The article discusses the relationship between torque and angular momentum of the rotating body and its solved problems.

**The torque and angular momentum are the rotational analogue of force and linear momentum respectively. The net torque on the rotating body produces its rate of change in angular momentum about the axis of rotation as per Newton’s laws. If torque is absent, then its angular momentum is conserved. **

Let’s consider a rigid body where a tangential force works on the point mass m at the distance r from its axis of rotation.

When a net force functions on the body that is fixed to an axis, its momentum (mv) varies and it starts moving. Since a force is applied away from its axis of rotation, the angular momentum (L) is built from *the product of the linear momentum (P) on the body and perpendicular distance (r) from the axis of rotation.*

[latex]\vec{L} = \vec{r} \times \vec{P}[/latex] ……………….. (1)

**The magnitude of angular momentum** is,

[latex]L = rPsin\theta[/latex] ……………..(2)

where [latex]\theta[/latex]

θ

is the angle between r and P.

**If internal particles are at the origin of the body or [latex]\vec{r}[/latex] **

**and [latex]\vec{P}[/latex] **

** are antiparallel 180 ^{o} or parallel 0^{o} to each other, the linear momentum [latex]\vec{P}[/latex] **

**and angular momentum [latex]\vec{L}[/latex] **

**become zero. **

**Torque and Angular Momentum Relationship**

Due to applied force at distance, a torque is generated on the body so that it can rotate about its axis. That’s how a torque sets the rotational motion on the body.

Like angular momentum formula, **the torque also equivalent to the applied force at distance.**

[latex]\vec{\tau } = \vec{r} \times \vec{F}[/latex] ……………….(3)

**The magnitude of torque** is,

[latex]\tau = rFsin\theta[/latex]

T=rFsinθ

The angle between r and F is zero. i.e., [latex]sin\theta[/latex] = sin90^{o} = 1

sinθ=sin90o = 1

So, [latex]\tau = rF[/latex] ……………………..(4)

T=rF1………………..(4)

**Newton’s laws of motion **says, F = ma

[latex]\tau = r(ma) [/latex]…………… (5)

T=r(ma)…………(5)

Note that the body is accelerated means the body’s motions change; so its momentum.

[latex]\tau = rm\frac{dv}{dt}[/latex]

T=rm*dv/dt

[latex]\tau = \frac{d}{dt}rmv[/latex]

T=d/dt*rmv

[latex]\tau = \frac{d}{dt}rP[/latex]

T=d/dt*rp

From equation (2),

[latex]\tau = \frac{dL}{dt}[/latex] …………………..(*)

** The relationship between torque and angular momentum is equivalent to the force and linear momentum described by Newton’s laws of motion.** The equation (*) is Newton’s law of motion formula in rotational motion. That’s how the torque and angular momentum enable us to transform the state of rotational motion.

**What is the torque acting on the spinning top that changes its momentum from 30 kgm/s to 50 kgm/s in 5 seconds?**

** Given**:

L_{1} = 30 kgm/s

L_{2} = 50 kgm/s

t_{1} = 0s

t_{2} = 5s

** To Find: **[latex]\tau[/latex] =?

T=?

__Formula:__

[latex]\tau = \frac{dL}{dt}[/latex]

T=dL/dt

__Solution:__

The torque acting on the top is calculated as,

[latex]\tau = \frac{dL}{dt}[/latex]

T=dL/dt

[latex]\tau = \frac{L_{2}-L_{1}}{t_{2}-t_{1}}[/latex]

T=L_{2}-L_{1}/t_{2}-t_{1}

Substituting all values,

[latex]\tau = \frac{50-30}{5-0}[/latex]

T=50-30/5-0

[latex]\tau = \frac{20}{4}[/latex]

T=20/4

[latex]\tau = 5[/latex]

T=5

**The torque acting on the top is 5Nm.**

**A rotating body having a radius of 1.5m moves at a momentum of 50 kgm/s. Calculate the torque acting on the body for 5 seconds which changes its momentum to 100 kgm/s.**

** Given**:

r = 1.5m

P_{1} = 50 kgm/s

t_{2} = 2s

t_{1 }= 0s

P_{2} = 100 kgm/s

** To Find**: [latex]\tau[/latex] =?

T=?

** Formula**:

L = r x P

[latex]\tau = \frac{dL}{dt}[/latex]

T=dL/dt

** Solution**:

The angular momentum of the body before torque induced is,

L_{1} = r x P_{1}

L_{1} = 1.5 x 50

L_{1} = 75kgm^{2}/sec

The angular momentum of the body after torque induced is,

L_{2} = r x P_{2}

L_{2} = 1.5 x 100

L_{2} = 150kgm^{2}/sec

The torque acting on the rotating body is calculated as,

[latex]\tau = \frac{dL}{dt}[/latex]

T=dL/dt

[latex]\tau = \frac{L_{2}-L_{1}}{t_{2}-t_{1}}[/latex]

π=L_{2}-L_{1}/t_{2}-t_{1}

Substituting all values,

[latex]\tau = \frac{150-75}{2-0}[/latex]

π=150-75/2-0

[latex]\tau = \frac{75}{2}[/latex]

π=75/2

[latex]\tau = 37.5[/latex]

π=37.5

**The torque acting on the body is 37.5Nm. **

**Find Torque from Angular Momentum**

The torque is found by differentiation of angular momentum.

Differentiate the equation (1),

[latex]\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r}\times \vec{P})[/latex]

[latex]\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{P} + \vec{r} \times \frac{d\vec{P}}{dt}[/latex]

The term [latex]\frac{d\vec{r}}{dt}[/latex]

is the linear velocity [latex]\vec{v}[/latex]

\ of the body.

[latex]\frac{d\vec{L}}{dt} = \vec{v} \times \vec{P} + \vec{r} \times \frac{d\vec{P}}{dt}[/latex]

The velocity and momentum is in the exact direction. So, [latex]\vec{v} \times \vec{P}[/latex] = [latex]vpsin\theta[/latex] = vpsin0^{o} = 0

[latex]\frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{P}}{dt}[/latex]

The term is [latex]\frac{d\vec{P}}{dt}[/latex] is the applied force [latex]\vec{F}[/latex] as per **Newton’s laws.**

[latex]\frac{d\vec{L}}{dt} = \vec{r} \times \vec{F}[/latex]…………………….(6)

**Torque and Angular Momentum Formula**

The term ‘[latex]\vec{r} \times \vec{F}[/latex]’ is the torque [latex]\vec{\tau }[/latex] acting on the body which changes angular momentum L.

[latex]\vec{\tau } = \frac{d\vec{L}}{dt} = \vec{r} \times \vec{F}[/latex] ……………….(%)

The position vector r and force F perpendicular to each other.

[latex]\vec{r} \times \vec{F} = rFsin\theta[/latex] = rFsin90^{o} = rF

Substituting above equation into equation (%),

[latex]\tau = \frac{dL}{dt} = rF [/latex]……………….(7)

[latex]\tau = \frac{dL}{dt} = rma[/latex]

mThe relation between **linear**** acceleration a and angular acceleration **[latex]\alpha[/latex] is, [latex]a = r\alpha[/latex]

[latex]\tau = \frac{dL}{dt} = rmr\alpha[/latex]

The torque delivers the required angular acceleration to the rigid body to accomplish the rotational motion. The direction of both [latex]\tau[/latex] and [latex]\alpha[/latex] along the rotation axis. If they are in the same direction, the body will accelerate angularly. But if they are in the opposite direction, the body will deaccelerate.

[latex]\tau = \frac{dL}{dt} = mr^{2}\alpha[/latex]

The term [latex]mr^{2}[/latex] is called **‘moment of inertia**’ (I) which describes *the body’s tendency to oppose angular acceleration.*

[latex]\tau = I\alpha[/latex] ………………….(8)

From equation (*), (7) and (8), ** the torque and angular momentum formula **is,

[latex]\tau = rF = I\alpha = \frac{dL}{dt} [/latex]……………….(@)

The above equation shows that **the**** torque working on the body as per the product of moment of inertia and angular acceleration changes its angular momentum.**

If there is no torque working on the body. i.e. [latex]\tau = 0[/latex], so [latex]\frac{dL}{dt}[/latex] is also zero. That means the angular momentum of the body does not vary or remain constant.** That’s how the angular momentum is conserved. **

**Read about Torque and Angular Velocity**

**What is the torque acting at 0.5m on a disc having a mass of 5kg which accelerates to 10 rad/s**^{2}?

^{2}?

** Given**:

r = 0.5m

m = 5kg

[latex]\alpha[/latex] = 10 rad/s^{2}

** To Find: **[latex]\tau[/latex] =?

** Formula: **[latex]\tau = I\alpha[/latex]

__Solution:__

The torque acting on an disc is calculated as,

[latex]\tau = I\alpha[/latex]

But the moment of inertia is I =mr^{2}

[latex]\tau = mr^{2}\alpha[/latex]

Substituting all values,

[latex]\tau = 5\times 0.5^{2}\times 10[/latex]

[latex]\tau = 5\times 0.25 \times 10[/latex]

[latex]\tau = 12.5[/latex]

**The torque acting the disc is 12.5Nm.**

**A force of 50N is applied at a distance of 2m on the rigid body of 5kg which accelerates angularly to 5 rad/s**^{2}. Calculate the torque acting on the body.

^{2}. Calculate the torque acting on the body.

** Given**:

F = 50N

r = 2m

m = 5kg

[latex]\alpha [/latex]= 5 rad/s^{2}

** To Find: **[latex]\tau[/latex] =?

** Formula: **[latex]\tau = rF = I\alpha[/latex]

__Solution:__

The torque on the rigid body is calculated as,

[latex]\tau = rF = I\alpha[/latex]

But I =mr^{2}

[latex]\tau = rF = mr^{2}\alpha[/latex]

Substituting all values,

[latex]\tau = 2 \times 50 = 5\times 2^{2} \times 5[/latex]

[latex]\tau = 100 = 100[/latex]

**The torque acting on the rigid body is 100Nm.**

**Torque and Angular Momentum for a System of Particles**

Suppose the system S contains the particle j having mass m_{j} and velocity v_{j}.

From equation (1) The **angular momentum of particle j** is given by,

[latex]\vec{L}_{j} = \vec{r}_{j} \times \vec{P}_{j}[/latex]

Hence, **the total angular momentum of the rotating system** is,

[latex]\vec{L}_{S} = \sum_{j=1}^{i=n}\vec{r}_{j} \times \vec{P}_{j}[/latex]

From equation (*), **the change in angular momentum of the system** is,

[latex]\frac{d\vec{L}_{S}}{dt} = \sum_{j=1}^{i=n}(\vec{r}_{j} \times \frac{d\vec{P}_{j}}{dt})[/latex]

The term [latex]\frac{d\vec{P}_{j}}{dt}[/latex] is the net force [latex]\vec{F}_{j}[/latex] acting on the system.

[latex]\frac{d\vec{L}_{S}}{dt} = \sum_{j=1}^{i=n}(\vec{r}_{j} \times \vec{F}_{j} )[/latex]

As per equation (%),

[latex]\frac{d\vec{L}_{S}}{dt} = \sum_{j=1}^{i=n}\vec{\tau }_{j}[/latex] …………………..(9)

In a close system, **the net torque is the sum of internal and external torques on individual particles within the system**.

[latex]\frac{d\vec{L}_{S}}{dt} = \tau _{j}^{ext} +\tau _{j}^{int}[/latex]

But all internal forces within the body are zero.

[latex]\frac{d\vec{L}_{S}}{dt} = \tau _{j}^{ext} [/latex]……………….(10),

From the above equation, we understand that,** when external torque acts on the body, its total angular momentum changes.**