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Science > Physics > Rotational Motion > You are Here |

#### Example – 01:

- A disc begins to rotate from rest with a constant angular acceleration of 0.5 rad/s
^{2}and acquires an angular momentum of 73.5 kg m^{2}/s in 15 s after the start. Find the kinetic energy of the disc in 20 s after the start. **Given:**Angular acceleration = α = 0.5 rad/s^{2}, Change in angular momentum = dL = 73.5 kg m^{2}/s, time taken = t = 15 s, initial angular speed = ω_{1 }= 0 rad/s**To find:**K.E. in 20 s after start.**Solution:**

We have, ω_{2 }= ω_{1 }+ αt = 0 + 0.5 x 20 = 10 rad/s

Change in angular momentum = dL = L_{2} – L_{1} = Iω_{2} – Iω_{1}

73.5 = I x 7.5 – I x 0

I = 73.5/7.5 = 9.8 kg m^{2}

Now, Kinetic energy is given by

K.E. = ½ I ω^{2} = ½ x 9.8 x 10^{2} = 490 J

**Ans:** The kinetic energy of the disc in 20 s after the start is 490 J

#### Example – 02:

- The angular momentum of a body changes by 80 kg m
^{2}/s when its angular velocity changes from 20 rad/s to 40 rad/s. Find the changes in its K.E. of rotation. **Given:**Change in angular momentum = 80 kg m^{2}/s. Initial angular speed = ω_{1 }= 20 rad/s, final angular speed = ω_{2 }= 40 rad/s.**To Find:**Change in kinetic energy = ?**Solution:**

Change in angular momentum = dL = L_{2} – L_{1} = Iω_{2} – Iω_{1}

80 = I x 40 – I x 20

∴ 80 = 20 I

∴ I = 80/20 = 4 kg m^{2}

Change in kinetic energy = K.E._{2} – K.E._{1} = ½ I ω_{2}^{2 }– ½ I ω_{1}^{2} = ½ I ( ω_{2}^{2 }– ω_{1}^{2})

Change in kinetic energy = ½ x 4 x ( 40^{2 }– 20^{2})

Change in kinetic energy = 2 x ( 1600^{ }– 400) = 2400 J

Ans: The change in kinetic energy of rotation is 2400 J

#### Example – 03:

- An energy of 500 J is spent to increase the speed of wheel from 60 r.p.m. to 240 r.pm. Calculate the moment of inertia of the wheel.
- Given: Change in kinetic energy = 500 J, Initial angular speed = N
_{1}= 60 r.pm., final angular speed = N_{2}= 240 r.p.m. - To Find: Moment of Inertia of the wheel.
- Solution:

ω_{1 }= 2πN_{1}/60 = 2π x 60/60 = 2π rad/s

ω_{2 }= 2πN_{2}/60 = 2π x 240/60 = 8π rad/s

Change in K.E. = K.E._{2} – K.E._{1}

∴ Change in K.E. = ½ Iω_{2}^{2} – ½ Iω_{1}^{2} = ½ I(ω_{2}^{2} – ω_{1}^{2})

∴ 500. = ½ I((8π)^{2} – (2π)^{2})

∴ 1000. = I(64π^{2} – 4π^{2})

∴ 1000. = I(60π^{2})

∴ I = 1000/60π^{2} = 1.688 kgm^{2}

**Ans: **The moment inertia of wheel is 1.688 kg m^{2}

#### Example – 04:

- A wheel of the moment of inertia 1 kgm
^{2}is rotating at a speed of 30 rad/s. Due to friction on the axis, it comes to rest in 10 minutes. Calculate i) total work done by friction, ii) the average torque of the friction iii) angular momentum of the wheel two minutes before it stops rotating. **Given:**Moment of inertia of wheel = 1 kg m^{2}, Initial angular speed = ω_{1}= 30 rad/s, final angular speed = ω_{2}= 0 rad/s, time = t = 10 min = 10 x 60 = 600 s**To Find:**Work done by friction = ?, Torque acting = ?, angular momentum of the wheel two minutes before it stops rotating.**Solution:**

Work done by friction = Change in K.E. = K.E._{1} – K.E._{2}

∴ Work done by friction. = ½ Iω_{1}^{2} – ½ Iω_{2}^{2} = ½ I(ω_{1}^{2} – ω_{2}^{2})

∴ Work done by friction. = ½ x 1 ((30)^{2} – (0)^{2}) = 0.5 x 900 = 450 J

Now angular acceleration = α = (ω_{2} – ω_{1})/t = (0 – 30)/(10 x 60) = -30/600 = – 0.05 rad/s^{2}

Negative sign indicates it is retardation.

The magnitude of angular acceleration = α = 0.05 rad/s^{2}

Average torque = I α = 1 x 0.05 Nm

It is retarding torque

To find the angular momentum of the wheel two minutes before it stops rotating.

t = 10 min – 2 min = 8 min = 8 x 60 = 480 s.

ω_{2} = ω_{1}+ αt = 30 – 0.05 x 480 = 30 – 24 = 6 rad/s

Angular momentum = Iω_{2} = 1 x 6 = 6 kg m/s^{2}

**Ans: **Work done by friction is 450 J, Retarding torque acting is 0.05 Nm,

angular momentum of the wheel two minutes before it stops rotating is 6 kg m/s^{2}.

#### Example – 05:

- A ceiling fan of a moment of inertia 30 kgm
^{2}rotates about its own axis at the speed of 120 r.p.m. under the action of an electric motor of power 62.8 watts. If electric power is cut off, how many revolutions will it complete before coming to rest? **Given:**Moment of inertia of fan = 30 kg m^{2}, Initial angular speed = N_{1}= 120 r.p.m., final angular speed = ω_{2}= 0 rad/s,**To Find:**Number of revolutions before coming to rest.**Solution:**

ω_{1 }= 2πN_{1}/60 = 2π x 120/60 = 4π rad/s

Power = τ ω_{1 }= I α ω_{1}

∴ 62.8 = 30 x α x 4π

∴ 62.8 = 120π x α

∴ α = 62.8/(120 x 3.14) = 1/6 rad/s^{2}.

ω_{2}^{2} = ω_{1}^{2} + 2α θ

∴ (0)^{2} = (4π)^{2} + 2(- 1/6) θ

∴ 0 = 16π^{2} – (1/3) θ

∴ (1/3) θ = 16π^{2}

∴ θ = 48π^{2}

No. of rotations = θ/2π = 48π^{2}/2π = 24π = 24 x 3.14 = 75.36

**Ans: **Number of revolutions before coming to rest are 24π or 75.35

**Example – 06:**

- When an angular velocity of a body changes from 20 rad/s to 40 rad/s, its angular momentum changes by 80 kgm
^{2}/s. Find the change in kinetic energy of rotation. **Given:**Initial angular speed = ω_{1}= 20 rad/s, final angular speed = ω_{2}= 40 rad/s, Change in angular momentum = dL = 80 kgm^{2}/s.**To Find:**Change in kinetic energy of rotation = ?.**Solution:**

Change in angular momentum = L_{2} – L_{1}

∴ dL = Iω_{2} – Iω_{1} = I(ω_{2} – ω_{2})

∴ 80 = I(40 – 20) = 20 I

∴ I = 80/20 = 4 kg m^{2}

Change in K.E. = K.E._{2} – K.E._{1}

∴ Change in K.E. = ½ Iω_{2}^{2} – ½ Iω_{1}^{2} = ½ I(ω_{2}^{2} – ω_{1}^{2})

∴ Change in K.E. = ½ x 4 ((40)^{2} – (20)^{2}) = 2 x (1600 – 400) = 2 x 1200 = 2400 J

**Ans: **the change in kinetic energy of rotation is 2400 J

**Example – 07:**

- A flywheel of mass 2 kg has the radius of gyration of 0.2 m. Calculate the K.E. of rotation when it makes 5 revolutions per second.
**Given:**Mass of disc = M = 2 kg; radius of gyration = K = 0.2 m, Angular speed = n = 5 r.p.s.**To Find:**Kinetic energy =?**Solution:**

I = MK^{2} = 2 x (0.2)^{2} = 2 x 0.04 = 0.08 kg m/s

Now, K.E. = ½ Iω^{2} = ½ I(2πn)^{2} = ½ x 0.08 x ( 2 x 3.142 x 5)^{2}

∴ K.E. = ½ Iω^{2} = ½ I(2πn)^{2} = 0.04 x ( 231..42)^{2 }= 39.49 J

**Ans:** Kinetic energy of disc is 39.49 J

#### Example – 08:

- A circular disc of mass 10 kg and radius 0.2 m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque 10 Nm. calculate the angular velocity of the disc at the end of 6 s from rest.
**Given:**Mass of disc = M = 10 kg; radius of disc = R = 0.2 m, Torque acting = τ = 10 Nm, initial angular velocity = ω_{1}= 0 rad/s, time = t = 6 s**To Find:**final angular velocity = ω_{2}= ?,**Solution:**

Moment of Inertia of a disk is given by

I = ½ MR^{2} = ½ (10)(0.2)^{2} = 5 x 0.04 = 0.2 kg m^{2}

Now, the torque acting on rotating body is given by

τ = I α

∴ 10 = 0.2 x α

∴ α = 10/0.2 = 50 rad/s^{2}

Now, ω_{2} = ω_{1}+ αt = 0 + 50 x 6 = 0 + 300 = 300 rad/s

**Ans: **The angular velocity of the disc at the end of 6 s from rest is 300 rad/s

#### Example – 09:

- A torque of 400 Nm acting on a body of mass 40 kg produces an angular acceleration of 20 rad/s
^{2}. Calculate the moment of inertia and radius of gyration of the body. **Given:**Mass of body = M = 40 kg; Torque acting = τ = 400 Nm, Angular acceleration = α = 20 rad/s^{2}**To Find:**Moment of inertia = I = ? and radius of gyration = K = ?**Solution:**

The torque acting on a rotating body is given by

τ = I α

∴ 400 = I x 20

∴ I = 400/20 = 20 kg m^{2}

Now, I = MK^{2}

∴ K^{2} = I/M = 20/40 = 0.5

∴ K = 0.707 m

**Ans: **Moment of inertia of the body is 20 kg m^{2} and the radius of gyration is 0.707 m.

#### Example – 10:

- If the radius of a solid sphere is doubled by keeping its mass constant, compare the moment of inertia about any diameter.
**Given:**For sphere R_{2}= 2R_{1}.**To Find:**Ratio of the moment of inertia = I_{1}/I_{2}= ?**Solution:**

The M.I. of a sphere in two cases about its diameter is given by

**Ans: **The ratio of the moment of inertia in two cases is 1:4

#### Example – 11:

- A flywheel in the form of a disc is rotating about an axis passing through its centre and perpendicular to its plane loses 100 J of energy, when slowing down from 60 r.p.m. to 30 r.p.m. Find the moment of inertia about the same axis and change in its angular momentum.
**Given:**lost in energy = 100 J. Initial angular speed = N_{1}= 60 r.p.m., final angular speed = N_{2}= 30 r.p.m.,**To Find:**Moment of inertia = I = ? and change in angular momentum = dL =?**Solution:**

ω_{1 }= 2πN_{1}/60 = 2π x 60/60 = 2π rad/s

ω_{2 }= 2πN_{2}/60 = 2π x 30/60 = π rad/s

Loss in K.E. = K.E._{1} – K.E._{2}

∴ Loss in K.E. = ½ Iω_{1}^{2} – ½ Iω_{2}^{2} = ½ I(ω_{1}^{2} – ω_{2}^{2})

∴ 100. = ½ I((2π)^{2} – (π)^{2})

∴ 200. = I(4π^{2} – π^{2})

∴ 200. = I(3π^{2})

∴ I = 200/3π^{2} = 6.753 kgm^{2}

Change in angular momentum = L_{2} – L_{1}

∴ dL = Iω_{1} – Iω_{2} = I(ω_{1} – ω_{2})

∴ dL = 6.753(2π – π) = 6.753 π = 6.753 x 3.142 = 21.22 kg m^{2}/s

**Ans:** Moment of inertia is 6.753 kg m^{2} and change in angular momentum is 21.2 kg m^{2}/s

#### Example – 12:

- A disc of diameter 50 cm and mass 2 kg rotates about an axis passing through its centre and at right angles to its plane with a frequency of 8 revolutions per sec. Find the angular momentum of the disc and the rotational K.E.
**Given:**Diameter of disc = D = 50 cm, Radius of disc = 50/2 = 25 cm = 0.25 m, Mass of disk = M = 2 kg, Number of revolutions per second = n = 8**To Find:**Angular momentum = L = ?, rotational kinetic energy = ?**Solution:**

ω_{ }= 2πn = 2π x 8 = 16π rad/s

Moment of Inertia of a disk is given by

I = ½ MR^{2} = ½ (2)(0.25)^{2} = 1 x 0.0625 = 0.0625 kg m^{2}

L = I ω = 0.0625 x 16π = 0.0625 x 16 x 3.142 = 3.142 kg m^{2}/s

K.E. = ½ Iω^{2} = ½ x 0.0625 x (16π)^{2} = ½ x 0.0625 x 256 x 3.142^{2}

K.E. = 78.98 J

**Ans:** Angular momentum is 3.142 kg m^{2}/s and rotational K.E. is 78.98 J.

#### Example – 13:

- A solid sphere of diameter 25 cm and mass 25 kg rotates about an axis through its centre. Calculate its moment of inertia if its angular velocity changes from 2 rad/s to 12 rad/s in 5 seconds. Also, calculate the torque applied.
**Given:**Diameter of sphere = D = 25 cm, Radius of sphere = 25/2 = 12.5 cm = 0.125 m, Mass of sphere = M = 25 kg, Initial angular speed = ω_{1}= 2 rad/s, final angular speed = ω_{2}= 12 rad/s,time = t = 5 s**To Find:**Moment of inertia = I = ?, Torque required = τ = ?**Solution:**

Moment of inertia of sphere is given by

**Ans:** Moment of inertia is 0.1562 kg m^{2}and torque acting is 0.3124 Nm

#### Example – 03:

- Assuming that the earth is a uniform homogeneous sphere of radius 6400 km and density 5500 kg/m
^{3}, calculate its M. I. about its axis of rotation. What is the rotational K. E. of the earth when spinning about its axis? **Given:**Radius of earth = 6400 km = 6.4 x 10^{6}m, Density = 5500 kg/m^{3}.**To Find:**M.I. of earth = ?, K.E. of the earth =?**Solution:**

**Ans:** Moment of inertia is 9.896 x 10^{37} kgm^{2} and kinetic energy is 2.612 x 10^{29} J

#### Example – 04:

- Assuming the earth to be a sphere, calculate its M. I. about the axis of rotation. Calculate the angular momentum and rotational K. E. of the earth about its axis. Mass of earth = 6 x 10
^{24}kg; R = 6400 km. **Given:**Mass of earth = M = 6 x 10^{24 }kg; radius of earth = R = 6400 km = 6.4 x 10^{6}m, Time period for earth = 24 hr = 24 x 60 x 60 s.**To Find:**Moment of inertia = I = ?, Angular momentum = L = ?, Kinetic energy = ?**Solution:**

.

**Ans : **Moment of inertia is 9.3 x 10^{37} kgm^{2} , angular momentum = 7.149 x 10^{33} kg m^{2}/s,

and kinetic energy is 2.612 x 10^{29} J

#### Example – 06:

- Find the ratio of the spin angular momentum of the earth to its orbital angular momentum. Distance of earth from the sun = 1.5 x 10
^{8}km; Radius of earth = 6400 km; Mass of earth = 6 x 10^{24 }kg. One year = 365 days.. - Given:
- Distance of earth from the sun = r = 1.5 x 10
^{8}km; Radius of earth = R =6400 km = 6.4 x 10^{6}m; Mass of earth = 6 x 10^{24 }kg, Time period for orbital motion = T_{o}= 365 days = 365 x 24 x 60 x 60 s, spin period = 24 hr = 24 x 60 x 60 s.

**Ans: **The ratio of the spin angular momentum of the earth to its orbital angular momentum is 2.658 x 10^{-7}.

Science > Physics > Rotational Motion > You are Here |

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