Question

A disc of moment of inertia 1.5 kilogram metre square is initially at rest. it is acted upon by a constant torque of 120 Newton metre for 5 second. find the final angular momentum and kinetic energy of the disc​

Answer 1

$\underline{\boxed{ \huge\purple{ \rm{Answer}}}}$

Given :

$\rm \: moment \: of \: inertia = 1.5 \: kg {m}^{2} \\ \rm \: torque = 120 \: Nm \\ \rm \: time = 5 \: s$

To Find :

$\rm \: final \: angular \: momentum \: and \: kinetic \: energy.$

Formula :

$\rm \: angular \: momentum \: \red{L = I \omega} \\ \\ \rm \: kinetic \: energy = \blue{\frac{1}{2} I { \omega}^{2} }$

Calculation :

$\implies \rm \: \tau = I \alpha = I( \frac{ \omega}{t}) \\ \\ \implies \rm \: \omega = \frac{ \tau{t}}{I} = \frac{120 \times 5}{1.5} = 400 \: \frac{rad}{s} \\ \\ \implies \rm \: L = i \omega = 1.5 \times 400 = \green{600 \: kg \frac{ {m}^{2} }{s} } \\ \\ \implies \rm \: KE = \frac{1}{2} I { \omega}^{2} = \frac{1}{2} \times 1.5 \times {400}^{2} = \orange{120 \: KJ}$

Answer 2

Answer:

• Angular Momentum (L) of the disc is 600 Kg-m² / sec.
• Kinetic Energy (K.E) of the disc is 120 Kilo Joules.

Given:

1. Moment of Inertia (I) = 1.5 Kg-m².
2. Torque (τ) = 120 N-m
3. Time Period (t) = 5 Seconds.

Explanation:

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From the relation we know,

⇒ τ = I α

Where,

• τ Denotes Torque.
• I Denotes Moment of inertia.
• α denotes Angular Acceleration.

Now,

⇒ τ = I α

Substituting the values,

⇒ 120 = 1.5 × α

⇒ α = 120 / 1.5

⇒ α = 1200 / 15

⇒ α = 80

⇒ ω / t = 80   ∵ [ α = ω / t ]

⇒ ω / 5 = 80

⇒ ω = 80 × 5

⇒ ω = 400

⇒ ω = 400 rad/sec ___[ 1 ]

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From the relation we know,

⇒ L = I ω

Where,

• L Denotes Angular momentum.
• I Denotes Moment of inertia.
• ω denotes Angular velocity.

Now,

⇒ L = I ω

Substituting the values,

⇒ L = 1.5 Kg-m² × 400 rad/sec  [ From Equation - 1 ]

⇒ L = 1.5 × 400

⇒ L = 15 × 40

⇒ L = 600

⇒ L = 600 Kg-m² / sec.

∴ Angular Momentum of the disc is 600 Kg-m² / sec.

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From the relation we know,

⇒ K.E = 1 / 2 I ω²

Where,

• K.E Denotes Kinetic Energy.
• I Denotes Moment of inertia.
• ω denotes Angular velocity.

Now,

⇒ K.E = 1 / 2 I ω²

Substituting the values,

⇒ K.E = 1 / 2 × 1.5 × ( 400 )²   [ From Equation - 1 ]

⇒ K.E = 1 / 2 × 1.5 × 160000

⇒ K.E = 1.5 × 80000

⇒ K.E = 15 × 8000

⇒ K.E = 120000

⇒ K.E = 120 × 10³

⇒ K.E = 120 K J

∴ Kinetic Energy (K.E) of the disc is 120 Kilo Joules.

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