Obtain an expression for total kinetic energy in terms of radius of gyration of the body

No, if you think about it, if that ball has a radius of 2m. So when the ball is touching the ground, it's center of mass will actually still be 2m from the ground. It's true that the center of mass is initially 6m from the ground, but when the ball falls and touches the ground the center of mass is again still 2m from the ground. That means the height will be 4m. If you work the problem where the height is 6m, the ball would have to fall halfway through the floor for the center of mass to be at 0 height. I have a question regarding this topic but it may not be in the video. Suppose a ball is rolling without slipping on a surface( with friction) at a constant linear velocity. We know that there is friction which prevents the ball from slipping. Why doesn't this frictional force act as a torque and speed up the ball as well?The force is present. It can act as a torque. Also consider the case where an external force is tugging the ball along. In this case, my book (Barron's) says that friction provides torque in order to keep up with the linear acceleration. In the first case, where there's a constant velocity and 0 acceleration, why doesn't friction provide torque? Can someone please clarify this to me as soon as possible? Thanks a lot!! #APphysicsCMechanics If the ball is rolling without slipping at a constant velocity, the point of contact has no tendency to slip against the surface and therefore, there is no friction. If the ball were skidding and rolling, there would hav...

\( \newcommand\) • • • • • • • • • • • • Learning Objectives By the end of this section, you will be able to: • Explain how the work-energy theorem leads to an expression for the relativistic kinetic energy of an object • Show how the relativistic energy relates to the classical kinetic energy, and sets a limit on the speed of any object with mass • Describe how the total energy of a particle is related to its mass and velocity • Explain how relativity relates to energy-mass equivalence, and some of the practical implications of energy-mass equivalence The tokamak in Figure \(\PageIndex\): The National Spherical Torus Experiment (NSTX) is a fusion reactor in which hydrogen isotopes undergo fusion to produce helium. In this process, a relatively small mass of fuel is converted into a large amount of energy. (credit: Princeton Plasma Physics Laboratory) Conservation of energy is one of the most important laws in physics. Not only does energy have many important forms, but each form can be converted to any other. We know that classically, the total amount of energy in a system remains constant. Relativistically, energy is still conserved, but energy-mass equivalence must now be taken into account, for example, in the reactions that occur within a nuclear reactor. Relativistic energy is intentionally defined so that it is conserved in all inertial frames, just as is the case for relativistic momentum. As a consequence, several fundamental quantities are related in ways not kno...

Rotational Motion of a Rigid Body Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. A rigid body is an object with a mass that holds a rigid shape, such as a phonograph turntable, in contrast to the sun, which is a ball of gas. Many of the equations for the mechanics of rotating objects are similar to the motion equations for linear motion. Angular velocity and angular acceleration The angular displacement of a rotating wheel is the angle between the radius at the beginning and the end of a given time interval. The SI units are radians. The average angular velocity (ω, Greek letter omega), measured in radians per second, is The angular acceleration (α, Greek letter alpha) has the same form as the linear quantity and is measured in radians/second/second or rad/s 2. The kinematics equations for rotational motion at constant angular acceleration are Consider a wheel rolling without slipping in a straight line. The forward displacement of the wheel is equal to the linear displacement of a point fixed on the rim. As can be shown in Figure , d = S = rθ Figure 1 A wheel rolling without slipping. In this case, the average forward speed of the wheel is v = d/ t = ( rθ)/ t = rω, where r is the distance from the center of rotation to the point of the calculated velocity. The direction of the velocity is tangent to the path of the point of rotation. The average forward acceleration of the wheel is a T = r(ω f − ω o)/...

A uniform circular disc of radius R lies in the X Y plane with the centre coinciding with the origin. The moment of inertia about an axis passing through a point on the X axis at a distance x = 2 R and perpendicular to the X Y plane is equal to its moment of inertia about an axis passing through a point on the Y axis at a distance y = d and parallel to the X axis in the X-Y plane. The value of d is:

Let M and R be the mass and radius of the body, V is the translation speed, ωis the angular speed and I is the moment of inertia of the body about an axis passing through the centre of mass. Kinetic energy of rotation,`E_R=1/2MV^2` Kinetic energy of translation,`E_r=1/2Iomega^2` Thus, the total kinetic energy 'E' of the rolling body is E = E R + E r `=1/2MV^2+1/2Iomega^2` `=1/2MV^2+1/2MK^2omega^2 ......(I=MK^2 " and K is the radius of gyration")` `=1/2MR^2omega^2+1/2MK^2omega^2 ........(V=Romega)` `:.E=1/2Momega^2(R^2+K^2)` `:.E=1/2MV^2/R^2(R^2+K^2)` `E=1/2MV^2(1+K^2/R^2)` Hence proved.

\( \newcommand\) • \(\require In engineering design, the radius of gyration is used to determine the stiffness of structural columns and estimate the critical load which will initiate column buckling. Question 10.6.1. How are \(k_x\text