Physics Help. A string is wrapped around a disk of mass m = 2.3 kg and radius R = 0.08 m.

Question

Starting from rest, you pull the string with a constant force F = 6 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance x = 0.12 m, your hand has moved a distance of d = 0.26 m. vcm (center of mass of velocity) = 0.78 m/s.

 

At this instant, how much rotational kinetic energy does the disk have relative to its center of mass? krot = ?

Answer 1

Hi Ab!

 

This problem is a little tricky, because the usual condition about rolling on the ground without slipping does not appear to apply (I will assume, however, that the rope comes off the disk without slipping; otherwise, all bets are off and the problem gets a *lot* more difficult with the given information).  If it did, then the disk would have rolled a linear distance identical to the amount your hand pulling the rope has moved.

 

Because we cannot (apparently) rely on rolling without slipping (and thus v_cm ≠ Rω), I will go to another principle:  the rotational work-energy principle.  This can be stated (in analogy to the linear W = ΔKE) as:

 

(W_rot)_net = ΔKE_rot

 

where 

 

(W_rot)_net= τ_netθ

 

with 

 

τ_net = net torque applied

θ = angular distance over which the torque is applied (in radians)

 

[compare to W_net = Fd for linear motion]

 

Now, the tension in the rope, which is what I am presuming is 6 N (constant), applies a torque to the disk.  Here is where I think a little hand-waving happens.  They need some friction for there to be rolling (if the surface were absolutely frictionless, the disk would just slide as we pulled the rope).  However, if there is slipping while rolling, the friction with the surface would also provide a torque.  However, that would be harder to deal with, and I think they mean for us to ignore that torque, and take the torque of the rope as the only one applied to the disk.  This is, I think, what they mean when they say the surface is "nearly frictionless:"  there is enough friction for the disk to roll a little, so that pulling the rope off the disk is possible, but not so much that we have to consider its torque on the disk.

 

So that leaves the tension's torque as the net torque on the disk.  Let's calculate a value for this torque, using the fundamental definition:

 

τ = Frsinθ

 

In this case, F is just the tension of 6 N.

r is the distance from the axis of rotation (which we choose to be at the center of mass, to get the rotational kinetic energy "relative to [the disk's] center of mass")

θ = the angle between F and r

 

Now, r is drawn from the axis of rotation (at the center of mass) to where the force is being applied to the disk (on the rim).  Hence, r = the radius of the disk (0.08 m).

 

Since r is in the radial direction (along the radius of the disk) and the tension force is being applied tangential to the disk at all times, the angle between F and r in this case is always 90^o, and sin(90^o) = 1.

 

So we have:

 

τ = (6 N)(0.08 m) = 0.48 N·m

 

Now, we have to get the angular displacement in of the disk in radians.  If we take the rope to be coming off the disk without slipping, then 0.26 m of rope coming off the outer rim of the disk corresponds to a point on the outer rim of the disk moving 0.26 m, as well.  What does this mean for how far the disk has rotated (its angular displacement, in radians)?

 

We can appeal to the foundational definition of an angle measure in radians:

 

θ = s/r

 

where

 

θ = angular distance rotated (in radians)

s = linear distance moved by a point on the rim

r = radius of the disk

 

So 

 

θ = (0.26 m)/(0.08 m) = 3.25 radiams

 

Then, bringing all this together, we have:

 

(W_rot)_net = τθ = (0.48 N·m)*(3.25 radians) = 1.56 J

 

So 1.56 J = ΔKE_rot = KE_rotf - KE_roti, and KE_roti = 0 [starts at rest]

 

Thus, 

 

1.56 J = KE_rotf  at the time in question.

 

I will check this again later to see if there are any errors, but I think the concept, at least, is correct for the information given.

 

Incidentally, the linear work done will equal the disks linear kinetic energy, which works out to give the v_cm indicated. 

 

If you have any questions about this, or want to look at details further, just let me know.  I hope this helps!

 

Answer 2

The rotational kinetic energy, Krot, is (1/2) I ω²  

For a disk  I =  (1/2) m R²  

For the case of rolling without frictions   ω =  vcm/R

 

Putting this together Krot = (1/4) m vcm²     This is one half the translational kinetic energy.

 

With the numerical values given Krot =  0.35 J