Chapter 7: System of Particles and Rotational Motion

The axis of the door is in the y-axis and the door is in the x-y plane. The door rotates in the positive as well as negative z-axis as the force is applied in the z-axis. The gravity acting on the door is in the direction of –y-axis. Therefore, the weight of the door cannot rotate the door along the y-axis.

Given,

Where,

r_cm is the place where mass m is placed at the nth vertex

r_cm = 0

(n-1)mb + ma = 0

b = -ma/(n-1)m

Where,
is the position vector. Also, the negative sign indicates that the centre of mass lies on the other side of the nth vertex.

Let M be the mass of the half-disc

Area of the half-disc = πR2/2

Mass per unit area m = 2M/πR2

a) when the disc is half, the centre of mass is (0, 4R/3π)

b) when the disc is quarter, the centre of mass is (4R/3π, 4R/3π)

a) The law of conservation of angular momentum can be applied as there is no external torque on the system and the gravitational and its normal reaction towards the external forces have net torque zero.

b) Using law of conservation of angular momentum,

L_f = L_i

I₁ω₁ + I₂ ω₂

ω = I₁ ω₁+ I₂ ω₂/I₁ + I₂

c) Final kinetic energy = rotational + translation energy

K_f = KE_R +KE_T

∆K = -I₁I₂/2(I₁ +I₂)( ω₁ – ω₂)2 < 0

d) K_f < K_i as there is lose in energy during friction between the moving surfaces.

a) The disc was in rotational motion before bringing it in contact with the table and v_cm = 0 was the rest.

b) There is a decrease in the linear velocity of the rotating disc when it is brought in contact with the table as there is a force of friction acting on the disc.

c) According to Newton’s third law, when the rotating disc is brought in contract with the table, the linear velocity of the disc changes.

d) The figure below explains the effects of b) and c). Frictional force is responsible.

e) v_cm = ω₀R is the condition that needs to be satisfied.

f) For rolling motion to take place without the slipping of the disc, the frictional force is responsible.

a) We know that v₁ = ωR

v₂ = ω²R

From the figure we can see that the direction of v₁ and v₂ are tangentially upwards and meet at point C. Therefore, the frictional forces on the surfaces is f₁₂ = -f₂₁.

b) The external forces f₁₂ and f₂₁ are equal and opposite and is given as

f₁₂ = -f₂₁

f₁₂ + f₂₁ = 0

|f₁₂| = |-f₂₁| = F

Therefore, the external torque is given as

(F)(3R)

So, it can be said that the velocity of the drum got doubled that is v₂ = 2v₁.

c) The anticlockwise and clockwise angular velocities of the drum are ω₁ and ω₂ respectively. When the velocities are equal, there is no force of friction and it is given as

ω₁/ ω₂ = 2/1

Area of square = area of rectangle

a) c² = ab since I = mr²

Because c > b we can say that c² > b2

Therefore, I_xR/I_xS < 1

b)

Therefore, I_yR/I_yS > 1

c) I_zR – I_zS = m(d_R/2)² – m(d_S/2)²

Solving the above equation we can prove that I_zR/I_zS > 1

Let a and α be the linear and angular acceleration respectively. Then

F – f = Ma

Where M is the mass of the disc

f is the force of friction which is applied at the centre O

Torque to disc, τ = I_D α

Moment of inertia of the disc, I_D = 1/2 MR²

fR = 1/2 MR².a/R

Ma = 2f

F – f = 2f

3f = F

f = F/3

F = 3μMg, which is the maximum force applied on the disc to roll it on the surface without slipping.