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Science > Physics > 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 R_{2}= ½ R_{1}, Present period = T_{1}= 24 hr**To Find:**New period = T_{2}=?

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 V_{2}= 1/27 V_{1}, Present period = T_{1}= 24 hr**To Find:**New period = T_{2}=?

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 = I_{d}=?

The angular speed after coupling is

Let M.I. of the disc be I_{d} and that of bob be I_{b}.

By the principle of conservation of angular momentum

**Ans: **M.I. of the disc is 2 x 10^{-4} kg m^{2}

** ****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 m^{2}, find the new frequency of rotation of the disc. **Solution:****Given:**Initial speed of rotation of disc = N_{d}= 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 = I_{d}= 2 x 10^{-4}kg m^{2}**To Find:**New frequency of rotation = N = ?

Let M.I. of the disc be I_{d} and that of bob be I_{b}.

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 = I_{d}=?

Let M.I. of the disc be I_{d} and that of bob be I_{b}.

By the principle of conservation of angular momentum

**Ans: **M.I. of the disc is 5.94 x 10^{-4} kg m^{2}

#### 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 %, I_{2}= I_{1}– 40% I_{1}= 0.6 I_{1},**To Find:**New angular speed = N_{2}=?

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 %, I_{2}= 0.75 I_{1},**To Find:**New angular speed = N_{2}=?

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 = N_{1}= 81 r.p.m., radius of gyration reduces by 10%, K_{2}= K_{1}– 10% K_{1}= 0.90 K_{1},**To Find:**increase in angular speed = N_{2}– N_{1 }=?

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 m
^{2}. He slowly folds his arms until the M. I. is reduced to 60 kg m^{2}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 = I_{1}= 150 kg m^{2}. Final M.I. of system = I_{2}= 60 kg m^{2}. Final angular speed = ω_{2}= 5 rad/s**To Find:**Initial angular speed w_{1 }=? 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 = N_{1}= 500 r.p.m.,**To Find:**New frequency of rotation = N = ?

M.I. of the first wheel = I_{1} and M.I. of the second wheel = I_{2},

Let N be their final common angular speed.

The two wheels are identical. I_{1 }= I_{2}.

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 m
^{2}and 2 kg m^{2}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 = I
_{1}= 4 kg m^{2}, Initial speed of rotation of first wheel = N_{1}= 120 r.p.m., M.I. of the second wheel = I_{2}= 2 kg m^{2}, Initial speed of rotation of second wheel = N_{2}= 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 m
^{2}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 = I
_{1}= 4 kg m^{2}, Initial speed of rotation of first wheel = N_{1}= 120 r.p.m., M.I. of the second wheel = I_{2}= 4 kg m^{2}, Initial speed of rotation of second wheel = N_{2}= 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 m
^{2}is set spinning at 600 r.p.m. Wheel B of M.I. 2 kg m^{2}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 = I_{A}= 0.5 kg m^{2}, Initial speed of rotation of first wheel = N_{A}= 600 r.p.m., M.I. of the second wheel = I_{B}= 2 kg m^{2}, Initial speed of rotation of second wheel = N_{B}= 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 > Physics > Rotational Motion > You are Here |

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