is thrown =mvR
Since the angular momentum will be conserved, the total angular momentum of the ball and the platform with the boy will also be zero.
(I+MR²)⍵+mvR = 0
→⍵ = -mvR/(I+MR²)
The negative sign denotes the instantaneous speed of the boy will be opposite to the ball.
58. Suppose the platform with the kid in the previous problem is rotating in an anticlockwise direction at an angular speed of ⍵. The kid starts walking along the rim with a speed v relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.
ANSWER: The initial angular momentum of the system = (I+MR²)⍵
Let the new angular speed of the platform = ⍵'
The speed of the rim = ⍵'R
Hence the speed of the boy with respect to the ground = ⍵'R+v
Angular momentum of the boy = M(⍵'R+v)R
The angular momentum of the platform = I⍵'
Since the total angular momentum will be conserved,
I⍵'+M(⍵'R+v)R = (I+MR²)⍵
→I⍵'+M⍵'R²+MvR = (I+MR²)⍵
→(I+MR²)⍵' = (I+MR²)⍵-MvR
→⍵' = {(I+MR²)⍵-MvR}/(I+MR²)
→⍵' = ⍵ - MvR/(I+MR²)
59. A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate
(a) the speed of the center of mass, (b) the angular speed of the rod about the center the center of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the center of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.
ANSWER: (a) The impulse of the force = Ft
The initial speed of the center of mass = u =0
If the final speed of the center of mass is v, then
Change of linear momentum = Impulse
→mv-m*0 = Ft
→mv = Ft
→v = Ft/m
(b) Let the angular speed of the rod about the center of mass =⍵
The moment of inertia of the rod about the CoM and perpendicular to the rod = I = ml²/12
Final angular momentum = I⍵
Initial angular momentum = zero
The torque of the force about the center of mass T = Fl/2
The angular impulse = Tt = Flt/2
The change in angular momentum = angular impulse
→I⍵-0 = Flt/2
→⍵ = Flt/2I = Flt/2(ml²/12)
=6Flt/ml²
=6Ft/ml
(c) The kinetic energy of the rod
Linear kinetic energy + Rotational kinetic energy
= ½mv²+½I⍵²
= ½m(Ft/m)²+½(ml²/12)(6Ft/ml)²
=F²t²/2m + 36F²t²*/24m
=(F²t²/m)*(½+3/2)
=2F²t²/m
(d) The angular momentum = I⍵ =(ml²/12)*(6Ft/ml)
=6Ftml²/12ml
=Ftl/2
60. A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance traveled by the center of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.
ANSWER: Let the mass of the particle and the rod be m and M respectively. If the velocity of the particle is v initially and the angular speed of the rod after the strike = ⍵.
Taking the rod and the particle as a system, there is no external force or torque on the system.
 |
| Figure for Q-60 |
The linear and angular momentum will be conserved. If the velocity of the center of mass after the strike is v', then
Mv' = mv
→v' = mv/M ...........................(i)
The angular momentum of the particle about the center = mvL/2
The angular momentum of the rod =I⍵ = ML²⍵/12
Equating the two,
ML²⍵/12 = mvL/2
→ML⍵/6 = mv
→⍵ = 6mv/ML
Time taken in turning through a right angle
t = π/2⍵ = πML/12mv
The distance traveled by the center of mass of the rod in this time
=v't
=(mv/M)*(πML/12mv)
=πL/12
When M = 4m
v' = mv/4m = v/4
And, ⍵ = 6mv/ML = 6mv/4mL = 3v/2L
Initial kinetic energy of the system = ½mv²
Final kinetic energy of the system = Translational K.E +Rotational K.E
=½Mv'²+½I⍵²
=½*4m*v²/16 + ½*ML²/12*(3v/2L)²
=mv²/8 +(4mL²/24)*(9v²/4L²)
=mv²/8 + 9mv²/24
=(1/8+9/24)mv²
={(3+9)/24}mv²
=12mv²/24
=½mv²
= Initial Kinetic energy of the system
Since the initial and final kinetic energy of the system is equal hence the collision is elastic.
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