Rolling without slipping magnitude

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Homework Help Overview

The problem involves a solid uniform disk mounted on an inclined plane, with a string wrapped around it and attached to another disk. The objective is to find the magnitude of the acceleration of the center of mass of the rolling disk while neglecting friction in the axles. The discussion centers around the dynamics of the system, including forces and torques acting on both disks.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using Newton's second law and rotational dynamics to set up equations for the disks. There is a focus on the role of static friction and tension in the system, as well as the relationship between linear and angular acceleration.

Discussion Status

Some participants have provided insights into the forces acting on the disks and the nature of static friction. There is an ongoing exploration of the direction of static friction and its impact on the torque and acceleration of the disks. The discussion reflects a mix of interpretations regarding the forces involved.

Contextual Notes

Participants are grappling with the implications of static friction not necessarily being at its maximum value and the relationship between linear and angular quantities. There is uncertainty about the correct direction of static friction and its effect on the system's dynamics.

REVIANNA
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Homework Statement


problems_MIT_boriskor_BKimages_rolling_disc_attached_to_rotating_pulley.png
[/B]
A thin light string is wrapped around a solid uniform disk of mass M and radius r, mounted as shown. The loose end of the string is attached to the axle of a solid uniform disc of mass m and the same radius r which can roll without slipping down an inclined plane that makes angle θ with the horizontal. Find the magnitude of the acceleration of the center of mass of the rolling disc, a. Neglect friction in the axles of the pulley and the rolling disk.

The Attempt at a Solution


1- I would use ##F=ma## for the disc and ##τ=Iα## for the pulley to write to equations and use the condition ##a=αr##.

2-friction b/w the incline causes torque in the disc

3-##I=0.5mr^2##

4-the net torque on the pulley will have a cos(θ) term (to have only perpendicular elements).

what I don't know-
will only the force of static friction cause net torque in the pulley (or T will cause torque in
the pulley)

edit:
assuming T causes torque
##Tr cos(θ)=0.5 M r^2 α##
##mgsin(θ)+μmgcos(θ)-T=m*a##
using ##α r= a##

I solved the equations but μ and cos(θ) terms don't get canceled
the answer is just in terms of M,m and g and θ

pleasez help!
 
Last edited:
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REVIANNA said:

The Attempt at a Solution


I would use ##F=ma## for the disc and ##τ=Iα## for the pulley to write to equations and use the condition ##a=αr##.
You'll need to use more than just F = ma for the disk on the incline. It is rotating as well as translating.

friction b/w the incline causes torque in the disc
Which disc?
the net torque on the pulley will have a cos(θ) term (to have only perpendicular elements).
The string comes off of the pulley in a direction that is tangent to the rim of the pulley.

what I don't know-
will only the force of static friction cause net torque in the pulley (or T will cause torque in
the pulley)
The static friction force acts on the disc on the incline, not on the pulley. T is the only force applying a torque to the pulley.

Also, the static friction is not necessarily at its maximum possible value. You cannot assume ##f = \mu mg\cos \theta##.
 
Last edited:
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TSny said:
Which disc?
the one on the incline
will static friction point down the incline?
 
REVIANNA said:
the one on the incline
OK
will static friction point down the incline?
You should be able to decide this based on the direction of the angular acceleration of the disc on the incline.
 
TSny said:
direction of the angular acceleration of the disc on the incline.
the disc is rolling down the incline therefore static friction should also point down the incline (because it is the one causing the rotation)
but the answer turns out to be correct if I consider the static friction up the incline (like T)
angular acceleration is pointing out of the screen.(counter clockwise rotation)
what is wrong?
 
Yes, angular acceleration is pointing out of the screen. So, which way does the torque have to act on the disc?
 
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TSny said:
torque have to act on the disc
so stupid of me obviously up the incline! I am confusing net F (linear acceleration) with net torque(angular)
thanks a lot
 
OK. Good work.
 

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