Physics Definition And Proof Based Problems

### Definition And Proof Based Problems

Definition And Proof Based Problems
Q 3114134959

(i) Define moment of inertia. Write the parallel and perpendicular axis theorem.
(ii) Derive an expression for moment of inertia of a disc of radius r, mass m about an axis along its diameter.

Solution:

(i) Moment of inertia is inertial equivalent in rotational motion. Moment of inertia of a rigid body is defined as the sum of the products of the constituent masses and the squares of the perpendicular distance from the axis of rotation. If m_1 m_2, ..... m_n, are the masses at perpendicular distances r_l> r_2, ..r_n then, moment of inertia

I = m_1r_(1)^(2) + m_2r_(2)^(2) + .... + m_n r_(n)^(2)

sum_(i=1)^(n) m_ir_(i)^(2)

It is measured in kgm^2 and has the
dimensions of ML^2

"Parallel Axis theorem"
The moment of inertia about an axis passing parallel to the axisthrough the centre of mass of a rigid body is the sum of the moment of inertia (I_(cm) of the body about the axis through centre of mass and the product of its mass M and square of the separation (a^2) between the parallel axis.

i.e I = I_(cm)+Ma^2

"Perpendicular Axis theorem"
If I_x and I_y are the moment of inertia about the x and y-axis of any rigid
mass, then the moment of inertia I.
about z-axis is the sum of I_x and I_r

(ii) Moment of in tertia of a uniform circular disc about its diameter : Using the theorem of perpendicular axes, we get I_d + I_d = Moment of inertia of the disc about perpendicular axis yoy'

i.e , 2I_d = 1/2 MR^2

or I_d = 1/4 MR^2
Q 3164534455

Establish a relation between angular momentum and moment of inertia of a rigid body. Define moment of inertia in terms of it.

Solution:

We know L = r x p r x mu
Since u = rw,
we have L = r x nromega) = .r2omega = Iomega
Q 3154634554

Define a Rigid Body. Name two kinds of motion which a rigid body can execute. What is the meant by the term Equilibrium ? For the equilibrium of a body, two conditions need to be satisfied. State them.

Solution:

Rigid body is one in which the separation between any two constituent masses remains constant. It can execute translatory and rotational motion. Equilibrium identifies the stability of any body. For equilibrium,

(i) sumf = 0 (i.e.,) the total unbalanced force should be zero.
(ii) sumr = 0 (i.e.,) the net torque acting on the body should be zero.
Q 3104834758

Define moment of inertia. Write any two factors on which it depends. When the diver leaves the diving boarp, why does he bring his hand and feet closer together in order to make a somersault ?

Solution:

Moment of inertia of a body about a given axis is defined as the sum of the products of masses of all the particles of body and squares of their respective perpendicular distances from axis of rotation. Moment of inertia depends on
(i) Distribution of Mass
(ii) Orientation of axis of rotation.
Diver does so, so that moment of inertia I of
his body decreases. As angular momentum
(I m) remains constant, therefore, angular
velocity w of his body increases.
Q 3134623552

Prove that the torque acting due to a force F in the xy plane is, t_z = x F_y- y F_x.

Solution:

We know vect = vecr xxvecF
If vecr = xhati + yhatj +zhatk and vecF = F_xhati + F_Zhatk we have

t = t_xhati + t_y hatj +t_zhatk= | ((hati,hatj, hatk), (x, y , z) (F_x,F_y F_z))|

= hati (yF_z_zF_y) -hatj(x(F_z-zF_x)+hatk(xF_y_yF_x)

Comparing the coefficients of i,] and k, we have

t_2 = xF_y-yF_x
Q 3174623556

State and prove the law of conservation of angular momentum.

Solution:

Torque vect = vecr xxvecF = vecr xxvec(dp)/(dt) (vecr xxvecp) = vec(dl)/(dt)

If no external torque acts on a body, , vect= 0

then vec(dL)/(dt) = 0 or vecL is conserved.
Q 3124112951

Prove the Kepler's law, that the line joining the sun and the planet sweeps equal areas in equal time, using the angular momentum conservation with the planet.

Solution:

When the planet moves along the line joining the sun and the planet it sweeps some area given by by A = 1/2r^2theta where theta is the angular displacement.

:. (dA)/(dt)=1/2r^2(dtheta)/(dt) = 1/2 r^2omega

(dA)/(dt) = 1/2m mr^2omega= L/(2m)
Since no torque acts, angular momentum L

is a constant so (dA)/(dt) is a constant i.e the line joining the sun and the planet sweeps equal intervals of time
Q 3104723658

State and prove the parallel axis theorem.

Solution:

The moment of inertia about an axis parallel to the axis through the centre of mass is the sum of the moment of inertia about the axis through centre of mass and the product of the mass M and the square of the separation between the axes, i.e.,
I= MI_(cm)+Ma^2

Moment of inertia is the product of mass and the square of the separation from the axis. The moment of inertia of the system of n masses shown in the a figure about the parallel is

I= m_1 (x_1+a)^2 + m_2 (a-x_2)^2 +m_3 (x_3-a)^2

+ .........for n masses.

In genereal, I = sum(i-1)^(n) m (x_x_i+a_i)^2

if all the n masses are considered equal.

I= sum(i-1)^(n) mx_(i)^(2) + sum(i-1)^(n) ma^(2)+2asum(i-1)^(n) mx_i

I= I_(cm)+Ma^2 :. summx_i = 0 if the centre of mass is considered as origain
Q 3154112954

What is torque ? Write its formula in vector form. Handle, to open the door, is always provided at the free edge of a door. Why ?

Solution:

Torque is the moment of force. It is written as the cross-product of the position vector of the point of application of the force and the force, i.e., vect = vecr xxvecF

the angle between vecr and vecF sin theta, where theta is It is measured in Nm and has dimensions torque is the product of the perpendicular distance between the axis of rotation and point of application of force in line with the force. It is the cause of rotational motion. To open the door, handle is to be rotated. So we need a torque. With less force one can exert more torque if the handle is at the edge.
Q 3114823750

State and prove the perpendicular axis theorem.

Solution:

According to perpendicular axis theorem, the sum of the moment of inertia about x and y axes is equal to the moment of inertia about z-axis. The mass m has co-ordinates (x, y).

The moment of inertia about x-axis,

I_x = my^2
about y - axis, I_y = mx^2
I_x+I_y=m(x^2+y^2)
= M(sqrt(X^2+Y^2)^2

I_x+L_y= M(r distance from z-axis))^2

I_x+i_y+i_2
Q 3174112956

What is torque ? Write its formula in vector form. Handle, to open the door, is always provided at the free edge of a door. Why ?
Hots
Solution:

Angular momentum L = I omega
Rotational K.E., E_k = Iomega^2

E_k = 1/2 ((Iomega)^2)/(I) = (L^2)/(2I)
Q 3154212154

What is the torque provided by a force acting through the centre of mass of a sphere ?

Solution:

Zero. Since r = r.LF and r.L = 0 for all points on the axis.
Q 3154312254

A projectile fired into the air suddenly explodes into several fragments. What can you say about the motion of the fragments after the collision ?

Solution:

No external force acts. The centre of mass will follow its original path with every particle scattered.
Q 3174834756

Prove that the rate of change of the angular momentum of particle is equal to the torque acting on it.

Solution:

Torque rotating a particle in xy plane is

t = xE_y - yF_x..(I)

P_x = mv_x and P_y = mv_y are x, y component
of linear momentum of body.

According to Newton's 2nd law of motion.

F_x= (dP_x)/(dt) = (d)/(dt)(mvb_x) = (mdv_x)/(dt)

F_y = (dP_Y)/(dt) = (d)/(dt) (mv_y) = (mdvy)/(dt)

Substituting in equation (i), we get

t = xm(dv_y)/(dt) - ym (dv_x)/(dt)

t = m[x(dv_y)/(dt) - y (dv_x)/(dt)]

t = m (d)/(dt) (xv_y- yv_x)

t = (d)/(dt) (xmv_y - ymmu_x)

t = (d)/(dt) (xmv_y- ymu_x)

t = (d)/(dt) (xP_y- yP_x)

Put (xP_y - yP_x) = L

t = vec(dL)/(dt)
Q 3154723654

What is the analogue of mass in rotational motion ? Derive the expression for the kinetic energy of a rotating body.

Solution:

Moment of inertia is the analogue of mass in rotational motion.
Let the body consists of particles of masses
m_1 m_2, m_3 at perpendicular distances r_1
r_2, r_3 respectively from axis of rotation.
If v1_ v_2, v_3 . are the respective linear velocities
of particles, then

u_1 = r_1 omega, v_2 = r_2omega, v_3 = r_2omega

K.E of mass m_1 is 1/2 m_1v_(1)^(2) = 1/2 m_1 (r_1omega)^2

Similarly K.E of other particles of the body are

1/2 m_2r_(2)^(2)omega^2 , 1/2 m_3r_(3)^(2) omega^2

= 1/2 |sum_(i=1)^(i=m) m_i r_(i)^(2) |omega^2

put summ_ir_(i)^(2) = I

K.E of rotation = 1/2 Iomega^2
Q 3124123051

Derive the relations

(i) L = Iomega (ii) t = Ialpha

Solution:

(i) To prove L= Iomega
We know,

L= r xx p = r xx mv

(ii) To prove t= Ialpha

We know t = r F = rma
= mr^2alpha= Ialpha

vect = vecrxxvecF
Q 3134034852

Derive an expression for moment of inertia of a thin circular ring passing through the centre and perpendicular to the plane of the ring.

Solution:

Consider a ring of mass M and radius r. Divide the ring into large number of small segments, each of length dl. For every segment dl, the moment of inertia is, m'r2 where m' is its mass.

m' = (M)/(2pir) dl

Moments of inertia I ' = (M)/(2pir) dl
Net moment of inertia for the ring about the perpendicular axis

I = int I'dl = int_(0)^(2pir) (M)/(2pir) M/(2pir)r^2dl
I = int I'dl = int_(0)^(2pir) dl = (Mr)/(2pi) |l|_(0)^(2pi) = Mr^2