Question

Moment of inertia of a wheel is 1000 kgm2. At a given instant, its angular velocity is 10 rad/sec. After the wheel rotates through an angle of 100 radians, the angular velocity of the wheel is 100 rad/sec. The torque applied on wheel in nm is

Answer 1

Torque= 4.95 × 10^4 N. m

_______________________________________________________  

And We know that,

ω² - ω1² = 2 ∝ Ф

So therefore,

∝ = ω² - ω1²/2Ф

∝ = (100)² - (10)²/2 * 100

∝ = 49.5 rad/s²

T=I * α  

T = 1000 *  49.5

T = 4.95 × 10^4 N. m

Answer 2

Answer:

The torque applied on wheel is  4.95 * 10⁴ Nm

Step-by-step explanation :

Given,

Moment of inertia = 1000 kg m²

Angular velocity 1 ( ω₁ ) = 10 rad / sec

Angle = 100 radians

Angular velocity 2 ( ω₂ ) = 100 rad / sec

Since we know,

ω₂² - ω₁² = 2 α Ф

where,

ω₁ is angular velocity 1

ω₂ is is angular velocity 2

α is angular acceleration

Ф is the angle formed

α = ω₂² - ω₁² / 2 Ф

Substituting the values,

α = (100 ) ² - ( 10 ) ² / 2 ( 100 )

α = (10000)  - ( 100 )  / 200

α = 49.5 rad/s²

Now we know,

T = I * α

where

T is the torques

I is the Inertia

α  is the angular acceleration

Substituting the values,

T = 1000 *  49.5

T = 4.95 * 10⁴ Nm

Hence,

The torque applied on wheel is  4.95 * 10⁴ Nm