NCERT Solution Class 11th Physics Chapter – 7 Systems of Particles and Rotational Motion Notes

NCERT Solution Class 11th Physics Chapter – 7 Systems of Particles and Rotational Motion

NCERT Solution Class 11th Physics Chapter – 7 Systems of Particles and Rotational Motion Notes Important Formulae discuss Systems of Particles and Rotational Motion Notes.

NCERT Solution Class 11th Physics Chapter – 7 Systems of Particles and Rotational Motion

Chapter – 7

Systems of Particles and Rotational Motion

Notes

C.M. of a body or a system may or may not lie inside the body.

The momentum of the C.M. of the system remains constant if the external force acting on it is zero.

C.M. of the system moves with a constant velocity if the external force on the system is zero.

Only the angular component of the force gives rise to torque.

Both angular momentum and torque are vector quantities.

The rotatory cum translatory motion of a ring, disc, cylinder, spherical shell, or solid sphere on a surface is called rolling.

The axis of rotation of the rolling body is parallel to the plane on which it rolls.

When the angular speed of all the particles of the rolling body is the same, it is called rolling without slipping.

The linear speed of different particles is different, although the angular speed is the same for all the particles.

K.E. is the same for all bodies having the same m, R, and ω.

Total energy and rotational kinetic energy are maximum for the ring and minimum for the solid sphere.

For ring K_r = K_t, K_r = 1/2 K_t for disc, K_r = 66%Kt for spherical shell and for solid sphere K_r = 40% of K_t.

The body rolls down the inclined plane without slipping only when the coefficient of limiting friction (p) bears the following relation:
µ ≥ (K²/K²+R²) tan θ

The relative values of p for rolling without slipping down the inclined plane are as follows :
μ_ring > μ_shell > μ_disc > μ_{solid sphere}

When a body roll Is without slipping, no work is done against friction.

A body may roll with slipping if friction is less than a particular value and it may roll without slipping if the friction is sufficient.

M.I. is not a scalar quantity because for the same body its values are different for different orientations of the axis of rotation.

M.I. is defined w.r.t. the axis of rotation.

M.I. is not a vector quantity because the clockwise or anticlockwise direction is not associated with it.

The radius of gyration depends on the mass and the position of the axes of rotation.

M.I. depends on the position of the axis of rotation.

The theorem of ⊥ar axes is applicable to thin laminae like a sheet, disc, ring, etc.

The theorem of || axes is applicable to all types of bodies.

M.I. about the axis in a particular direction is least when the axis of rotation passes through the C.M.

A pair of equal and opposite forces with different lines of action is known as a couple.

A body may be in partial equilibrium i.e. it may be in translational. equilibrium and not in rotational equilibrium or vice-versa.

If the sum of forces is zero, it is said to be in translational equilibrium. 0 If the sum of moments of forces about C.G. is zero then it is said to be in rotational equilibrium.

Important Formulae:
Position vector of C.M. of a system of two particles is
R_cm = m₁r₁+m₂r₂/m₁+m₂

Position vector of C.M of a system of two particles of equal masses is
R_cm = r₁+r₂/2

Torque acting on a particle is given by
τ = r × p

Angular momentum is given by
L = r × p
or
L = mv r = Iω = mr² ω

τ = dL/dt

τ = Iα

I₁ω₁ = I₂ω₂
or
I₁/T₁=I₂/T₂

K.E. of rotation, K_t = 1/2Iω²

Power in rotational motion, P = τω

According to theorem of perpendicular axes,
I_z = I_x + I_y

According to theorem of || axes, 1 = I_c + mh²

K.E. of a body rolling down an inclined plane is given by
E = 1/2mv² + 1/2 Iω² = K_t + K_r

K_r/K_t = ½Iω²/½mv² = ½ mK²ω²/ ½ mv²

= K²ω²/R²ω² = K²R²

K_rE = 1/2mK²ω²/1/2 m(R²+K²)ω² = K²/K²+R²

K_t/E = R²/R²+K²

If inclined plane is smooth, then the body will slide down and on reaching the bottom, its sliding velocity (V_s) is given by
V_s = √2gh and acceleration is a_s = g sin θ.

For rough inclined plane :
V_r = √2gh/√1+K²/R²

The acceleration of the body rolling down the inclined plane is
a_r = g sin θ √1+K²/R²

Time taken to reach the bottom is t_s = √2l/a_s and t_r = √2l/a_r

If a particle of mass m is moving along a circular path of radius r with acceleration ‘a’, then
τ = mr² α
Where α = a/r

The value of K²/R² for different bodies are as follows