Register to reply 
Rolling/sliding with friction 
Share this thread: 
#1
Mar507, 06:45 AM

P: 296

1. The problem statement, all variables and given/known data
A sphere(mass m) is sliding on a frictionless surface until it meets a surface with friction(coefficient mu). This causes the sphere to start rotating, while it still slides. Calculate the speed of the sphere while it is both sliding and rotating on the surface with friction. 3. The attempt at a solution My first attempt was just to consider a constant frictional force acting on the sphere, and hence the speed is V=V0mu*g*t But then I heard that I also had to take into account that the sphere starts to rotate, and then the above is incorrect. I'd say that the frictional force on the sphere is the same no matter how fast it is rotating. Can anyone help me? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Mar507, 10:17 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

Don't worry about the individual forces look at the energy.
As long as the sphere is sliding, it's energy is its tranlational kinetic energy: 1/2 mv^{2}. When the sphere starts rolling without slipping, it has both translational kinetic energy and rotational kinetic energy which must add to give the above. When "rolling without slipping, note that when the sphere has made one complete rotation, which takes [itex]t= 2\pi/\omega[/itex] seconds, where [itex]\omega[/itex] is the angular speed, in radians per second, the center of the sphere has moved a distance equal to the circumference of the sphere, [itex]2\pi R[/itex]. At speed v, that will take time [itex]t= 2 \pi R/v[/itex] so that [itex]\omega= v/R[/itex]. Write the rotational kinetic energy in terms of v, set the sum of the two kinetic energies equal to the initial kinetic energy and solve for v as a function of the original sliding speed. You can't calculate both without some additional information. 


#3
Mar507, 11:56 AM

Mentor
P: 41,568

(A) The sphere slides on the frictionless surface (no rotation) (B) The sphere meets the frictioned surface and begins rotating as well as sliding (C) The sphere reaches a speed such that it rolls without slipping I assume that you only care about phase B? If so, the question is as easy as you think. (Usually one is asked to find the speed at which phase Crolling without slippingbegins; that takes a little more effort.) Be sure you have described the problem exactly as given. The sphere loses total KE due to friction as it slips along the surface in phase B. 


#4
Mar507, 01:02 PM

P: 296

Rolling/sliding with friction
Okay. Actually there is a second part of the question where I have to determine how much heat is generated during phase B, when the sphere is sliding. In this question, do I have to take into account that the sphere is rotating? Otherwise I think the solution is the following:
The angular speed of the sphere is: omega=5/(2*R)*g*mu*t Which means that phase C starts when v=R*omega: v0mu*g*t=R*(5/(2*R)*g*mu*t) => t=2/7*v0/(mu*g) Which means that(in phase B) the sphere slides a distance of x = v0*t  1/2*mu*g*t^2 = v0*(2/7*v0/(mu*g))  1/2*mu*g*(2/7*v0/(mu*g))^2 = 12/49*v0^2/(mu*g) The total work due to friction is Wf = Ff*x = m*g*mu*x = 12/49*v0^2*m Or am I missing that the sphere is rotating? Because I guess if the sphere was spinning really fast, it would generate more heat. 


#5
Mar507, 01:32 PM

Mentor
P: 41,568

But rather than try to calculate the work directly, why not just calculate the change in KE? The mechanical energy lost will equal the amount of thermal energy generated by friction. That's the easy way! 


#6
Mar507, 01:45 PM

P: 296

So the kinetic energy before phase B is
Kb = 1/2*m*v0^2 After phase C there is kinetic energy due to the translation and due to the rotation Ke = 1/2*m*v^2+1/2*I*omega^2 where the moment of inertia is 2/5*m*R^2 for the sphere. From my previous equations the speed v in phase C is v=5/7*v0 and omega = v/R = 5/(7R)*v0 hence Ke = 1/2*m*(5/7*v0)^2+1/2*2/5*m*R^2*(5/7*v0/R)^2 = 5/14*m*v0^2 Which makes the work done by friction W=1/7*m*v0^2 Right? Thank you so far, you are great ! 


Register to reply 
Related Discussions  
Direction of Rolling/Sliding Friction  Introductory Physics Homework  5  
Sliding friction  Introductory Physics Homework  5  
Sliding, and rolling, w/ w/o friction  Classical Physics  1  
Sliding to Rolling motion problem.  Introductory Physics Homework  8  
Sliding/rolling billiard balls  General Physics  8 