Physics |
Chemistry |
Biology |
Mathematics |

Science > Physics > Rotational Motion > You are Here |

#### Example – 1:

- A thin uniform rod of length 1 m and mass 1 kg is rotating about an axis passing through its centre and perpendicular to its length. Calculate the moment of inertia and radius of gyration of the rod about an axis passing through a point midway between the centre and its edge perpendicular to its length.
**Solution:****Given:**Mass of rod = M = 1 kg, length of rod =*l*= 1m.**To****Find:**M.I. = I =? and radius of gyration = K =? at h =*l*/4

Moment of inertia of a thin uniform rod about a transverse axis

passing through its centre is given by

Now I = MK^{2}

K^{2 }= I/M = 0.1458/1 = 0.1458

K = 0.3818 m

**Ans:** M.I. about the required axis is 0.1458 kg m^{2}, and radius of gyration is0.3818 m

#### Example – 2:

- Calculate the M. I. of a thin uniform rod of mass 100g and length 60 cm about an axis perpendicular to its length and passing through (1) its centre and (2) one end.
**Solution:****Given:**Mass of rod = M = 100 g = 0.1 kg, length of rod =*l*= 60 cm = 0.6 m.- Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is given by

I = M*l ^{2}*/12 = 0.1 x (0.6)

^{2}/12 = 0.003 kg m

^{2}

- Moment of inertia of the thin uniform rod about a transverse axis passing through its end is given by

I = M*l ^{2}*/3 = 0.1 x (0.6)

^{2}/3 = 0.012 kg m

^{2}

**Ans:** Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is 0.003 kg m^{2 }and the moment of inertia of the thin uniform rod about a transverse axis passing through its end is 0.012 kg m^{2}.

#### Example – 3:

- Find the M. I. of a thin uniform rod 1 m long and weighing 0.024 kg about a transverse axis passing through (1) its centre (2) one end (3) a point 20 cm from one end.
**Solution:****Given:**Mass of rod = M = 0.024 kg, length of rod =*l*= 1 m.- Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is given by

I = M*l ^{2}*/12 = 0.024 x (1)

^{2}/12 = 0.002 kg m

^{2}

- Moment of inertia of the thin uniform rod about a transverse axis passing through its end is given by

I = M*l ^{2}*/3 = 0.024 x (1)

^{2}/3 = 0.008 kg m

^{2}

- Moment of inertia of the thin uniform rod about a transverse axis passing through a point 20 cm from one end.

**Ans: **Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is 0.002 kg m^{2}, Moment of inertia of the thin uniform rod about a transverse axis passing through its end is 0.008 kg m^{2}, Moment of inertia of the thin uniform rod about a transverse axis passing through a point 20 cm from one end is 4.16 x 10-3 kg m^{2}

#### Example – 4:

- A thin ring has mass 0.25 kg and radius 0.5 m. Calculate the moment of inertia about i) an axis passing through its centre and perpendicular to the plane and ii) an axis passing through a point on its circumference, perpendicular to its plane iii) diameter. iii) an axis tangent to ring and its plane
**Solution:****Given :**Mass of Ring = M = 0.25 kg, Radius of ring = R = 0.5 m

Moment of inertia about a transverse axis passing through its centre is given by

I = MR^{2} = 0.25 x (0.5)^{2} = 0.25 x 0.25 = 0.0625 kgm^{2}

Moment of inertia about a transverse axis passing through its circumference is given by

I = 2MR^{2} = 2 x 0.25 x (0.5)^{2} = 2 x 0.25 x 0.25 = 0.125 kgm^{2}

Moment of inertia about its diameter is given by

I = MR^{2}/2 = 0.25 x (0.5)^{2}/2 = 0.25 x 0.25/2 = 0.03125 kgm^{2}

Moment of inertia about an axis tangent to the ring and its plane is given by

I = 3/2MR^{2} = 3/2 x (0.25 x (0.5)^{2}) = 3/2 x (0.25 x 0.25) = 0.09375 kgm^{2}

Ans: Moment of inertia about a transverse axis passing through its centre is 0.0625 kgm^{2}

Moment of inertia about a transverse axis passing through its circumference is 0.125 kgm^{2}

Moment of inertia about its diameter is 0.03125 kgm^{2}

Moment of inertia about an axis tangent to the ring and its plane is 0.09375 kgm^{2}

#### Example – 5:

- Calculate the moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation coinciding with its diameter and tangent perpendicular to its plane.
**Solution:****Given:**Mass of Ring = M = 500 g = 0.5 kg, Radius of ring = R = 0.5 m

Moment of inertia about its diameter is given by

I = MR^{2}/2 = 0.5 x (0.5)^{2}/2 = 0.5 x 0.25/2 = 0.0625 kgm^{2}

Moment of inertia about an axis tangent to ring and perpendicular to its plane is given by

I = 2MR^{2} = 2 x 0.5 x (0.5)^{2} = 0.25 kgm^{2}

Moment of inertia about its diameter is 0.0625 kgm^{2}

Moment of inertia about an axis tangent to ring and perpendicular to its plane is 0.0625 kgm^{2}

^{ }Example – 6:

- Calculate the M.I. of a disc of mass 0.5 kg and radius 10 cm about an axis passing through its centre and at right angles to its plane. What would be the M. I. if the axis passes through a point at a distance from the centre equal to half the radius of the disc?
**Solution:****Given :**Mass of disc = M = 0.5 kg, Radius of disc = R = 10 cm = 0.1 m

Moment of inertia about a transverse axis passing through its centre is given by

I = MR^{2}/2 = 0.5 x (0.1)^{2}/2 = 0.25 x 0.25 = 0.0025 kgm^{2 } = 2.5 x 10^{-3} kg m^{2}

By parallel axes theorem

Ans: M.I. about the axis passes through a point at a distance

from the centre equal to half the radius of the disc is 3.75 x 10^{-3} kg m^{2}

**Example – 7:**

- Calculate the M. I. of a solid brass sphere of radius 12 cm about a tangent to the sphere. Density of brass = 8500 kg/m
^{3} **Solution:****Given:**Radius of sphere = R = 12 cm = 0.12 m, density = r = 8500 kg/m^{3}

Moment of inertia of solid sphere about its tangent is given by

#### Example – 8:

- A solid sphere has a radius ‘R’. If its radius of gyration of this sphere about its diameter is √2/5R. show that the radius of gyration about the tangential axis of rotation is √7/5R.

By parallel axes theorem

#### Example – 9:

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

Moment of inertia of solid sphere about its diameter is given by

**Ans:** The ratio of initial M.I. to the Final M.I. is 1:4

#### Example – 10:

- A solid cylinder of uniform density of radius 2 cm has a mass of 50 g. If its length is 10 cm, calculate the moment of inertia about i) its own axis of rotation passing through its centre ii) an axis passing through its centre and perpendicular to the length.
**Solution:****Given:**Mass of cylinder = M = 50 g = 0.05 Kg, Radius of cylinder = R = 2 cm = 0.02 m, length of cylinder =*l*= 10 cm = 0.1 m**To Find: Moment Of Inertia =?**

I = MR^{2}/2 = 0.05 x (0.02)^{2}/2 = 0.05 x 0.0004/2 = 10^{-5} kgm^{2}

- Moment of inertia of the cylinder about an axis passing through its centre and perpendicular to the length

**Ans:** Moment of inertia of the cylinder about its own axis passing through its centre is 10^{-5} kgm^{2}

Moment of inertia of the cylinder about an axis passing through its centre and perpendicular to the length is 4.67 x 10^{-5} kgm^{2}

#### Example – 11:

- 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 torque applied.
**Solution:****Given:**Diameter of sphere = 25 cm, Radius of sphere = R = 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 seconds.**To Find:**Moment of inertia = I =? and torque = τ = ?

Moment of inertia of solid sphere about its diameter by

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

#### Example -12:

- 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.
**Solution:****Given:**Mass of disc = M = 10 kg, Radius of Disc = R = 0.2 m, Torque applied = t = 10 Nm, Initial angular speed w_{1}= 0 rad/s, time = t = 6 s.**To Find:**final angular speed = ω_{2}=?

Moment of inertia of disc about an axis passing through its centre

and perpendicular to its plane is given by

**Ans:** Final angular speed is 300 rad/s

#### Example – 13:

- 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. **Solution:****Given:**Torque acting = τ = 400 Nm, Mass = M = 40 kg, angular acceleration = α = 20 rad/s^{2}.**To Find:**Moment of Inertia =?, Radius of gyration =?

Torque acting on body is given by

τ = I α

I = τ/α = 400/20 = 20 kg m^{2}.

Now the moment of inertia is given by

I = MK^{2}

K^{2} = I/M = 20/40 = 0.5

K = √50

K = 0.707 m

**Ans:** The moment of inertia is 20 kg m^{2} and the radius of gyration is 0.707 m

#### Example – 14:

- The M.I. of a uniform circular disc about an axis passing through its centre and perpendicular to its plane is ½ MR
^{2}. Find the distance of a parallel axis from the centre of mass about which the M.I. of the disc is MR^{2}. Radius of disc = √2 cm.

By parallel axes theorem

**Ans: **The distance of the axis is 1 cm

#### Example – 15:

- A disc has a radius of gyration of 0.02 m when rotating about an axis passing through its centre and at right angles to its plane. What would be its radius of gyration when rotating about an axis coincident with a diameter of its face?
**Solution:****Given:**Radius of gyration = K = 0.02 m for axis passing through centre.**To Find:**Radius of gyration = K = ? for axis passing through diametre

The M.I. in the first Case

M.I. of the disc about diameter is

**Ans: **The radius of gyration is 1.414 x 10^{-2} m

#### Example = 16:

- Radius of gyration of a body about an axis at a distance of 0.12 m from its centre of mass is 0.13 m. Find its radius of gyration about a parallel axis through the centre of mass.
**Solution:****Given:**Distance between axes = h = 0.12 m, radius of gyration = K = 0.13 m**To Find:**Radius of gyration = K = ? for axis passing through centre.

**Ans: **Radius of gyration is 0.05 m for axis passing through centre.

#### Example – 17:

- A homogeneous rod XY of length L and mass M is pivoted at the centre C such that it can rotate freely in the vertical plane. Initially, the rod is in a horizontal position. A bob of wax of mass M falls vertically on the rods midway between point C and Y. If the rod rotates with angular speed w what will be the angular speed in terms of V and L?

Science > Physics > Rotational Motion > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |