**what is the net torque on the pulley about the axle? (figure 1)** This is a topic that many people are looking for. **amritsang.org** is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, ** amritsang.org ** would like to introduce to you **Pulley With Two Hanging Masses – Rotational Dynamics, Inertia, Torque, Acceleration & Net Force**. Following along are instructions in the video below:

This problem. We have a pulley. Which is basically a solid disc with a mass mass of 20 kilograms.

And its attached to two hanging. Masses. A 5 kilogram and a 10 kilogram block.

So once this system is released from rest. What will be the acceleration of the system now because the 10 kilogram mass is heavier. We know the whole system is going to rotate in this direction.

Lets call this m2 and this ones gonna be m1 and the mask of the dis slash pulley its capital f. And lets say that the radius of the pulley is 2 meters. So with this information how can we calculate the acceleration of the system.

So first lets do it the easy way the acceleration is going to be the net force divided by the total mass of the system. So the force thats driving a system in motion is the weight of this block. Its causing the whole system to move in this direction.

So thats going to be m2 g. Now theres another force that impedes that motion. And thats the weight force of block.

One that slows. The system down and once the system to go in the opposite direction. So thats gonna be m1 g.

So the net force is the difference between the weight of this block and the weight of that block now lets divide by the total mass of the system. So thats gonna be the mass of the first block. Plus.

The mass of the second block and the inertial mass of the disk now we cant just use the mass of the disk. But we have to multiply by a constant c. The inertia of a disc is one half m r squared.

So in this problem. C is one half. And if you want to get the right answer you need to incorporate that value otherwise you wont get the right answer.

So now lets go ahead and calculate the acceleration of the. System so its gonna be. M2 g which is 10.

Times. 98. Minus.

M1g which is 5 times. 98. Divided by m1.

Which is 5 plus. M2 thats 10 plus. One half of the mass of the solid.

Disc which is 20 so 10. Times. 98 thats going to be 98 and 5.

Times. 98. Thats half of s.

Thats 49 half of 20 is 10 plus. Another 10 thats 20 plus. 5.

Thats gonna be. 25. So.

98 49 is 49 and 49 over 25 thats. 196 so the acceleration of the. System its 196.

Meters per second squared. So this is the answer now lets see if we can get the same answer using freebody diagrams that involve tension and torque. So now lets focus on all the forces that are in play.

In this object. Now block. 2.

Has an upward tension force that prevents the block from falling. And were free fall. Now that same tension force is what drives the pulley to turn in the clockwise direction.

So t2 creates a torque that causes it to rotate in the clockwise direction and torques that rotate in the clockwise direction are defined to be negative torque. So this is going to be negative torque. Now theres an upward tension force that pulls this block upward.

Thats t1 and t1 is also this force here so t1 impedes the rotation of the pulley going in this direction. So t1 wants to cause the pulley to turn in a counterclockwise direction. So therefore it creates a positive torque.

So this is gonna be positive torque 1. I know theres a lot of stuff in this picture. Hopefully you can see the important things so basically the important stuff is those torques and these two tension forces.

So now lets write an expression that will give us the net torque acting on the pulley so the net torque is the difference between these two torques so t1 is positive and torque 2 is negative based on the directions that theyre going now. The whole system the whole pulley is going to rotate in this direction. Because this mass is larger.

So therefore the net torque of the pulley because its moving in a clockwise direction the net torque has to be negative. So the sum of all torques is going to be the inertia times alpha and that has to be a negative quantity because the net torque is clockwise meaning that causes the disc to move in the clockwise direction. So thats what we need this negative sign.

So thats equal to torque. One tor. Now inertia specifically inertia of the disc thats one half m r.

Squared. Linear. Acceleration.

Is alpha times r. So alpha is a divided by r. So lets replace that alpha with a over r.

Now the tour created by t1 torque is force times. The lever arm so torque one is t1 times the radius of the circle. So heres a circle.

The tension force multiplied by the distance between where the force acts and the axis rotation. Which is the radius of the circle. So torque one is t 1.

Times r. And torque 2 is t 2. Times r in our next.

Step. Lets cancel the radius. So what we have left over is negative.

1 2 m. R. Times.

A and thats equal to t. 1. Times.

R. Minus t. 2.

Times r. And then lets divide every term by r. So we could.

Cancel r. So now. We have this equation.

Negative. 1 2. Ma.

Is equal to t1 minus t2. So im going to rewrite this equation above so t1 minus. T2 is equal to negative 1 2.

Ma. Next. We need to write an expression for t 1.

And t 2. So lets focus on blog to the net force acting on block. 2 is equal to the upward positive tension force t 2.

Minus the negative downward weight force which is m2 g now this block is moving in the negative y direction. So the net force acting on that block is going to be negative m2. A so we need to move this term to get t2 by itself so negative m.

2. A plus m2 g is equal to tension force so were gonna save this equation for now well get back to it later now lets focus on this block. The sum of all forces in the y direction.

Acting on m1 is going to be the upward tension force t1 minus the downward weight force m1 g. Now this block is accelerated in the positive. Y direction.

So the net force is b positive m1a as opposed to negative mta and thats going to be t1. Minus m1 g. So we need to move this to the other side to get t1 by itself so m.

1. A plus m 1 g. Is equal to t1 so now lets replace t1 with what we have here first let me get rid of this stuff.

So t1 is gonna be m. 1. A plus m 1 g.

And then minus t2. Which is what we have here and thats negative m. 2.

A plus m 2 g. And so all of that thats equal to negative 1 2 m. A.

So now i dont need this stuff anymore. Thats just taking up space. So lets distribute the negative sign.

So. This is gonna be m. 1.

A plus m 1 g. Plus m. 2.

A. Minus m2 g. And thats equal to negative.

1 2. M. A.

So everything with an a lets move it to the left side. So lets take this term move it to the left side. Everything that doesnt have an a on the left side.

Lets move it to the right so that includes these two terms lets get on that side so on the left. Im gonna have em. 1.

A plus em. 2. A now as you move this term to the left its going to change from negative to positive so its gonna be plus 1 2.

Ma. And thats equal to when we move that term to the right side. Its gonna be positive m 2 g.

And this term is positive on the left which means its gonna be negative on the right so minus m1g next. Well need to factor out the gcf the greatest common factor. Which is a so its going to be a times m.

1. Plus m. 2.

Plus. 1 2 m. And.

Thats equal to the difference. In the weights of these two objects. So its m2 g.

Minus. M1g. So now lets get the acceleration.

The acceleration is going to be what we have here divided by this term. So its m2 g. Minus.

M1. G. Divided by m1.

Plus. M2 plus. 1 2.

M. So. As you can see is the net force.

Which is the difference in the weights of these two objects divided by the total mass and notice that you need to see value. Thats associated with the inertia of this disc. Its always going to work out that way so now lets go ahead and calculate how much this is gonna be so m2 is ten and.

G is. 981. Is 5 and g.

Is the same and then its gonna be m1. Plus. M2 plus 1.

2. The mass of the. Pulley so this will give you the same answer of 196.

Meters per second squared. So its good to understand both ways to solve this problem. If you have a multiple choice test.

The first method might be more efficient. But if you have a free response problem then you may have to show you work you may have to show all the steps involved where you could derive this equation and come to this answer. .

Thank you for watching all the articles on the topic **Pulley With Two Hanging Masses – Rotational Dynamics, Inertia, Torque, Acceleration & Net Force**. All shares of amritsang.org are very good. We hope you are satisfied with the article. For any questions, please leave a comment below. Hopefully you guys support our website even more.