Find the Kinetic energy ,
When Relative Force achieve on rolling ball up to 3 cm from 1 cm , Is the constant is 3 Joule (J) Acting Ball of Mass 500 g Relative Force were is universally proportional .
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Answers
Explanation:
By the end of this section, you will be able to:
Describe the differences between rotational and translational kinetic energy
Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis
Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy
Use conservation of mechanical energy to analyze systems undergoing both rotation and translation
Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces
So far in this chapter, we have been working with rotational kinematics: the description of motion for a rotating rigid body with a fixed axis of rotation. In this section, we define two new quantities that are helpful for analyzing properties of rotating objects: moment of inertia and rotational kinetic energy. With these properties defined, we will have two important tools we need for analyzing rotational dynamics.
Rotational Kinetic Energy
Any moving object has kinetic energy. We know how to calculate this for a body undergoing translational motion, but how about for a rigid body undergoing rotation? This might seem complicated because each point on the rigid body has a different velocity. However, we can make use of angular velocity—which is the same for the entire rigid body—to express the kinetic energy for a rotating object. (Figure) shows an example of a very energetic rotating body: an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are generated as the grindstone does its work. This system has considerable energy, some of it in the form of heat, light, sound, and vibration. However, most of this energy is in the form of rotational kinetic energy.
Figure 10.17 The rotational kinetic energy of the grindstone is converted to heat, light, sound, and vibration. (credit: Zachary David Bell, US Navy)
Energy in rotational motion is not a new form of energy; rather, it is the energy associated with rotational motion, the same as kinetic energy in translational motion. However, because kinetic energy is given by

, and velocity is a quantity that is different for every point on a rotating body about an axis, it makes sense to find a way to write kinetic energy in terms of the variable

, which is the same for all points on a rigid rotating body. For a single particle rotating around a fixed axis, this is straightforward to calculate. We can relate the angular velocity to the magnitude of the translational velocity using the relation

, where r is the distance of the particle from the axis of rotation and

is its tangential speed. Substituting into the equation for kinetic energy, we find

In the case of a rigid rotating body, we can divide up any body into a large number of smaller masses, each with a mass

and distance to the axis of rotation

, such that the total mass of the body is equal to the sum of the individual masses:

. Each smaller mass has tangential speed

, where we have dropped the subscript t for the moment. The total kinetic energy of the rigid rotating body is
and since
for all masses,
The units of (Figure) are joules (J). The equation in this form is complete, but awkward; we need to find a way to generalize it.
Moment of Inertia
If we compare (Figure) to the way we wrote kinetic energy in Work and Kinetic Energy,
, this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. The quantity
is the counterpart for mass in the equation for rotational kinetic energy. This is an important new term for rotational motion. This quantity is called the moment of inertia I, with units
For now, we leave the expression in summation form, representing the moment of inertia of a system of point particles rotating about a fixed axis. We note that the moment of inertia of a single point particle about a fixed axis is simpl
, with r being the distance from the point particle to the axis of rotation. In the next section, we explore the integral form of this equation, which can be used to calculate the moment of inertia of some regular-shaped rigid bodies.
The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. Similarly, the greater the moment of inertia of a rigid body or system of particles, the greater is its resistance to change in angular velocity about a fixed axis of rotation. It is interesting to see how the moment of inertia varies with r, the distance to the
Energy in rotational motion is not a new form of energy; rather, it is the energy associated with rotational motion, the same as kinetic energy in translational motion. However, because kinetic energy is given by
\[K=\frac{1}{2}m{v}^{2}\]
, and velocity is a quantity that is different for every point on a rotating body about an axis, it makes sense to find a way to write kinetic energy in terms of the variable
\[\omega\]
, which is the same for all points on a rigid rotating body. For a single particle rotating around a fixed axis, this is straightforward to calculate. We can relate the angular velocity to the magnitude of the translational velocity using the relation
\[{v}_{\text{t}}=\omega r\]
, where r is the distance of the particle from the axis of rotation and
\[{v}_{\text{t}}\]
is its tangential speed. Substituting into the equation for kinetic energy, we find
\[K=\frac{1}{2}m{v}_{\text{t}}^{2}=\frac{1}{2}m{(\omega r)}^{2}=\frac{1}{2}(m{r}^{2}){\omega }^{2}.\]