Physics ROLLING MOTION AND KINETIC ENERGY OF ROLLING MOTION

### Topic covered

star ROLLING MOTION
star KINETIC ENERGY OF ROLLING MOTION

### ROLLING MOTION

● All wheels used in transportation have rolling motion. For specificness we shall begin with the case of a disc, but the result will apply to any rolling body rolling on a level surface. We shall assume that the disc rolls without slipping.
● This means that at any instant of time the bottom of the disc which is in contact with the surface is at rest on the surface.

We have remarked earlier that rolling motion is a combination of rotation and translation. We know that the translational motion of a system of particles is the motion of its centre of mass.

=> Let v_(cm) be the velocity of the centre of mass and therefore the translational velocity of the disc. Since the centre of mass of the rolling disc is at its geometric centre C (Fig. 7. 37), v_(cm) is the velocity of C.
● It is parallel to the level surface. The rotational motion of the disc is about its symmetry axis, which passes through C.
● Thus, the velocity of any point of the disc, like P_0, P_1 or P_2, consists of two parts, one is the translational velocity v_(cm) and the other is the linear velocity v_r on account of rotation. The magnitude of v_r is v_r = rω, where ω is the angular velocity of the rotation of the disc about the axis and r is the distance of the point from the axis (i.e. from C).

● The velocity v_r is directed perpendicular to the radius vector of the given point with respect to C. In Fig. 7.37, the velocity of the point P_2 (v_2) and its components v_r and v_(cm) are shown; v_r here is perpendicular to CP_2 . It is easy to show that v_z is perpendicular to the line P_OP_2.

● Therefore the line passing through P_O and parallel to ω is called the instantaneous axis of rotation.

● At P_o, the linear velocity, v_r, due to rotation is directed exactly opposite to the translational velocity v_(cm). Further the magnitude of v_r here is Rω, where R is the radius of the disc. The condition that P_o is instantaneously at rest requires v_(cm) = R_ω. Thus for the disc the condition for rolling without slipping is ..................... (7.47)

\color{green} ✍️ Incidentally, this means that the velocity of point P_1 at the top of the disc (v_1) has a magnitude v_(cm)+ R_ω or 2 v_(cm) and is directed parallel to the level surface. The condition (7.47) applies to all rolling bodies.

### Kinetic Energy of Rolling Motion

\color{green} ✍️ The kinetic energy of a rolling body can be separated into kinetic energy of translation and kinetic energy of rotation. ● This is a special case of a general result for a system of particles, according to which the kinetic energy of a system of particles (K) can be separated into the kinetic energy of motion of the centre of mass (translation) (MV^2//2) and kinetic energy of rotational motion about the centre of mass of the system of particles (K′ ). Thus,

color{blue}{ K = K′ + MV ^2 //2 ....................(7.48)}

● We assume this general result (see Exercise 7.31), and apply it to the case of rolling motion. In our notation, the kinetic energy of the centre of mass, i.e., the kinetic energy of translation, of the rolling body is mv_( cm)^2 //2, where m is the mass of the body and v_(cm) is the centre of the mass velocity.
● Since the motion of the rolling body about the centre of mass is rotation, K′ represents the kinetic energy of rotation of the body; K′ = Iω^2 //2 , where I is the moment of inertia about the appropriate axis, which is the symmetry axis of the rolling body. The kinetic energy of a rolling body, therefore, is given by

\color{red} ✍️  \ \ \ \ color{blue}{K= 1/2 I omega^2+1/2 mv_(cm)^2...................(7.49 a)}

● Substituting I = mk^2 where k = the corresponding radius of gyration of the body and v_(cm)= R ω, we get

K= 1/2 (mk^2 v_(cm)^2)/(R^2) +1/2 mv_(cm)^2

or

 \ \ \ \ color{blue}{K=1.2 mv_(cm)^2 (1+ k^2/R^2)}

..............(7.49 b)

Equation (7.49b) applies to any rolling body: a disc, a cylinder, a ring or a sphere.
Q 3260756615

Three bodies, a ring, a solid cylinder and a solid sphere roll down the
same inclined plane without slipping. They start from rest. The radii of the bodies are
identical. Which of the bodies reaches the ground with maximum velocity?

Solution:

We assume conservation of energy of the rolling body, i.e. there is no loss of energy
due to friction etc. The potential energy lost by the body in rolling down the inclined plane
color{blue}{(= mgh)} must, therefore, be equal to kinetic energy gained. (See Fig.7.38) Since the bodies
start from rest the kinetic energy gained is equal to the final kinetic energy of the bodies. From Eq. (7.49b),

color{blue}{K= 1/2 mv^2 (1+ k^2/R^2)} where v is the final velocity of (the centre of mass of) the body.
Equating color{blue}{K} and color{blue}{mgh},

color{blue}{mgh= 1/2 mv^2 (1+ k^2/R^2)}

or color{blue}{v^2 =((2gh)/(1+k^2//R^2))}

Note is independent of the mass of the rolling body;

For a ring, color{blue}{k^2 = R^2}

color{blue}{v_("ring")= sqrt((2gh)/(1+1))}

color{blue}{= sqrt(gh)}

For a solid cylinder color{blue}{k^2 = R^2//2}

color{blue}{v_("disc")= sqrt((2gh)/(1+1//2))

color{blue}{= sqrt((4gh)/3)}

For a solid sphere color{blue}{k^2 = 2R^2//5}

color{blue}{v_("sphere")= sqrt((2gh)/(1+2//5))}

color{blue}{= sqrt((10 gh)/7)}

From the results obtained it is clear that among the three bodies the sphere has the greatest and
the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
Suppose the bodies have the same mass. Which body has the greatest rotational kinetic energy while
reaching the bottom of the inclined plane?