# How Does Torque Affect the Motion of a Disc Pulled Across a Table?

• ago01
In summary: It can be seen from a combination of Newton's second and third laws. Internal forces cancel leaving the acceleration of the centre of mass to obey ##F_{ext} = ma_{com}##.
ago01
Homework Statement
A uniform cylindrical spool of string has radius R=3 cm, and mass M=500g. You accelerate the spool along a table by applying a constant force F=2N to the string. The spool rolls along the table without slipping and you can ignore the mass of the string. What is the acceleration of the center of mass of the spool? What is the magnitude of the static friction?
Relevant Equations
Torque, angular acceleration, force
I started out by drawing a diagram:

So I thought I would try torque with the axis of rotation in the center:

##T_1R = \frac{Mr^2}{2}\alpha##

and given that ##T_1## is equal to ##F## in the positive direction.

Then knowing the relationship between angular acceleration and tangential acceleration:

##\alpha = \frac{a_t}{r}##
##a_t = \alpha*r##
##a_t = 7.998 \frac{m}{s^2}##

But this is not correct. The center of mass is translating and the rest of the object is rotating around it. ##a_t## would be the acceleration of a particle on the edge of the spool. Without this I cannot solve the second part so I am stuck.

So what if instead I just consider the translational portion?

##F_x = Ma##
##T_1 = Ma##
##T_1/M = a##
## a = 4 \frac{m}{s^2}##

But this is also incorrect. This would be like dragging the cylinder with slipping. It seems I am missing the connecting equations for rotational and translational motion.

What does Fg represent?

Your first equation does not consider all the torques on the cylinder. Look again at your diagram.

haruspex said:
What does Fg represent?

Your first equation does not consider all the torques on the cylinder. Look again at your diagram.

I finally got it with that hint.

For some reason I remember the professor in the video saying that in rolling motion static friction is the reason the object moves forward. Reviewing my book, the diagrams of a bicycle wheel only show a vector for static friction (like normal) opposite the direction of the force. I guess I manufactured a vector for no reason to represent the forward motion... it makes sense "static friction is the reason it moves" because the static friction applies additional torque:

##\sum{F_x} = T_1 - F_s##

##\sum{\unicode{x03A4}} = T_1R + F_sR = I\alpha##

Solving for ##F_s## in the first equation and plugging it into the second equation I get the answer.

But where I am confused still is - this is the tangential acceleration right (since we have ##\alpha = \frac{a_t}{R}##)? How is this also the acceleration of the CM moving forward?

ago01 said:
But where I am confused still is - this is the tangential acceleration right (since we have ##\alpha = \frac{a_t}{R}##)? How is this also the acceleration of the CM moving forward?
It can be seen from a combination of Newton's second and third laws. Internal forces cancel leaving the acceleration of the centre of mass to obey ##F_{ext} = ma_{com}##.

Most mechanics textbooks should go through this calculation.

## 1. How does pulling a disc across a table create friction?

When you pull a disc across a table, the disc's surface comes into contact with the table's surface. This creates a force called friction, which is the resistance between two surfaces when they are in contact and moving relative to each other.

## 2. What factors affect the amount of friction when pulling a disc across a table?

The amount of friction when pulling a disc across a table is affected by the type of surfaces in contact, the force applied to the disc, and the smoothness of the surfaces. Rougher surfaces and heavier forces will create more friction, while smoother surfaces and lighter forces will create less friction.

## 3. How does friction impact the motion of the disc when pulled across a table?

The friction between the disc and the table's surface will act in the opposite direction of the applied force, slowing down the disc's motion. This is because the force of friction works to resist the motion between the two surfaces in contact.

## 4. Can the amount of friction be reduced when pulling a disc across a table?

Yes, the amount of friction can be reduced by using a lubricant, such as oil or wax, between the disc and the table's surface. This reduces the roughness of the surfaces in contact and allows the disc to slide more easily.

## 5. How does the weight of the disc affect the amount of friction when pulling it across a table?

The weight of the disc does not directly affect the amount of friction when pulling it across a table. However, a heavier disc may require a greater force to overcome the friction and move across the table, while a lighter disc may require less force.

• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
97
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
266
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
474
• Introductory Physics Homework Help
Replies
335
Views
9K
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
781