ROTATIONAL MOTION

6 )

From a uniform circular disc of radius R, a circular disc of radius R/6 and having centre at a distance + R/2 from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc.

[(â€“ R) / 70]

[(+ R) / 70]

[(â€“ R) / 7]

[(+ R) / 7]

View Solution

Answer :

Let mass per unit area of disc = m

Mass of disc = M = πR² ∙ m

Mass of removed disc = M' = π(R / 6)² ∙ m = [(πR²m) / 36]

from figure

M × 0 = M' × (R / 2) + (M – M') x --- here point P is new centre of mass at

distance x

O = [(M'R) / 2] + (M – M')x

x(M – M') = [(– M'R) / 2]

x = [(– M'R) / {2(M – M')]

= [{{(– πR²m) / (36)} ∙ R} / {2m[πR² – {(πR²) / (36)}}]

= [{(– πR²mR) / {{(36 × 2m × {(35πR²) / (36)}}]

= [(– R) / (70)]

hence centre of mass [(– R) / 70]

7 )

A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from + ve x edge of the plate. Find the position of centre of mass of the remaining portion with respect to centre of mass of whole plate.

â€“ 7 cm

+ 9 cm

â€“ 9 cm

+ 7 cm

View Solution

Answer :

Let mass of plate per unit area = m

Mass of whole Plate = M = π(56 / 2)² m = π(28)²m

Mass of removed part = M₁ = π(42 / 2)² m = π(21)²m

Mass of remaining Part = M – M₁

C.M of whole disc = O i.e at origin

C.M of removed Plate = r₁ = 28 – 21 = 7cm

C.M of remaining Portion = r₂

hence

M × O = M₁r₁ + (M – M₁)r₂

M₁r₁ = – (m – m₁)r₂

r₂ = [(M₁r₁) / {– (m – m₁)}]

= [{π(21)²m} / {– {π(28)² – π(21)²}}m] × 7

= [(21)² / (– 343)] × 7

r₂ = – 9 cm

8 )

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is :

30 m/s

20 m/s

10 m/s

15 m/s

View Solution

Answer :

The Velocity of centre of mass = V_cm = [(m₁v₁ + m₂v₂) / (m₁ + m₂)]

= [{(10)(14) + (4)(0)} / (14)] = 10 m/s

9 )

A particle performing uniform circular motion has angular momentum L., its angular frequency is doubled and its K.E. halved, then the new angular momentum is :

L/2

L/4

View Solution

Answer :

E = (1 / 2)Iω² = (1 / 2)Iω ∙ ω = (1 / 2)Lω . . . . . . as L = Iω

∴ L = (2E / ω)

i. e. (L₁ / L₂) = (E₁ / ω₁) ∙ (ω₂ / E₂)

given E₂ = (1/2) E₁

ω₂ = 2 ω₁

(L₁ / L₂) = 2 × 2 = 4

L₂ = (L₁ / 4) I. e. L₂ = (L/4)

10 )

A circular disc of radius R is removed from a bigger disc of radius 2R. such that the circumferences of the disc coincide. The centre of mass of the remaining portion is αR from the centre of mass of the bigger disc. The value of α is.

[(â€“ 1) / 3]

1/6

1/4

1/2

View Solution

Answer :

Let m be mass per unit area

M = mass of big disc = π(2R)²m

M₁ = mass of removed disc = πR²m

M₂ = M – M₁ = Mass of remaining Portion

Let G be the Central of mass of remaining Portion

M× O = M₁(OO') + (M – M₁)(OG)

M₁ ∙ R = – (M – M₁) (∝R)

M₁ = – α(M – M₁)

πR²m = – α(π ∙ 4R²m – πR²m)

R² = – α(3R²)

α = [(– 1) / 3]

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