Help in understanding sum of torque equation

In summary, the formula ∑τ = Iα is derived from the concept of torque, where τ represents torque, I represents moment of inertia of a rigid body, and α represents angular acceleration. The derivation involves an arbitrary particle on the rigid body with an applied external force tangent to the rotation. This results in the equation τ1 = Ftan * r1 = m1 * a1 * r1 = m1 * r12 * α, where m is the particle's mass, r is the particle's distance from the center of rotation, and a is the particle's linear acceleration. The sum of all these individual torques is equal to α∑miri2 = Iα. The confusion arises when considering the torque of
  • #1
Elias Waranoi
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I've been reading my physics book and there they derived the formula ∑τ = Iα where τ is torque, I is moment of inertia of a rigid body and α is the angular acceleration. They did by taking an arbitrary particle on the rigid body with an applied external force tangent to the rotation. τ1 = Ftan * r1 = m1 * a1 * r1 = m1 * r12 * α where m is the particles mass and r is the particles distance from the center of rotation and a is the particles linear acceleration. They sum this into ∑τ = α∑miri2 = Iα

What I don't understand is why the torque beside the first particle matters. In the image I attached an external force is applied to the blue particle so its torque is Ftan * r1 but if you take any other particle there is no external force applied there so Ftan * r1 should be zero? I mean the internal force of particles in a rigid body cancel out each other right? And the force that the blue particle exert on other particles is an internal force right? I understand that for an infinitesemal mass m1 the angular acceleration must be super high to become Ftan so I am quite puzzled.
 

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  • #2
I think the blue dot isn't a particle, but a point on the disk. And if it is rigid, then: yes, the sum simplifies to this one single term.
 
  • #3
Elias Waranoi said:
if you take any other particle there is no external force applied
You might be confusing 'external to the body' and 'external to the particle'. The external torque is distributed internally, under the constraint that all the parts have to undergo the same angular acceleration.
 
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  • #4
Compare with lifting a weight: your force acts at some point, but the weight moves in its entirety
 

1. What is torque and why is it important in understanding rotational motion?

Torque is a measure of the force that can cause an object to rotate around an axis. It is important in understanding rotational motion because it determines how much an object will rotate and in what direction.

2. How is torque calculated and what are its units of measurement?

Torque is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied. Its units of measurement are typically Newton-meters (N⋅m) or foot-pounds (ft⋅lbs).

3. What is the equation for calculating torque and how is it related to other equations in rotational motion?

The equation for torque is T = Fd, where T is torque, F is force, and d is distance. This equation is related to other equations in rotational motion, such as the equations for angular velocity and angular acceleration, through the concept of inertia.

4. How does the direction of the force affect the torque on an object?

The direction of the force affects the torque on an object by determining the direction of the rotation. A force applied perpendicular to the object's center of mass will produce the maximum torque, while a force applied parallel to the center of mass will produce no torque.

5. Can you provide an example of calculating torque in a real-world scenario?

Yes, for example, if you are trying to loosen a bolt with a wrench, the force you apply to the wrench handle will create a torque on the bolt. The longer the handle of the wrench, the greater the torque will be, making it easier to loosen the bolt.

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