ROTATIONAL MOTION-RPS-05
1. Ans: () 2rad
L = I𝜔 = (1.0 kg m2 ) ( )
Power, 𝑃 = 𝜏 × 𝜔 s
𝑃 1200 2 kg m2
So, 𝜏=𝜔= = 2𝑁−𝑚 =
600 s
2. Ans: () 8. Ans: ()
60
f = 60rpm = rps
60
=1 Since torque of string on stone is zero, the angular
I = 2 kg − m 2 momentum will be conserved
𝜔 = 𝜔0 + 𝛼t 9. Ans: ()
0 = 2𝜋 + 𝛼(60)
2𝜋 𝜋 2𝜋 𝜋
𝛼 = − 60 = − 30 ∴ (Retardation) 𝜔1 = = rad/s.
10 5
𝜏 = I𝛼 By conservation of Angular momentum
𝜋 𝜋
𝜏 = 2 × 30 = 15
I1 𝜔1 = I2 𝜔2
3. Ans: () 𝜋
⇒ 100 ( ) = [100 + 50(2)2 ]𝜔
I = mR2 = 10(0.2)2 = 0.4 5
1200×2𝜋 𝜋 2𝜋
𝜔= 𝑟𝑎𝑑/sec. ⇒ 𝜔= or
60 15 30
ω = 40π rad/s 10. Ans: ()
Angular Momentum L = I ω = 16π J-s
≈ 50.25 J-s
4. Ans: ()
𝑓 = 0.5
𝜔 = 2𝜋𝑓 = 𝜋 Initially as ant move towards axis I decreases so
𝐿 = 𝐼𝜔 w increases. After crossing axis I increases and w
𝑚2
= 0.6𝜋 𝑘𝑔 × decreases. so, first w increases then w decreases.
𝑆
5. Angular momentum is an axial vector 11. As the angular momentum is conserved
6. Relation between, kinetic energy (K) and angular ∴ I1ω1 – I2 ω2
momentum (L) is :- Or, mr22 ω = m(r/2)2 ω2 (∵ ω1 = 2ω)
𝐿 = √2𝐼𝐾 {where, I = moment of inertia} ω2 = 8 ω
𝐿1 𝐼 12. If torque is absent, Angular momentum will be
So, = √𝐼1 = 1: √2
𝐿2 2 conserved
7. Ans: () 𝐿2 1
𝐸 = 2𝐼 ⇒𝐸∝ 𝐼
Here, Mass, M = 1.0 kg
I∝m&I∝ K2
Diameter, D = 2.0 m
𝐼 𝑚 𝐾 2 1
∴ Radius, R =
D
= 1.0 m So 𝐼2 = 𝑚2 × (𝐾2 ) = 2 ⇒ 𝐸2 = 2 𝐸1 = 0.5𝐸
2 1 1 1
The moment of inertia of the body, 13. Ans: ()
1 1
I = MR2 = (1.0 kg)(1.0 m)2 = 1.0 kg m2 As, 𝐾 = 2 𝐼𝜔2 = 2 (𝐼 × 𝜔) × 𝜔
1 2𝐾
⇒ 𝐾 = 2 𝐿𝜔 or 𝐿= 𝜔
The angular velocity of the body,
𝐿2 𝐾2 𝜔1 1 1
10 So, = 𝐾 ×𝜔 = 2×2
𝜔 = 2𝜋𝑣 = 2 × 3.14 × 31.4 rad/s 𝐿1 1 2
𝐿2 1 𝐿
= 2rad/s ⇒ =4 or 𝐿2 = 4
𝐿1
The angular momentum of the body, 14. Ans: ()
𝑚 𝑚
𝐼𝑐𝑚 = 𝜇ℓ2 = 𝑚 1+𝑚2 ℓ2
1 2
15. Ans: ()
Wall = ∆KE
𝑚2 ℓ 𝑚1 ℓ 12 𝐾2
𝑟1 = 𝑚 , 𝑟2 = 𝑚 ⇒ 𝑊 = 0 − 2 𝑚𝑣𝑐𝑚 (1 + 𝑅2 )
1 +𝑚2 1 +𝑚2
𝑚 𝑚 ⇒ W = - 3J
𝐼𝑐𝑚 = 𝑚1 𝑟12 + 𝑚2 𝑟22 = 𝑚 1+𝑚2 ℓ2
1 2
OR
