Rotational Mechanics -- A solid sphere is rolled on a rough surface

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 1K views
Ayesha02
Messages
49
Reaction score
5
Homework Statement
A uniform solid sphere of mass M radius R is placed on a rough surface of given an initial linear velocity ##v_0## and angular velocity ##w_0##, where ##w_0## = ##v_0## /2R. plot the graph of angular speed about the center v/s time t.
Relevant Equations
##T## = I* alpha , where T= torque, alpha= angular acceleration
and
##w_0(f)## = ##w_0(i)##- alpha*t
I found out the time when rotation ceases to be 4 ##v_0## /5*mew*g, where mew=coefficient of friction of surface but I am unable to plot the graph post that time
 
Physics news on Phys.org
Ayesha02 said:
Homework Statement:: A uniform solid sphere of mass M radius R is placed on a rough surface of given an initial linear velocity ##v_0## and angular velocity ##w_0##, where ##w_0## = ##v_0## /2R. plot the graph of angular speed about the center v/s time t.
Relevant Equations:: ##T## = I* alpha , where T= torque, alpha= angular acceleration
and
##w_0(f)## = ##w_0(i)##- alpha*t

I found out the time when rotation ceases to be 4 ##v_0## /5*mew*g, where mew=coefficient of friction of surface but I am unable to plot the graph post that time
Is the initial angular velocity lower or greater than the one when the sphere rolls? The friction decreases linear velocity and either increases or decreases angular velocity, till rolling occurs. Determine both as functions of time, Show your work in detail.
 
Last edited:
ehild said:
The friction decreases linear velocity and increases angular velocity. Determine both as functions of time, Show your work in detail.

I just did
##T## = 2/5 M R^2 * α
which gives α =5μg/ 2R

hence
##w_0(f)## = ##w_0(i)## - α t
i.e 0= ##v_0##/2R - 5μg/ 2R *t

hence i got t=4##v_0##/5μg (time when rotation ceases)
 
Ayesha02 said:
I just did
##T## = 2/5 M R^2 * α
which gives α =5μg/ 2R

hence
##w_0(f)## = ##w_0(i)## - α t
i.e 0= ##v_0##/2R - 5μg/ 2R *t

hence i got t=4##v_0##/5μg (time when rotation ceases)
You kick a ball. After a short time, does it skid or does it roll?
 
ehild said:
You kick a ball. After a short time, does it skid or does it roll?

Skid initially, after some time rolls.
Im assuming the surface is rough
 
Ayesha02 said:
Skid initially, after some time rolls.
Im assuming the surface is rough
Yes. And what is the angular speed in case of pure rolling with respect to the linear velocity of the CM?
 
Last edited:
ehild said:
Yes. And what is the angular speed in case of pure rolling with respect to the linear velocity of the CM?

Angular momentum conservation

I got the answer:bow: