Definitions [1]
Define torque and mention its unit.
Torque is defined as the moment of the externally applied force about a point or axis of rotation. The expression for torque is, `vectau = vecr xx vecF`.
Theorems and Laws [1]
State and prove the theorem of the parallel axis about the moment of inertia.
A body's moment of inertia along an axis is equal to the product of two things: Its moment of inertia about a parallel axis through its centre of mass and the product of the body's mass and the square of the distance between the two axes. This is known as the parallel axis theorem.
Proof: Let ICM represent a body of mass M moment of inertia (MI) about an axis passing through its centre of mass C and let I stand for that body's MI about a parallel axis passing through any point O. Let h represent the separation of the two axis.
Think about the body's minuscule mass element dm at point P. It is perpendicular to the rotation axis through point C and to the parallel axis through point O, with a corresponding perpendicular distance of OP. CP2 dm is the MI of the element about the axis through C. As a result, `I_(CM) = int CP^2 dm` is the body's MI about the axis through the CM. In a similar vein, `I = int OP^2 dm` is the body's MI about the parallel axis through O.
Draw PQ perpendicular to OC produced, as shown in the figure. Then, from the figure,
`I = int OP^2 dm`
= `int (OQ^2 + PQ^2) dm`
= `int [(OC + CQ)^2 + PQ^2] dm`
= `int (OC^2 + 2OC.CQ + CQ^2 + PQ^2) dm`
= `int (OC^2 + 2OC.CQ + CP^2)dm` ...(∵ CQ2 + PQ2 = CP2)
= `int OC^2 dm + int 2OC.CQ dm + int CP^2 dm`
= `OC^2 int dm + 2OC int CQ dm + int CP^2 dm`
Since OC = h is constant and `int dm = M` is the mass of the body,
`I = Mh^2 + 2h int CQ dm + I_(CM)`
The integral `int CQ dm` now yields mass M times a coordinate of the CM with respect to the origin C, based on the concept of the centre of mass. This position and the integral are both zero because C is the CM in and of itself.
∴ I = ICM + Mh2
This proves the theorem of the parallel axis.
Important Questions [27]
- Obtain an Expression for Total Kinetic Energy of a Rolling Body
- The Radius of Gyration of a Body About an Axis, at a Distance of 0.4 M from Its Centre of Mass is 0.5 M. Find Its Radius of Gyration About a Parallel Axis Passing Through Its Centre of Mass.
- Define Radius of Gyration. Write Its Physical Significance.
- The Kinetic Energy of a Rotating Body Depends Upon
- If ‘L’ is the Angular Momentum and ‘I’ is the Moment of Inertia of a Rotating Body, Then L^2/(2i)
- The Kinetic Energy of Emitted Photoelectorns is Independent of ............ .
- The Body is Rotating with Uniform Angular Velocity (W) Having Rotational Kinetic Energy (E). Its Angular Momentum (L) is
- A Solid Sphere of Mass 1 Kg Rolls on a Table with Linear Speed 2 M/S, Find Its Total Kinetic Energy.
- Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping. Hence find kinetic energy for a solid sphere.
- If a Rigid Body of Radius ‘R’ Starts from Rest and Rolls Down an Inclined Plane of Inclination ‘θ’ Then Linear Acceleration of Body Rolling Down the Plane is
- A Stone of Mass 2 Kg is Whirled in a Horizontal Circle Attached at the End of 1.5m Long String. If the String Makes an Angle of 30° with Vertical, Compute Its Period
- State the Theorem of Parallel Axes About Moment of Inertia.
- A solid cylinder of uniform density of radius 2 cm has mass of 50 g. If its length is 12 cm, calculate its moment of inertia about an axis passing through its centre and perpendicular to its length.
- A Body Starts Rotating from Rest. Due to a Couple of 20 Nm It Completes 60 Revolutions in One Minute. Find the Moment of Inertia of the Body.
- Explain the Physical Significance of Radius of Gyration
- A uniform solid sphere has radius 0.2 m and density 8 x 10^3 kg/m^3. Find the moment of inertia about the tangent to its surface.
- A Body of Moment of Inertia 5 Kgm2 Rotating with an Angular Velocity 6 Rad/S Has the Same Kinetic Energy as a Mass of 20 Kg Moving with a Velocity of
- A Thin Wire of Length L and Uniform Linear Mass Density R is Bent into a Circular Coil. M. I. of the Coil About Tangential Axis in Its Plane is
- A Uniform Solid Sphere Has a Radius 0.1 M and Density 6 X 103 Kg/M3• Find Its Moment of Inertia About a Tangent to Its Surface.
- A Ballet Dancer Spins About a Vertical Axis at 2.5\u0001 Rad/S with His Both Arms Outstretched. with the Arms Folded, the Moment of Inertia About the Same Axis of Rotation Changes by 25%. Calculate the New Speed of Rotation in R.P.M
- A Thin Ring Has Mass 0.25 Kg and Radius 0.5 M. Its Moment of Inertia About an Axis Passing Through Its Centre and Perpendicular to Its Plane is _______.
- The Moment of Inertia of a Thin Uniform Rod of Mass M And Length L, About an ·Axis Passing Through a Point,Midway Between the Centre and One End, Perpendicular To Its Length is . (A)`48/7ml^2`
- A wheel of moment of inertia 1 Kgm2 is rotating at a speed of 40 rad/s. Due to friction on the axis, the wheel comes to rest in 10 minutes.
- State and prove the theorem of the parallel axis about the moment of inertia.
- State an Expression for the Moment of Intertia of a Solid Uniform Disc, Rotating About an Axis Passing Through Its Centre, Perpendicular to Its Plane.
- Prove the Theorem of Parallel Axes About Moment of Inertia
- State the Theorem of Perpendicular Axes About Moment of Inertia.
