How Do Torque and Force Affect Acceleration and Friction in a Rolling Disc?

Try to be careful with signs. They'll get you every time.In summary, we have a disc with mass m, radius R, and moment of inertia about the center of mass of (mR^2)/3. An external force F is applied directly upwards at the right edge of the disc, causing it to roll without slipping on a horizontal surface. Using the equations F*R=I*alpha and -f=ma, we can find the acceleration a=2F/3m and the static frictional force f=2F/3 acting in the same direction as the acceleration. It is important to carefully consider the signs and draw a diagram to correctly identify all forces and accelerations.
  • #1
AdamP
For a disc with mass m, radius R, and moment of inertia about center of mass of (mR^2)/3, and applied external force is F.

A wheel is being pulled by a force F directly upwards, which causes it to roll without slipping (due to static friction), on a horizontal surface. The upward force F is applied at the most right tip or edge of the disc.

Find the acceleration, a, and the static frictional force, f.


I'm not sure how to set up the torque or force equations etc...

So, if F is directly upward, and on the edge, is it like F*cos90=0, so
0-f=ma?
and second equation,
F*(2R)=(moment of inertia about point of contact of disc and surface) * (angular speed)
where
moment of inertia about point of contact = (moment of inertia about center of mass, given) + m*R^2
and
angular speed=a/R (because no slip?)

so in this case, solving the F*2R= --- etc. equation, we get
a=3F/2m
and using 0-f=ma,
f=-3F/2

Does this sound right at all or what should I do? Any suggestion is much appreciated!
 
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  • #2
Welcome to PF!

AdamP said:
For a disc with mass m, radius R, and moment of inertia about center of mass of (mR^2)/3, and applied external force is F.
Is the disk nonuniform? For a uniform disk [itex]I_{cm} = 1/2 m R^2[/itex].
So, if F is directly upward, and on the edge, is it like F*cos90=0, so
0-f=ma?
The only horizontal force on the disk is the frictional force, so f = ma.
and second equation,
F*(2R)=(moment of inertia about point of contact of disc and surface) * (angular speed)
where
moment of inertia about point of contact = (moment of inertia about center of mass, given) + m*R^2
and
angular speed=a/R (because no slip?)
Both friction (f) and the upward force (F) will exert a torque about the center of mass. Set the net torque equal to [itex]I \alpha[/itex]. (Yes, [itex]a = \alpha R[/itex], since it doesn't slip.)

so in this case, solving the F*2R= --- etc. equation, we get
a=3F/2m
and using 0-f=ma,
f=-3F/2
Redo your solution with my suggestions.
 
  • #3
Thank you!

So...

Equation 1:
(F-f)*R=I*alpha
(where I is the the moment of inertia going through center of mass, i.e. 1/2mR^2) <--- so we just choose the torque equation to go through the center of mass, is that the easiest case? cos i remember in class we talked about picking the point of contact and use the moment of inertia with respect to that point, like using parallel axis theorem or something... that doesn't apply here does it? on what cases then do we do that?)

Equation 2:
-f=ma=m(r*alpha) <--- do we need the negative sign here? (cos that will affect our result a bit wouldn't it?)

So just solve the two equations two unknowns to get the answer.

(If I used it withOUT the negative sign, I get a=2F/3m, and f=2F/3, does this sound right? the acceleration and friction should be in the opposite directions right, or no..? do we just state it out or what do I do?
cos otherwise, WIth the negative sign, I would have gotten a=-2F/m, and f=-2F, which can't be right... right?)

Thanks!
 
  • #4
AdamP said:
Equation 1:
(F-f)*R=I*alpha
(where I is the the moment of inertia going through center of mass, i.e. 1/2mR^2) <--- so we just choose the torque equation to go through the center of mass, is that the easiest case? cos i remember in class we talked about picking the point of contact and use the moment of inertia with respect to that point, like using parallel axis theorem or something... that doesn't apply here does it? on what cases then do we do that?)
You can certainly choose the contact point if you like. That makes the equation even easier: [itex]FR = I_{contact} \alpha = 3/2 m R^2 \alpha[/itex].

Equation 2:
-f=ma=m(r*alpha) <--- do we need the negative sign here? (cos that will affect our result a bit wouldn't it?)
The negative sign just depends on your choice of coordinates. I chose positive to be to the left, so both a and [itex]\alpha[/itex] are positive. (Otherwise I can't use [itex]a = I \alpha[/itex].)

So just solve the two equations two unknowns to get the answer.
Right.

(If I used it withOUT the negative sign, I get a=2F/3m, and f=2F/3, does this sound right? the acceleration and friction should be in the opposite directions right, or no..? do we just state it out or what do I do?
These answers look good to me. The acceleration must be in the same direction as the frictional force, since the friction causes the acceleration! (The moral of this story: always draw a careful diagram and identify all the forces and accelerations.)

cos otherwise, WIth the negative sign, I would have gotten a=-2F/m, and f=-2F, which can't be right... right?)
No good. I'll bet you got your signs mixed up somewhere.
 

Related to How Do Torque and Force Affect Acceleration and Friction in a Rolling Disc?

What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis to the point where the force is applied.

What is acceleration?

Acceleration is the rate of change of an object's velocity. It is calculated by dividing the change in velocity by the change in time.

How are torque and acceleration related?

Torque and acceleration are related through Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. The net force can be calculated by adding up all the individual torques acting on the object.

How does torque affect rotational motion?

Torque is responsible for causing rotational motion. The greater the torque applied to an object, the greater the acceleration and the faster the object will rotate.

What are some real-life applications of torque and acceleration?

Torque and acceleration are essential concepts in many fields, including physics, engineering, and biomechanics. They are used to understand and design machines, vehicles, and sports equipment, among other things. For example, understanding torque and acceleration is crucial for designing a car engine that can produce enough power to move the vehicle efficiently.

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