# 11th Class Physics Chapter 5 Circular Motion Short Question Answers

## 11th Class Physics Chapter 5 Circular Motion Short Question Answers Below

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_{t}.Angular velocity:The angular velocity is rate of change of angular displacement of an object moving along a circle.It is vector quantity and denoted by its direction is found out by right hand rule stated as follow:

Curl the fingers of right and around the rotation axis in the direction of rotation then the thumb points towards the direction of angular velocity

^{ɷ→}is represented by a line drawn parallel to the axis of rotation.

I =mr

^{2}…………….. (1)

If the rigid body is made of n-particles then different particles are at different distances from axis of rotation.Thus:

I =m

_{1}r

_{2}+m

_{2}r

_{2}

^{2}……………m

_{n}r

^{2}

_{n}

Where m

_{1}m

_{2}+……………are masses of n particles at distances r

_{1}r

_{2}…………r

_{n }from axis of rotation respectively.So equation (1) can be written as

I=

^{n}∑

_{i=1 }m

_{i }

**r**

_{i}^{2}…………..(2)L

_{ᵒ}

^{→}.=r

^{→}×p

^{→}

Its magnitude is given by

L

_{ᵒ}=rp sinΘ……………. (1)

where Θ the angle between position vecteor r

^{→}and velocity of linear momentum p

^{→}

Equation (1) can be written as :

L

_{ᵒ}=mrv sinΘ ………. (2)

In case of circular orbital motion such as satellite of mass m’

revolving around the Earth in a circle of radius r the angle between radius ‘r’ and tangential velocity ‘v’ is always 90

^{ᵒ}.Therefore

L

_{ᵒ}=mrv sin90

^{ᵒ }Hence L

_{ᵒ}=mrv

Direction of Angular Momentum:-

As we know that teh angular momentum is defined by

L^{→}r^{→}×p^{→}

The angular momentum is a vector quantity.The direction of angular momentum L^{→}is perpendicular to the plane containing r^{→ }and p^{→} as show in fig (b)

Its direction is determined by right hand rule.If an object is rotating along certain axis then direction of L^{→} will be along axis of rotation,the angular momentum is directed outward along the axis of rotation.

Direction of Angular velocity:Angular velocity is a vector quantity.Its direction also can be found by the right hand rule.Curl the fingers of the right hand around the rotation axis in the direction of rotation then thumb points towards the direction of angular velocity ‘w’as shown in fig.

(b) Whenever a frame of reference of free falling object is moving under gravity the weight of the body in that frame of reference will be zero.In other words it will be weightless.As the relative acceleration of an inside body with respect to its frame of reference is zero because both are falling together with the same acceleration.Therefore it appears weightless.

Hence if the satellite is in free fall all the objects inside it appear to be weightless.

_{1}about an axis.Thus his angular velocity ɷ

_{1}decreases.When he pulls his legs and arms into closed tuck position his moment of inertia is reduced to a new value I

_{2}.So the value of his angular velocity ɷ

_{2}increases.As the angular momentum is conserved so

I

_{1}ɷ

_{1}=I

_{2}ɷ

_{2}= constant

Hence the diver spins faster when moment of inertia becomes smaller and angular velocity increases to conserve angular momentum.In this way he can make more somersaults.

Work =F

^{→}.v

^{→}=Fv cos90

^{ᵒ}=0

Thus the work done by the centripetal force on the body is zero.

When crow takes away a piece of bread the mass of bread on the rotating plate decreases and its moment of inertia also decreases.But since the angular velocity of the plate increases that is

I_{1}ɷ_{1}= I_{2}ɷ_{2}

W-T=F=ma

or T=W-ma

The apparent weight is given by

W

^{/}= T=W-ma

=mg-ma

W

^{/}=m(g-a)……………..(1)

When the chain supporting the elevator breaks the elevator falls freely with an acceleration to ‘g’

Thus a=g

Putting it in equation we get

W

^{/}=T=m(g-g)=0

W

^{/}=0

Hence the apparent weights of both the bodies inside the elevator is zero or the bodies seem to be weightless.It is the state of weightless.

S=r Θ

This equation is valid only if angle Θ is measured in radians not in degrees r.As the angle used in S.I unit is in radians

No the centripetal force does not do any work because this force is always acting at right angle to the motion of the body or to the velocity of the body along the tangent.

Therefore, work=F^{→}.v^{→}=Fv cos 90=0

Therefore the work done by the centripetal force is zero.

If the body of mass ‘m’is taken to the Moon its weight is given W_{m}=mg_{m} =acceleration due to gravity of Moon.It has been found that acceleration due to gravity of Moon ‘gm’ is one sixth of the value of acceleration due to gravity of earth ‘g’i.e.

gm=g/6

Therefore the weight of the body at the Moon will reduce 1/6 of its value on earth.

g=0

but W=mg

Here g =0

W=m×0=0

or W=0

Hence the weight of a body becomes zero at the center of the earth.

Suppose the elevator is descending with an acceleration ‘a’ it means that weight ‘w’ is greater than the tension ‘T’ on the body.Net force acting on the body =W-t=F

Or W-T=ma

Or T=W-ma =mg-ma=m(g-a)

Thus T=m(g-a).

It is defined as the product of mass of a particle and square of its perpendicular distance from the axis of rotation.It is denoted by I.

Mathematically it can be written as

I=mr^{2}

Where m is the mass of the particle and r is the perpendicular distance of particle from axis of rotation.

Its Uniy: The SI unit of moment of inertia is kgm^{2}.

A spaceship in an orbit is a freely falling bidy having an acceleration ‘a’ towards the center of Earth equal to the value of the acceleration due to gravity ‘g’ at the radius of the orbit.The apparent weight W^{/} is expressed as

W^{/}=W-ma

Where W is its actual weight thus

W^{/}=mg-ma=m(g-a)

But in the case of free fall,a=g

Therefore W^{/} =m (g-g)=0

Hence W^{/}=0

Thus the spaceship will be called weightless with the condition of falling freely with an acceleration a=g.

The energy due to the spinning of a body about its axis is called rotational kinetic energy.

Mathematically it can be written as

K.E._{max}=1/2 Iɷ^{2}

Practical use of Rotational K.E.:This energy is put to practical use by flywheels which are essential parts of many engines.A flywheel stores energy between the power strokes of the pistons so that the energy is distributed over the full revolution of the crankshaft and hence the relation remains smooth.

^{→}is along the radius and velocity v

^{→}along tangent.Thus they become perpendicular to each other.It means that acceleration a

^{→}and velocity v

^{→}have different directions.

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