Problem#1
A ballet dancer has a moment of inertia of 4 kg.m2 when his arms reach his body and 16 kg.m2 when his arms are stretched. When the two arms are pressed against his body, the speed of the dancer's rotation is 12 rounds/s. If then the arms are stretched, the rotation speed becomes. . . .?
By using momentum conservation law we can solve this problem easily,
When his hand is closed the ballet dancer has a moment of inertia, I = 4 kgm2 and angular velocity, ω = 12 put / s.
When his hand is stretched the ballet dancer has a moment of inertia, I '= 16 kgm2, then the angular velocity (ω') is
L = L'
Iω = I'ω'
4 kg.m2 x 12 rad/s = 16 kg.m2 x ω '
ω'= 3 rad/s
Problem#2
A thin circular ring m and R finger rotate against its axis with angular velocity ω. Two objects each with mass M are fastened to the opposite ends of the ring diameter. The ring now rotates with speed ω '=. . . .?
Answer
The moment of inertia of the ring is Icincin = mR2,
When at the ends of the ring diameter objects are placed with mass M, then the moment of ring + object inertia is
IRing + object = mR2 + 2MR2
So by using the law of conservation of angular momentum, we get it
Icincinω = (Iring + object) ω’
mR2ω = (mR2 + 2MR2)ω’
ω’= mω/(m + 2M)
Problem#3
Wheel A has a mass of 2 kg and radius of 0.20 m, and the initial angular velocity is 50 rad / s (approximately 500 rpm). Wheel A is coupled (one axis) with chip B which has a mass of 4 kg and radius 0.10 m, and the initial angular velocity is 200 rad / s. Determine the final angular velocity after both are pushed so that they touch. Is kinetic energy eternal during this process?
Assume both of the initial rotation directions are the same, the law of conservation of angular momentum gives,
ΔL = 0
L1 = L2
I₁ω₁ + I₂ω₂ – (I₁ + I₂)ω = 0
½m₁r₁²ω₁ + ½ m₂r₂²ω₂ - ½ (m₁r₁² + m₂r₂²)ω = 0
½ (2 kg)(0.2 m)²(50 rad/s) + ½ (4 kg)(0.1 m)²(200 rad/s) – ½ ((2 kg)(0.2 m)² + (4 kg)(0.1 m)²)ω = 0
ω = 100 rad/s
Problem#4
Dish 1 has a mass of 440 g, a mesh of 3.5 cm and rotates ω0 = 180 rpm with a perfectly slippery axle and can be ignored. At first the disk 2 whose mass is 270 g and the 2.3 cm track does not rotate, but then falls freely on the disc below, and due to friction, both move at angular velocities equal to angular velocity ω, the kinetic energy lost due to friction is . . . .?
The inertial moment of plate 1,
I1 = ½m1r12 = ½ (0,44 kg)(3,5 x 10-2)2 = 2,695 x 10-4 kgm2
The inertial moment of plate 2,
I2 = ½m2r22 = ½ (0,27 kg)(2,3 x 10-2)2 = 0,714 x 10-4 kgm2
Initial momentum before the two plates are united,
Li = I1ω1 + I2ω2 = 2,695 x 10-4 kgm2 x 180 x 2π rad/60 s + 0 = 5,08 x 10-3 kgm2/s
Final momentum (both plates are united),
Lf = (I1+ I2)ω = (2,695 x 10-4 kgm2 + 0,714 x 10-4 kgm2)ω = 3,409 x 10-4 kgm2 x ω
The law of conservation of angular momentum gives,
Li = Lf
5,08 x 10-3 kgm2/s = 3,409 x 10-4 kgm2 x ω
ω = 14,9 rad/s
Initial kinetic energy,
EKi = ½ I1ω02 = ½(2,695 x 10-4 kgm2)(180 x 2π rad/60 s)2 = 478,77 J x 10-4 J
Final kinetic energy,
EKf = ½ (I1 + I2)ω2
EKf = ½(2,695 x 10-4 kgm2 + 0,714 x 10-4 kgm2)(14,9 rad/s)2 = 378,42 x 10-4 J
Change in kinetic energy of rotation of both wheels,
ΔEK = EKf – EKi = –100.35 x 10-4 J (energy lost)
In %,
Lost energy = (ΔEK/EKi) x 100% ≈ 21%
Problem#5
In a nightmare you dream that you are a hamster running in an exercise wheel. Typical hamsters are 300 g and can run at speeds of 3.2 m/s. A typical exercise wheel has a moment of inertia about its centre of 0.250 kgm2. How fast should the wheel have been rotating in your dream? The radius of the wheel is 12.0 cm. Treat the hamster as a point mass. Hint what was the angular momentum of the system before the hamster started running?
Before the hamster starts running, the exercise wheel is not rotating. Considered as a system, angular momentum must be conserved.
Lf = Li
For this particular problem,
Lwheel + Lhamster = 0
The wheel is a rotating object so its angular momentum is given by Lwheel = ─Iω, where the minus sign indicates that it is into the paper. For a point particle, the angular momentum is Lhamster = Rmv out of the paper. Thus we have
Iω+ Rmv = 0
So the angular velocity of the wheel is
ω = Rmv/I = (0.3 kg)(0.12 m)(3.2 m/s)/(0.25 kgm2/s) = 0.461 rad/s
Problem#6
A door with width L = 1.0 m and mass M = 15 kg is hinged on one side so that it can rotate freely. A bullet, as shown, is fired into the exact centre of the door. The bullet has mass 25 g and a speed of 400 m/s. What is the angular velocity of the door with respect to the hinge just after the bullets embeds itself in the door? The door may be treated as a thin rectangular sheet. The bullet may be treated as a point mass.
Answer
Since we have a collision in which there is a change in rotation, we apply the Law of Conservation of Momentum,
Lf = Li (*)
Initially the bullet is traveling in a straight line so its angular momentum is bmv, where b is the distance of closest approach to the point of rotation. Since everything will rotate about the door hinge, we take the hinge as the point of rotation. Hence b = ½L. After the collision, the bullet is stuck in the door and rotates with the door in a circle. The angular momentum of an object moving in a circle is r2mω, where r is the radius of rotation. Clearly r = ½L.
Initially the door is not rotating and thus has no angular momentum. Afterwards, it is rotating and thus has an angular momentum given by Iω. Note that the door is not rotating about its centre of mass, so we need to use the parallel axis theorem. Consulting a table, we find Icm = (1/12)ML2. The centre of mass is d = ½L from the hinge.
Thus equation (*) applied to this problem is
(½L)2mω + [(1/12)ML2 + M (½L)2]ω = ½Lmv
Dividing through by ½L and solving for ω, we find
ω = mv/[½m + (2/3)M]L = (0.025)(400)/[½(0.025) + (2/3)15] = 0.999 rad/s
Problem#7
A 150g piece of playdough slides across a frictionless table at v = 5.50 m/s. It collides with a disk of radius R = 35.0 cm and mass M = 2.50 kg which has a fixed frictionless axle. The playdough stick to the disk. Treat the playdough as a point mass. (a) If the disk is not rotating initially, what is its angular velocity after the collision? (b) What angular velocity would the disk need to have initially, if the disk stopped completely after the collision?
Answer
(a) Since we have a collision in which there is a change in rotation, we apply the Law of Conservation of Momentum,
Lf = Li (*)
Initially the playdough is traveling in a straight line so its angular momentum is bmv, where b is the distance of closest approach to the point of rotation. Since everything will rotate about the centre of the disk, we take the centre as the point of rotation. Thus b = 0.20 m. After the collision, the playdough is stuck on the disk and rotates with the door in a circle. The angular momentum of an object moving in a circle is r2mω, where r is the radius of rotation. Clearly r = R, the radius of the disk.
Initially the disk is not rotating and thus has no angular momentum. Afterwards, it is rotating and thus has an angular momentum given by Iω. Consulting a table of moments of inertia, we find I = ½MR2.
Thus equation (*) applied to this problem is
R2mω + ½MR2ω = bmv
Solving for ω, we find
ω = bmv/[m + ½M]R2 = (0.2)(0.15)(5.5)/[(0.15) + ½(2.5)](0.35)2 = 0.962 rad/s .
(b) If everything stops, Lf = 0. Equation (*) becomes
0 = bmv + ½MR2ωinitial
Solving for ωinitial, we get
ωinitial = 2bmv/MR2 = 2(0.2)(0.15)(5.5)/(2.5)(0.35)2
The minus sign indicates that the disk would have to rotate clockwise.
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