Question

A wheel has moment of inertia 2.5 kgm² has an initial angular velocity of 40 rad/s. A constant torque of 10N acts on the wheel. The time during which the wheel is accelerated to 60 rad/s is:
1. 4s
2. 5s
3. 6s
4. 2.5s

Answer 1

Correct Answer - Option 2 : 5s

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 

• Equations of Motion under rotational motion is written in similar way as that of linear motion, by keeping acceleration as angular acceleration, velocity as angular velocity, and displacement as angular displacement.

 ω_v = ω _u + α t

ω u is initial angular velocity, ω _v is final angular velocity

Calculation: 

Given, moment of inertia I = 2.5 kgm2

Torque τ = 10 N m

Angular acceleration \(\alpha = \frac{\tau }{I}\) 

\(\implies \alpha = \frac{10}{2.5} = 4 rad / s^2\)

initial angular velocity ω u = 40 rad/s.

Final angular velocity ω v = 60 rad/s.

ωv = ω u + α t

⇒ 60 = 40 + 4t

\(\implies t = \frac{60 - 40}{4} = 5 \hspace{0.1cm} s\)

So, the correct option is 5 second.