What is the acceleration of a cylinder when a torque is applied to it?

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SUMMARY

The discussion focuses on calculating the acceleration of a cylinder when a torque is applied, specifically a 5 kg cylinder with a radius of 0.1 meters and a static friction coefficient of 0.6. An external torque of 2 Newton meters is applied, leading to a calculated acceleration of 2.67 m/s². The solution involves using the equations of motion and the relationship between torque, friction, and angular acceleration, confirming that the initial assumption of rolling without slipping is valid.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with the moment of inertia formula, I = 0.5MR²
  • Knowledge of static and kinetic friction coefficients
  • Basic principles of Newton's second law for rotation and translation
NEXT STEPS
  • Study the relationship between torque and angular acceleration in rotational motion
  • Learn about the conditions for rolling without slipping
  • Explore the concept of static friction and its role in rotational dynamics
  • Review the equations of motion for both linear and rotational systems
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Students in physics or engineering, particularly those studying mechanics, as well as educators looking for examples of rotational motion and torque applications.

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Homework Statement


Got a spicy one for you today.
There is a cylinder of mass 5kg (M) with no angular velocity and no velocity, on a surface of static friction constant μ=0.6 . It's radius is .1 meters (R). Its baricentric moment of inertia is characterized by I=.5MR^2 . Gravity is 10m/s. Kinematic μ=0.4 is also given of the flat surface.

At time t=0s, torque is applied by an external force of a magnitude of η=2 Newton meters so that it begins to roll.

Find the acceleration of the center of mass of the cylinder.
The answer in the book is a=2.67m/s^2
Capture.PNG

Homework Equations


Rototranslation
Σmoment of inertia = Iα
ma=Σforces
αR=a

The Attempt at a Solution


Alright, so I began assuming it would begin rolling without sliding, thus we use static μ.
I sum up the inertias like so to find the angular acceleration of the cylinder.
η-μR=Iα ⇒ R(η-μR)/I=a
This gives me a negative number. I am confused. The answer in the book is a=2.67m/s^2

Why is it that simply using inertias does not work? I hear it might be done setting a fixed point and using angular momentum, but the problem is in the Rototranslation chapter.

Edit:
I believe I managed to solve it:

Ffriction=ma α=a/r
Mcupla-Ff*r=mr^2*α/2 ⇒ Mc-mar=mr*a/2 ⇒ Mc= 3mar/2 ⇒ a=2Mc/(3mr) ⇒ a=2.67m/s^2
 
Last edited:
Physics news on Phys.org
With regard to your first attempt, if the coefficient of static friction is ##\mu_s## and the normal force is N, what is the force of static friction? Think carefully before answering.
 

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