- Rotational Motion
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- Angular velocity
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- Angular acceleration
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- Rotation with constant angular acceleration
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- Kinetic energy of Rotation
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- Calculation of moment of inertia
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- Parallel Axis Theorem|Theorems of Moment of Inertia
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- Perpendicular Axis Theorem
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- Torque
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- work and power in rotational motion
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- Angular acceleration
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- Relationship between Angular momentum and torque
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- Conservation of Angular momentum
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- Radius of gyration
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- Rolling Motion|Kinetic Energy of rolling bodies
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- Rotational Motion problems with solutions

- Let us now understand the motion of rolling body. For this consider a body with circular symmetry for example cylinder, wheel, disc , sphere etc.

- When such a body rolls on a plane surface , the motion of such a body is a combination of translational motion and rotational motion as shown below in the figure.

- At any instant the axis normal to the digram through the point of contact P is the axis of rotation. If the speed of the centre of mass relative to an observer fixed on the surface is V
_{cm}then the instantaneous angular speed about an axis through P would be

ω=V_{cm}/R

where R is the radius of the body.

- To explain this consider the figure given below

At any instant different particles of the body have different linear speeds. The point P is at rest V_{cm}=0 instantaneously , the centre of mass has speed V_{cm}=Rω and the highest point on the circumference p' has speed V_{cm}=2Rω relative to point P.

- Now again consider the first figure the top of the cylinder has linear speed V
_{cm}+ Rω=V_{cm}+V_{cm}=2V_{cm}, which is greater than the linear speed of any other point on the cylinder. We thus note that the center of mass moves with linear speed V_{cm}while the contact point between the surface and rolling object has a linear speed of zero.

- Therefore at that instant all particles of the rigid body are moving with the same angular speed ω about the axis through P and the motion of the body is equivalent to pure rotational motion.

- Now let us calculate the kinetic energy of the rolling bodes
- From above paragraph, it is clear that at any instant all particles of the rigid body are moving with the same angular speed ω about the axis through P and the motion of the body is equivalent to pure rotational motion
- Thus total kinetic energy is

K=½(I_{P}ω^{2})

where I_{P}is the moment of inertia of the rigid body about point P.

- From parallel axis theorem

I_{P}=I_{cm}+MR^{2}

where I_{cm}is the moment of inertia of the body of mass M about parallel axis through point O.

Therefore

K=½(I_{cm}ω^{2})+½(MR^{2}ω^{2})=½(I_{cm}ω^{2})+½(M(V_{cm})^{2})

here the first term represents the rotational kinetic energy of the cylinder about its center of mass, and the second term represents the kinetic energy the cylinder would have if it were just translating through space without rotating. Thus, we can say that the total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.

- If k is the radius of gyration of the body about a parallel axis through O then I=Mk
^{2}and total kinetic energy would then be,

a. Up the incline while ascending and down the incline while descending

b.Up the incline while ascending and descending

c. down the incline while ascending and up the incline while descending

d.down the incline while ascending and descending

Imagine the cylinder to be moving on a frictionless surface.In both the cases the acceleration of the CM of the cylinder is gsinθ.This is also the acceleration of the point of contact

of the cylinder and the inclined plane..Also no torque (about the center of the cylinder) is acting on the cylinder since we assumed the surface to be a frictionless and the forces

acting on the cylinder is mg and N which passes through the center of the cylinder.Therefore the net movement of the point of contact in both the cases is in downward direction

Therefore frictional force will act in upward direction in both the cases

a. 28.6 m

b 30 m

c. 28 m

d. none of these

Let h be the height

The rotational and translational KE of the ball at the bottom will be changed to Gravitational energy when the sphere stops .

We therefore writes

(1/2)Mv

For a solid sphere I=(2/5)Mr

So (1/2)Mv

or

v

or h=28.6 m

a. (2/3)gsinθ

b.(2/3)gcosθ

c gsinθ

d none of these

Net force on the cylinder

F

or ma=mgsinθ -f

Where f is the frictional force

Now τ=fXR=Iα

Now in case of pure rolling we know that

a=αR => α=a/R

So f=Ia/R

From 1 and 2

a=mgsinθ /[m+(I/R

Now I=mR

So a=(2/3)gsinθ

Class 11 Maths Class 11 Physics Class 11 Chemistry

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