The article discusses the relation between torque and moment of inertia of the rotating body and its solved problems.

**The torque and moment of inertia maintain the body under rotatory motion. When torque is induced on the body, it begins accelerating inversely proportional to its moment of inertia. That’s why the torque yielded on the body is the product of its moment of inertia and angular acceleration. **

**Newton’s laws of motion **express that the *body stays stationary or moves from one point to another with a distinct velocity; unless any external force acts on it*. **That means the body obtains the acceleration depending on its total mass and the strength of the applied external force. **

Employing Newton’s principle in rotational motion, when torque or moment of force is generated on the body at rest or moving, it initiates accelerating angularly. That’s why every rigid body executes rotational motion about its axis bears an angular acceleration when torque is induced.

We have understood in the previous articles that* inertia is the property of the body, which represents the body’s tendency to oppose the motion*. **That’s why inertia is inversely proportional to the acceleration of the body**. Consequently, Newton’s first law of motion has also termed the **Law of Inertia.**

Each particle within such a rotating body has its masses, and they all revolve about the body’s center axis of rotation. Hence, the amount of torque required to accelerate the particles within the body depends on the mass distribution of the whole body. **The quantity of the body that expresses the mass distribution is called the ‘ moment of inertia’. **

In rotational motion, the inertia quantity is considered as the moment of inertia of a body, determined by integrating the total masses M of the particles and their distances R from its axis of rotation.

Therefore, **the body’s moment of inertia (I) is** I = MR^{2}.

**Torque and Moment of Inertia Relation**

Newton’s laws of motion relate the torque and moment of inertia in rotational motion.

When we switch on the fan, we induce a torque to it. Now the acceleration of the fan will depend on how much the moment of inertia the fan has and how much torque we need to induce.

The moment of inertia is the rotational mass of the body, whereas torque is the rotational force functioning on it. The torque τ needed to be induced on the body is proportional to both angular acceleration and moment of inertia. But the moment of inertia I decreases the angular acceleration α of the body.

**The statement yields a relation between the torque and moment inertia as**,

τ=1α

**Read about Torque and Angular Momentum**

**How much torque is required to rotate at 15 rad/s**^{2} for the body to have a moment of inertia of 5 kgm^{2}?

^{2}for the body to have a moment of inertia of 5 kgm

^{2}?

** Given**:

I = 5 kgm^{2}

α = 15 rad/s^{2}

** To Find**: τ =?

** Formula**:

τ = I α

** Solution**:

The torque required to induce the body to rotate it at 15 rad/s^{2} is calculated using **the relation between torque and moment of inertia**,

τ= I α

Substituting all values,

τ = 5 x 15

τ= 75

**The torque required to rotate the body is 75 Nm.**

**How much torque must be induced at 2m on a body of 4kg to rotate at 5 rad/s**^{2}?

^{2}?

** Given**:

M = 4 kg

R = 2 m

α = 5 rad/s^{2}

** To Find: **τ =?

** Formula**:

τ = I α

** Solution**:

The torque required for the body is calculated as,

τ = Iα

But the moment of inertia of the disc is, I = MR^{2}.

τ = MR^{2}α

Substituting all values,

τ = 4 x 2^{2} x 5

τ = 4 x 4 x 5

τ = 80

**The torque required to rotate the body is 80 Nm.**

**Torque and Moment of Inertia Formula**

The torque and moment of inertia formula are easily determined by substituting the linear equivalent in Newton’s law of motion formula with its angular equivalent.

**The torque τ is the angular equivalent to the applied force F, and the moment of inertia I is the angular equivalent of the mass m. Hence, Newton’s law of motion (F = ma) for rotational motion becomes,**

τ = Iα

**How to Find Torque from the Moment of Inertia**

The torque from the moment of inertia is derived from the change in the angular momentum.

**The rotational motion also follows Newton’s laws of motion. So when torque works on the body, its angular momentum varies due to acceleration. Since the angular momentum is the product of inertia and angular velocity, we can derive the needed torque from its moment of inertia.**

The** angular**** momentum** of the body when torque is induced is given by L = r x P

Where P is **linear momentum**. i.e., P = mv

L = r x mv

**The relation between linear velocity v and angular velocity ω is** (r x ω )

L = r x m (r x ω)

L = mr^{2}ω

But mr^{2} term is thebody’s **moment of inertia (I).**

L = I ω

We have learned that **the torque induced on the body is the rate of angular momentum change**.

τ = dL/dt

Substituting** angular momentum formula**,

τ = dIω/dt

τ = Idω/dt

The term dω/dt is the** angular acceleration** α of the body. i.e., α= dω/dt

**Finally, the torque is obtained from the moment of inertia is, **

τ = lα

**The disc of 0.1kg having a radius of 1m is rotating at 2 rad/s**^{2.} The moment of inertia of the disc is I = 1/2mr^{2}. Calculate the torque induced on the disc.

^{2.}The moment of inertia of the disc is I = 1/2mr

^{2}. Calculate the torque induced on the disc.

** Given**:

m = 0.1 kg

r = 1 m

α = 2 rad/s^{2}

** To Find: **τ=?

** Formula**:

τ = I α

** Solution**:

The torque induced on the disc is calculated as,

τ= I α

For disc, the moment of inertia is I = 1/2mr^{2}** **……………..(Given)

τ= 1/2mr^{2} α

Substituting all values,

τ = 1/2 x 0.1 x 1^{2} x2

τ= 0.2/2

τ = 0.2/2

τ = 0.1

**The torque induced on the disc is 0.1 Nm.**

**The thin rod of 100vkg having a length of 6m is rotating at 20 rad/s**^{2}. The moment of inertia of the thin rod is I =1/12mr^{2}. Calculate the torque induced on the thin rod.

^{2}. The moment of inertia of the thin rod is I =1/12mr

^{2}. Calculate the torque induced on the thin rod.

** Given**:

m = 100 kg

r = 6 m

α= 20 rad/s^{2}

** To Find: ** τ =?

** Formula**:

τ = I α

** Solution**:

The torque induced on the thin rod is calculated as,

τ = I α

For thin rod, the moment of inertia is I = 1/12mr^{2}** **……………..(Given)

τ= 1/12mr^{2}α

Substituting all values,

τ = 1/12 x 100 x 6^{2} x 20

τ= 72000/12

τ = 6000

**The torque induced on the disc is 6000 Nm.**