__Servo Torque Calculator:__

__Servo Torque Calculator:__

Enter the values of moment of inertia, I_{(kg.m2),} angular acceleration, α_{(rad/s2),} mass of the object m_{(kg)}, acceleration due to gravity, g and distance, r_{(m)}.

__Servo Torque Formula:__

__Servo Torque Formula:__

The servo torque formula calculates the torque required by a servo motor to move or rotate an object. This calculation is essential in robotics, mechanical design, and control systems to ensure that the servo motor can provide sufficient torque for the application’s demands.

Servo Torque, T_{s(N.m )} in Newton metres is equal to the sum of :

(i) the product of moment of inertia, I _{(kg.m2) }in kilogram metre square, angular acceleration, α_{(rad/s2)} in radian per second square and

(ii) the product of mass of the object m_{(kg)} in kilogram, acceleration due to gravity, g, distance, r_{(m)} from the pivot point to the center of mass of the object in meters (m) and divided by 2.

Servo Torque, T_{s(N.m ) }= I_{(kg.m2)} * α_{(rad/s2)} + (m_{(kg) }* g * r_{(m)}) / 2

T_{s(N.m ) }= Servo Torque in Newton metres, N.m.

I_{(kg.m2) }= Moment of Inertia of the object in kilogram square metres, kg.m^{2}.

α_{(rad/s2)} = Angular acceleration in radians per second square, rad/s^{2}.

m_{(kg) }= Mass of the object in kilograms, kg.

g = Acceleration due to gravity, approximately 9.81 m/s^{2} on the surface of the Earth..

r_{(m)} = Distance from the pivot point to the center of mass of the object in meters, m.

__Servo Torque Calculation:__

__Servo Torque Calculation:__

1. A servo motor is used to rotate a disk that has a moment of inertia of 0.02 kg.m^{2} and requires an angular acceleration of 4 rad/s^{2}. The disk’s mass is 0.5 kg, and its center of mass is 0.25 metres from the pivot point. Calculate the required servo torque.

Given: I_{(kg.m2)} = 0.02 , α_{(rad/s2) }= 4rad/s^{2}, m_{(kg) }= 0.5kg , g = 9.81m/s^{2}, r_{(m)} = 0.25m

Servo Torque, T_{s(N.m ) }= I_{(kg.m2)} * α_{(rad/s2)} + (m_{(kg) }* g * r_{(m)}) / 2

T_{s(N.m )} = 0.02 * 4 + (0.5 * 9.81 * 0.25) / 2

T_{s(N.m )} = 0.08+0.61275

T_{s(N.m ) }= 0.69275N.m

2. A servo motor is required to provide a torque of 5 N.m to achieve an angular acceleration of 10 rad/s² for an application. The arm attached to the servo motor has a mass of 2 kg and is 0.5 metres long (distance from the pivot point to the center of mass). Calculate the moment of inertia, I of the arm.

Given: T_{s(N.m )} = 5N.m, α_{(rad/s2)} = 10rad/s^{2}, m_{(kg) }= 2kg , g = 9.81m/s^{2}, r_{(m)} = 0.5m

Servo Torque, T_{s(N.m ) }= I_{(kg.m2)} * α_{(rad/s2)} + (m_{(kg) }* g * r_{(m)}) / 2

I_{(kg.m2)} = ( T_{s(N.m )} – (m_{(kg) }* g * r_{(m)}) / 2 ) / α_{(rad/s2)}

(m_{(kg) }* g * r_{(m)}) / 2 = ( 2 * 9.81 * 0.5 ) / 2

(m_{(kg) }* g * r_{(m)}) / 2 = 4.905N.m

I_{(kg.m2)} = 5 – 4.905 / 10

I_{(kg.m2)} = 0.0095kg.m^{2}.