`star` ROLLING MOTION

`star` KINETIC ENERGY OF ROLLING MOTION

`star` KINETIC ENERGY OF 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.

`●` 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.

`\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

`●` 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

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

`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?

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?

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?