Question

A wheel has moment of inertia 5 x 10^-3 kg m² and is making 20 revolutions per second. The magnitude of torque needed to stop it in 10s is:
1. 4 π x 10^-2 Nm
2. 4.5 π x 10^-2 Nm
3. 2 π x 10^-2 Nm
4. 2.5 π x 10^-2 Nm

Answer 1

Correct Answer - Option 3 : 2 π x 10^-2 Nm

Concept:

• Moment of Inertia: Moment of inertia is a property of an object related to its mass distribution. It plays the same role in rotational motion dynamics as played by mass in linear motion dynamics.
• Moment of inertia of a body r distance from the center of axis and having mass m is given as

I = mr2​

• Angular velocity (ω): The angular displacement per unit time gives angular velocity.
• Angular acceleration (α): The rate of change of angular velocity is called angular acceleration.
• Torque: The cross product of force and the distance of force from point of action is called torque.
  • Torque is also defined as the product of the moment of inertia and angular acceleration in the same way as force is defined as the product of mass and linear acceleration.

τ = Iα -- (1)

• First Equation of Motion for Rotational motion with constant torque applied is given as

ω_f = ω_i + αt -- (2)

Final angular speed ω_f, initial angular speed  ωi, α is the angular acceleration, t is time

Calculation: 

Given

Moment of inertia I =  5 x 10-3 kg m2 

torque τ  = 

Angular acceleration α 

time given t = 10 sec

Initial angular speed ω_i = 20 rev sec ^-1 = 40 π sec -1

(1 revolution = 2 π )

Final angular speed ω _f = 0 (it will be at rest)

0 = 20 + α (10)

⇒ α = - 2 rev  sec -2

Now using eqution (1)

Torque τ = 5 x 10-3 kg m2 × 2 rev  sec -2 = 5 x 10-3 kg m2 ×  40 π  sec -2 

⇒ τ = 2 π x 10-2 Nm

So, the correct option is 2 π x 10-2 Nm