Torque and Angular Acceleration: Every Facts You Need to Know


The angular acceleration of the object is due to the rotational motion of the object about its axis from the point of the center of gravity and torque is responsible for the rotational motion of the object.

As the force is applied tangentially to the body, the equivalent force is acted on the point situated opposite to it and acts in the opposite direction that tends to rotate it with angular acceleration, and hence torque and angular acceleration both come into the picture in the case of a rotating body.

Relation between Torque and Angular Acceleration

The net torque acting on the object is directly proportional to the angular acceleration of the object and inversely related to the inertia of rotations about its axis of rotation.

[latex]\alpha \propto \tau _{NET}[/latex]

The speed acquired by the object depends upon the torque applied to the body and angular acceleration is the change in the angular velocity with the time of the object rotating about an axis.

Torque and Angular Acceleration Formula

We know that the torque is a product of the force applied to the object and how far it is displaced from the applied force.

[latex]\tau =Force\times displacement[/latex]

And [latex]F=ma[/latex]

Consider a circular disc of radius ‘r’ and a force F is applied on the disc to rotate it about an axis exerting a torque τ moving with angular acceleration α.

Torque on a disc

θ is an angular displacement of a disc on the application of torque, ω is angular velocity and a is the acceleration of the disc.

Let the displacement be equal to ‘r’ from the axis of rotation then we get

[latex]\tau =ma\times r[/latex]

If the object is rotating with angular velocity ω then angular velocity is related to the angular acceleration as [latex]\alpha =\frac{\omega }{t}[/latex] and acceleration of the object is related to the angular acceleration as

[latex]a=r\alpha [/latex]

Using this in the above equation, we have

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

The term [latex]mr^2[/latex] is nothing but the moment of inertia of the object extended in all dimensions of the object. Hence,

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

This is the equation denoting the relationship between the torque and the angular acceleration of the object.

Torque and Angular Acceleration Graph

Referring to the equation giving the relation of torque and the angular acceleration of the object that we have found above, we can plot a graph of torque and angular acceleration.

Suppose the moment of inertia is I=0.67kg.m2, then

If [latex]\alpha =10[/latex] then

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

[latex]=0.67\times 10=6.7N.m[/latex]

If [latex]\alpha =20[/latex] then

[latex]\tau = 0.67\times 20=13.4 N.m[/latex]

If [latex]\alpha =30[/latex] then

[latex]\tau = 0.67\times 30=20.1 N.m[/latex]

Hence we get a graph of torque v/s angular acceleration as plotted below:-

torque and angular acceleration
Graph of torque v/s angular acceleration

The slope of the graph of torque v/s angular acceleration will obviously give us the moment of inertia of the body and it is seen that the angular acceleration increases linearly with increasing torque.

Hence, if we found out the torque given to the object and the angular acceleration that the object acquires then can find out the moment of inertia of the object by plotting a graph of torque v/s angular acceleration.

Torque and Angular Acceleration Direction

If we look into the below diagram carefully, we can understand that the force is applied on the object with is tangent to it and corresponding to it the object starts rotating with the angular acceleration perpendicular to the force applied on the object.

The torque created at the axis of rotation is shown in the diagram which is perpendicular to the force as well as the angular acceleration of the object.

The direction of torque and angular acceleration of the body

The same we can depict on the three axis as shown below:-

The direction of Force, Angular Acceleration, and Torque

In this diagram, we can clearly see that, force, angular acceleration, and torque all lie perpendicular to each other.

If the force is applied on the x-axis then the direction of the angular acceleration of the object will be perpendicular to the force in the y-direction and the corresponding torque will be applied in the azimuthal axis that is z-axis perpendicular to both.

The same concept can be memorized using the right-hand thumb rule.

The direction of Force, Angular Acceleration, and Torque using the RHT Rule

If you hold your right hand as shown in the above picture, the thumb denotes the direction of the torque exerted on the object, the curve fingers represent the direction of the angular acceleration of the object and the force which is perpendicular to both is denoted by the palm of your hand.

How to Find Torque with Angular Acceleration?

The angular acceleration of the object is a resultant of the exertion of torque on its body.

Torque can easily be found by knowing the angular acceleration of the object and the moment of inertia using the formula [latex]\tau =I\alpha [/latex], where τ is a torque on a body, I is the moment of inertia and alpha is an angular acceleration of the object.

The moment of inertia is the product of the sum of all the masses of the particle constituting the object and the square of the distance from the point of the angular acceleration of the edge of the object and the axis of rotation and is the tendency of the object to lower the angular acceleration.

If a solid disc of weight 1kg and radius 12cm is rotated giving an angular acceleration of 2π rad/s2 then what is the torque applied on a disc?

Given: m =1kg

r =12cm =0.12m

a =2π rad/s2

The moment of inertia of the disc is

[latex]I=\frac{1}{2}mr^2 [/latex]

[latex]=\frac{1}{2}\times 1\times (0.12)^2 [/latex]

[latex]=\frac{0.0144}{2}=0.0072kg.m^2[/latex]

Now we can calculate the torque exerted on the disc as

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

[latex]=0.0072\times 2\pi [/latex]

[latex]=0.045 N.m[/latex]

Hence the torque applied to the object is 0.045 N.m.

How to Find Angular Acceleration from Torque?

As soon as the torque is applied to the body, it will start rotating with some angular acceleration depending upon the moment of inertia of a body.

The angular acceleration can be calculated from how much is the torque applied to the object using the formula [latex]\alpha =\frac{\tau }{I}[/latex]. The torque applied will produce the angular acceleration while the moment of inertia of the body will try to oppose this angular acceleration at the same time.

What is the angular acceleration of a bowling ball of mass of 800 grams and having a radius of 12cms if the torque of 3.5 times 10-4 N.m is applied to the ball?

Given: [latex]τ=3.5 \times 10-4 N.m[/latex]

r =12cm =0.12m

m = 800gm =0.8kg

Now let us first calculate the moment of inertia of the bowling ball. Since the bowling ball is spherical in shape

[latex]I=\frac{2}{5}mr^2[/latex]

[latex]=\frac{2}{5}\times 0.8\times (0.12)^2[/latex]

[latex]=0.0046kg.m^2[/latex]

Hence, the angular acceleration of the bowling ball is

[latex]\alpha =\frac{\tau }{I}[/latex]

[latex]=\frac{3.5\times 10^{-4}}{0.0046}[/latex]

[latex]=0.076m/s^2[/latex]

The angular acceleration of the bowling ball is 0.076 m/s2.

Is angular acceleration produced by torque?

The angular acceleration is a result of a rotational motion of the object which is obtained by giving a torque to the body.

The torque is created by applying the force perpendicular to the axis of rotation of the body and the body starts rotating on its axis of rotation making a 90-degree angle to the direction of torque applied.

Torque and Angular Acceleration of a Flywheel

A flywheel is a machine used to store the energy within it and generates a high amount of electric power when it is given a torque to accelerate.

Flywheel on a steam machine; Image Credit: pixabay

Consider a flywheel rotating clockwise as force F is exerted on it as shown in the figure below. The radius of the flywheel is ‘r’ and its rotational axis is located at the center.

Free body diagram of the flywheel

The torque acting on the flywheel is

[latex]\tau =Force\times displacement[/latex]

The force due to gravity experienced on the flywheel is F=mg and the radial displacement of the flywheel is along its radius ‘r’.

Hence, we get the expression for torque as

[latex]\tau =mgr[/latex]

Since the flywheel is raised to a height ‘h’ the potential energy loss in the machine is equal to mgh.

The kinetic energy of a rotating flywheel rotating with angular velocity ω is

[latex]K.E =\frac{1}{2}I\omega ^2[/latex]

Where I is a moment of inertia and ω is the angular velocity of the object

The angular acceleration of the object is the variation in the angular velocity with respect to time and is given by

[latex]\alpha =\frac{d\omega }{dt}[/latex]

The torque on the flywheel is

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

Substituting for alpha

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

This equation is independent of angular acceleration.

We know that the torque is directly proportional to the angular acceleration by the equation

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

Hence, angular acceleration is a ratio of torque and moment of inertia of the object.

[latex]\alpha =\frac{\tau }{I}[/latex]

Now, we have found the equation of torque for a flywheel, let us substitute it here in this equation to find the angular acceleration.

[latex]\alpha =\frac{mgr }{I}[/latex]

Multiplying ‘r’ in numerator and denominator we get

[latex]\alpha =\frac{mr^2g}{Ir}[/latex]

Since [latex]I=mr^2[/latex], using this in the equation above

[latex]\alpha =\frac{Ig}{Ir}[/latex]

[latex]\alpha =\frac{g}{r}[/latex]

The angular acceleration is inversely related to the radius of the object; this implies that, if the diameter of the object is greater, the angular acceleration of the object will be smaller.

Torque and Angular Acceleration for a Rigid Body

A rigid body is a solid object which does not deform in any sequence and the mass is continuously distributed in a rigid body.

The moment of inertia of the rigid body is constant and directly proportional to the angular momentum of the rotating object. It is given by the relation as

[latex]I=\frac{L}{\omega }[/latex]

Hence, torque on the rigid body is

[latex]\tau =\vec{r}\times \vec{F}[/latex]

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

And torque is related to the moment of inertia by the equation

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

Substituting the equation of angular momentum in this equation here, we get

[latex]\tau =\frac{L}{\omega }\times \alpha [/latex]

The ω is nothing but the angular velocity and is equal to the angular acceleration by time.

[latex]\tau =\frac{L}{\alpha t}\times \alpha [/latex]

[latex]\tau =\frac{L}{t}[/latex]

Hence, we get that, the torque acting on a rigid body is also equal to the angular momentum of the object per unit of time.

Hence, now we can find the angular acceleration of the object as

[latex]\alpha =\frac{\tau }{I}[/latex]

[latex]\alpha =\frac{L}{It}[/latex]

We can find the angular acceleration of a rigid body using this formula.

What is the angular acceleration and torque on a cylindrical rod of rigid mass?

Consider a cylindrical rod of length ‘L’ and is rotated clockwise, then the torque acting on the cylindrical rod of mass m is

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

The force due to gravity is [latex]F=mg[/latex] and r is half of the length of the rod, the distance from the axis of rotation to the point where force is acted.

Since the angle made by the rotating axis with the angular acceleration of the object is 90 degrees,

[latex]Sin90^0=1[/latex]

Hence, [latex]\tau =mg\times \frac{L}{2}\times 1[/latex]

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

Torque to angular acceleration of the rod is

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

For cylindrical rod, moment of inertia is [latex]I=\frac{1}{3}mL^2[/latex].

[latex]\tau =\frac{1}{3}mL^2\alpha [/latex]

Now equating both the equations,

[latex]\frac{mgL}{2}=\frac{1}{3}mL^2\alpha [/latex]

[latex]\frac{g}{2}=\frac{1}{3}L\alpha [/latex]

[latex]\alpha =\frac{3g}{2L}[/latex]

Hence we got the angular acceleration of the rigid cylindrical rod.

Frequently Asked Questions

What is torque and angular acceleration produced when a child of weight 21 kg sits on a merry-go-round of weight 60kg having a radius of 1.2 meters and a force of 230 N is applied to rotate it?

Given: Weight of a child is m =21kg

Weight of merry-go-round is M =60kg

Radius of merry-go-round is r =1.2m

F =230N

Moment of inertia of a merry of round is

[latex]I_1=\frac{1}{2}Mr^2[/latex]

[latex]=\frac{1}{2}\times 60\times 1.2^2[/latex]

[latex]=\frac{86.4}{2}[/latex]

[latex]=43.2 kg.m^2[/latex]

Moment of inertia of a child when it sits on a merry of round is

[latex]I_2=\frac{1}{2}mr^2[/latex]

[latex]=\frac{1}{2}\times 21\times 1.2^2[/latex]

[latex]=\frac{30.24}{2}[/latex]

[latex]=15.12 kg.m^2[/latex]

Hence, the total moment on inertia is

[latex]I=I_1+I_2[/latex]

[latex]=43.2+15.12[/latex]

[latex]=58.32 kg.m^2[/latex]

The torque acting on the merry-go-round is

[latex]\tau =Fr[/latex]

[latex]\tau =230\times 1.2=276N.m[/latex]

Hence, the angular acceleration of the merry-go-round due to the torque of 276 N.m is

[latex]\alpha =\frac{\tau }{I}[/latex]

[latex]=\frac{276}{58.32}[/latex]

[latex]=4.73 m/s^2[/latex]

The angular acceleration of the merry-go-round is 4.76 m/s2.

What is the angular acceleration and torque of a ceiling fan of mass 5.4kg having 3 blades of length 1m and a center disc with a radius of 12cm? The mass of each blade is 800 grams and the center mass weighs 3kg.

Given: Mass of blades is m =800g =0.8kg

Mass of disc M =3kg

Length of 3 blades L = 1m

Radius of disc r =12cm =0.12m

Total mass of fan is 5.4kg

The torque of blades of a fan is

[latex]\tau =Fr[/latex]

[latex]=mgL[/latex]

[latex]=5.4\times 9.8\times 1=52.92N.m[/latex]

Moment of inertia of the blade is

[latex]I_1=\frac{1}{3}mL^2[/latex]

[latex]=\frac{1}{3}\times 0.8\times 1^2=0.267kg.m^2[/latex]

Hence, the moment of inertia of 3 blades is

[latex]=3\times 0.267=0.801kg.m^2[/latex]

Moment of inertia of disc is

[latex]I_2=\frac{1}{2}Mr^2[/latex]

[latex]=\frac{1}{2}\times 3\times (0.12)^2=0.0216kg.m^2[/latex]

Hence, the total moment of inertia is

[latex]I=I_1+I_2[/latex]

[latex]=0.801+0.0216=0.8226 kg.m^2[/latex]

Hence, now we can calculate the angular acceleration of a fan

[latex]\alpha =\frac{\tau }{I}[/latex]

[latex]=\frac{52.92}{0.8226}[/latex]

[latex]=64.33 rad/s^2[/latex]

The angular acceleration of a ceiling fan is 64.33 rad/s2.

AKSHITA MAPARI

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess. Connect me on LinkedIn - linkedin.com/in/akshita-mapari-b38a68122

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