# Problems Based on Conservation of Angular Momentum

 Science > Rotational Motion > You are Here

#### Example – 1:

• If the earth had its radius suddenly decreased by half when spinning about its axis, what would the length of the day be?
• Solution:
• Given: Radius of earth R2 = ½ R1, Present period = T1 = 24 hr
• To Find: New period = T2 =?

By the principle of conservation of angular momentum

Ans: New length of the day would be 6 hours

#### Example -2:

• What will be the duration of the day if the earth suddenly shrinks to 1/27 th of its original volume? The mass being unchanged.
• Solution:
• Given: Volume of the earth V2 = 1/27 V1, Present period = T1 = 24 hr
• To Find: New period = T2 =?

By the principle of conservation of angular momentum

Ans: New duration of the day would be 2.67 hours

#### Example – 3:

• A disc is rotating in a horizontal plane about a vertical axis at the rate of 5π/3 rad/s. A blob of wax of mass 0.02 kg falls vertically on the disc and adheres to it at a distance of 0.05 m from the axis of rotation. If the speed of rotation thereby becomes 40 rev/min, calculate the M. I. of the disc.
• Solution:
• Given: Initial speed of rotation of disc = 5π/3 rad/s,  Mass of bob = M = 0.02 kg, Distance from axis of rotation = r = 0.05 m, New angular speed = N = 40 rev/min
• To Find: M.I. of disc = Id =?

The angular speed after coupling is

Let M.I. of the disc be Id and that of bob be Ib.

By the principle of conservation of angular momentum

Ans: M.I. of the disc is 2 x 10-4 kg m2

#### Example – 4:

• A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 r.p.m. A blob of wax of mass 20 g falls on the disc and sticks to it at a distance of 5 cm from the axis. If the M. I. of the disc about the given axis is 2 x 10-4 kg m2 , find the new frequency of rotation of the disc.
• Solution:
• Given: Initial speed of rotation of disc = Nd = 100 r.p.m.,  Mass of bob = M = 20 g = 0.02 kg, Distance from axis of rotation = r = 5 cm = 0.05 m, : M.I. of disc = Id = 2 x 10-4 kg m2
• To Find: New frequency of rotation = N = ?

Let M.I. of the disc be Id and that of bob be Ib.

By the principle of conservation of angular momentum

Ans: New angular speed = 50 r.p.m.

#### Example – 5:

• A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 180 r.p.m. A blob of wax of mass 1.9 g falls on the disc and sticks to it at a distance of 25 cm from the axis. If the speed of rotation now is 60 r.p.m. Calculate the moment of inertia of the disc.
• Solution:
• Given: Initial speed of rotation of disc = 180 r.p.m.,  Mass of bob = M = 1.9 g = 0.0019 kg, Distance from axis of rotation = r = 25 cm = 0.25 m, New angular speed = N = 60 r.p.m.
• To Find: M.I. of disc = Id =?

Let M.I. of the disc be Id and that of bob be Ib.

By the principle of conservation of angular momentum

Ans: M.I. of the disc is 5.94 x 10-4 kg m2

#### Example – 6:

• A ballet dancer spins about a vertical axis at 120 r.p.m. with her arms outstretched. With her arms folded, the M. I. about the axis of rotation decreases by 40%. Calculate the new rate of revolution.
• Solution:
• Given: Initial angular speed = 120 r.p.m., M.I. decreases by 40 %, I2 = I1 – 40% I1 = 0.6 I1,
• To Find: New angular speed = N2 =?

By the principle of conservation of angular momentum

Ans: New rate of revolution is 200 r.p.m.

#### Example – 7:

• A ballet dancer spins about a vertical axis at 90 r.p.m. with her arms outstretched. With her arms folded, the M. I. about the axis of rotation changes to 75%. Calculate the new rate of revolution.
• Solution:
• Given: Initial angular speed = 90 r.p.m., M.I. decreases to 75 %, I2 = 0.75 I1,
• To Find: New angular speed = N2 =?

By the principle of conservation of angular momentum

Ans: New rate of revolution is 120 r.p.m.

#### Example – 8:

• A man is standing on a rotating chair with his arms fully outstretched. The chair rotates with a speed of 81 r.p.m. On bringing arms close to his body, the radius of gyration is reduced by 10%. What is the increase in the speed of rotation (neglect friction)
• Solution:
• Given: Initial angular speed = N1 = 81 r.p.m., radius of gyration reduces by 10%, K2 = K1 – 10% K1 = 0.90 K1,
• To Find: increase in angular speed = N2 – N1 =?

By the principle of conservation of angular momentum

Ans: The Increase in angular speed = 100 – 81 = 19 r.p.m.

#### Example – 9:

• A man standing with outstretched arms on a frictionless rotating turntable has a mass 0.2 kg in each hand. The M. I. of the system in this position is 150 kg m2. He slowly folds his arms until the M. I. is reduced to  60 kg m2 The angular velocity of the system is now 5 rad/s. Find the initial angular velocity and the new K.E. of the system.
• Solution:
• Given: Initial M.I. of system  = I1 = 150 kg m2. Final M.I. of system = I2 = 60 kg m2. Final angular speed = ω2 = 5 rad/s
• To Find: Initial angular speed w1 =? Final K.E. =?

By the principle of conservation of angular momentum

Final K.E. = ½ I w2 = ½ x 60 x (5)2 = 750 J

Ans: Initial angular speed = 2 rad/s, Final kinetic energy = 750 J.

#### Example – 10:

• A wheel is rotating with an angular speed of 500 r.p.m. on a shaft. A second identical wheel initially at rest is suddenly coupled to the same shaft. What is the angular speed of the combination? Assume that the M.I. of the shaft is negligible.
• Solution:
• Given: Initial speed of rotation of first wheel  = N1 = 500 r.p.m.,
• To Find: New frequency of rotation = N = ?

M.I. of the first wheel = I1 and M.I. of the second wheel = I2,

Let N be their final common angular speed.

The two wheels are identical. I1 = I2.

By the principle of conservation of angular momentum

Ans: The angular speed of combination is 250 r.p.m.

#### Example – 11:

• Two wheels of moments of inertia 4 kg m2 and 2 kg m2 rotate at the rate of 120 rev/min and 240 rev/min respectively and in the same direction. If the two are coupled so as to rotate with a common angular velocity, find the speed of revolution.
• Solution:
• Given: M.I. of the first wheel = I1 = 4 kg m2, Initial speed of rotation of first wheel  = N1 = 120 r.p.m., M.I. of the second wheel = I2 = 2 kg m2, Initial speed of rotation of second wheel  = N2 = 240 r.p.m.,
• To Find: New frequency of rotation = N = ?

By the principle of conservation of angular momentum

Ans: The angular speed of combination is 160 r.p.m.

#### Example – 12:

• Two wheels of moments of inertia 4 kg m2 each rotate at the rate of 120 rev/min and 240 rev/min respectively and in the opposite direction. If the two are coupled so as to rotate with a common angular velocity, find the speed of revolution.
• Solution:
• Given: M.I. of the first wheel = I1 = 4 kg m2, Initial speed of rotation of first wheel  = N1 = 120 r.p.m., M.I. of the second wheel = I2 = 4 kg m2, Initial speed of rotation of second wheel  = N2 = 240 r.p.m.,
• To Find: New frequency of rotation = N = ?

As the two wheels are rotating in opposite direction, the total initial angular momentum is

By the principle of conservation of angular momentum

Ans: The angular speed of combination is 60 r.p.m.

#### Example – 13:

• Two wheels A and B can rotate side by side on he same axle. Wheel A of M.I. 0.5 kg m2 is set spinning at 600 r.p.m. Wheel B of M.I. 2 kg m2 is initially stationary. A clutch now acts to join A and B so that they must spin together. At what speed will they rotate? How does the rotational K.E. before joining compare with the rotational K.E. afterwards?
• Solution:
• Given: M.I. of the first wheel = IA = 0.5 kg m2, Initial speed of rotation of first wheel  = NA = 600 r.p.m., M.I. of the second wheel = IB = 2 kg m2, Initial speed of rotation of second wheel  = NB = 0 r.p.m.,
• To Find: New common frequency of rotation = N = ?

By the principle of conservation of angular momentum

Kinetic energy before coupling

Kinetic energy after coupling

Dividing equation (2) by (1)

Ans: The new angular speed is 120 r.p.m.

the ratio of initial K.E. to Final K.E.is 5

 Science > Rotational Motion > You are Here