## Chapter Chosen

System of Particles and Rotational Motion

## Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

## Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.

Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, ω=10 π rad s–1

Coefficient of kinetic friction, μk = 0.2

Initial velocity of both the objects, u = 0

The frictional force causes motion between two objects.

As per Newton’s second law of motion,

Frictional force,
f =ma

μkmg= ma

where,

a = Acceleration produced in the objects

= Mass

Therefore,

a = μkg                                           … (i)

As per the first equation of motion,

Final velocity of the objects can be obtained as,

v = u + at

= 0 + μkgt

= μkgt                                             … (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.

Torque, τ= – I α

where,

α = Angular acceleration

μkmgr = –I α

α =                                    ...(iii)

Using the first equation of rotational motion to obtain the final angular speed,

ω = ω0 + α

= ω0 + (-μkmgr / I)t                          ...(iv)

Rolling starts when linear velocity,

v = rω

Therefore,

v = r (ω0 μkmgrt / I)                         ...(v)

Equating equations (ii) and (v), we get

μkgt = r (

= rω0 -                           ...(vi)

For the ring,

I = mr

∴ μkgt = rω0 -
= rω0 - μkgt

2μkgt = rω

∴    t =

=

=0.80 s                                       ...(vii)

For the disc:

I = mr2

∴    μkgt = rω0 - μkmgr2t / mr

= rω0 - 2μkgt

3μkgt = rω

Therefore,

t =

=       ...(viii)

Since td > tr, the disc will start rolling before the ring.
146 Views

Is it necessary that there should be matter at the centre of mass of system?

No, it is not necessary that there be matter at the centre of mass of the system.

For e.g., if two equal point masses are separated by certain distance, the centre of mass lies at the mid point of two point masses and there is no mass at that point.
730 Views

What is the significance of defining the center of mass of a system?

The motion of n particle system can be reduced to one particle motion.

An equivalent single point object would enable us to discuss the overall motion of the system.
794 Views

What is the need of centre of mass?

Newton’s second law of motion is strictly applicable to point masses only. To apply the Newton's law of motion to rigid bodies, the concept of centre of mass is introduced.

The concept of centre of mass of a system enables us to discuss overall motion of the system by replacing the system by an equivalent single point object.
911 Views

Is it necessary for centre of mass to lie within the body?

No, centre of mass needs not to lie within the body. It is not necessary that the total mass of the system be actually present at the centre.

The position of the centre of mass is calculated using the usual Newtonian type of equations of motion.
943 Views

Define centre of mass.

Centre of mass of a body or a system of bodies is a point at which the entire mass of the body or system is supposed to be concentrated.
1449 Views