What is the Final Angular Velocity of a Cylinder?

Click For Summary
The discussion centers on calculating the final angular velocity of a solid cylinder when a rope is pulled with a force of 100 N. The cylinder has a radius of 20 cm and a mass of 20 kg, and the moment of inertia is given as MR²/2. Initial calculations yield a torque of 20 Nm and an incorrect angular velocity of 22.36 rad/s, while the correct answer is 18.4 rad/s. Participants suggest using work-energy principles and emphasize the importance of understanding rolling motion equations. The conversation highlights the need for practice with dynamics problems, particularly those involving rotational motion.
James50
Messages
18
Reaction score
1
I've got a synoptic paper coming up and for the life in me I can't remember how to do some basic dynamics!

Homework Statement



A massless rope is wrapped several times around a solid cylinder of radius R = 20 cm, and mass M = 20 kg, which is at rest on a horizontal surface. Someone pulls 1 m or the rope with a constant force of 100 N, setting the cylinder in motion. Assuming that the rope neither stretches nor slips, and that the cylinder rolls without slipping, what is the final angular velocity of the cylinder of mass M and radius R. The moment of inertia of the cylinder is MR2/2.

Homework Equations



T = F.R
L = Iw
T = dL/dt

T is torque, F force, R radius, I moment of inertia, W angular speed, L angular momentum, Y is the angle

The Attempt at a Solution



Torque is 100x0.2 = 20 Nm. Moment of inertia is 0.4 kg m2. Now rearranging torque as a function of angular momentum, you get T = I dw/dt.
Some chain rule... dw/dt = w dw/dY. A little bit of integrating... 0.5 w2 = (T/I)Y -> using Y=0 as the lower boundary.

Okay, now because 1m is pulled from the rope, and the circumference of the cylinder is 1.256m, it works out that the sphere has rotated by 5 rads. Put that, along with T=20 and I=0.4 into the above equation, you come out with 22.36 rad /s. The actual answer is 18.4 rad /s. Ugh!
 
Physics news on Phys.org
Welcome to PF!

Hi James50! Welcome to PF! :smile:

(have a theta: θ and an omega: ω :wink:)
James50 said:
I've got a synoptic paper coming up and for the life in me I can't remember how to do some basic dynamics!

eek!

You clearly need to do a lot of worked examples on rolling motion. :redface:

(btw, what's a synoptic paper? :confused:)

With rolling, you need two types of equation, one for rotational motion and one for linear motion …

you also need a rolling constraint (usually v = rω).

And in this case, since you're given distance (instead of time), mightn't a work energy equation save you … erm … both work and energy? :wink:
 


tiny-tim said:
Hi James50! Welcome to PF! :smile:

(have a theta: θ and an omega: ω :wink:)



Thanks! I saw some symbols hanging around in the latex reference things but then it messed up all my paragraphing!

You clearly need to do a lot of worked examples on rolling motion. :redface:

(btw, what's a synoptic paper? :confused:)


You're telling me! Nuclear physics and baryon decuplets I can do... But rotating bodies!
A synoptic paper is a general one, it lasts for something like 4 hours and has questions on all the modules in my degree (I've done over 30, ugh!)

And in this case, since you're given distance (instead of time), mightn't a work energy equation save you … erm … both work and energy? :wink:

I can't believe I didn't think of doing it that way. Got the right answer straight away, thanks! Although I still can't figure out the normal way of getting there!
 
James50 said:
A synoptic paper is a general one, it lasts for something like 4 hours and has questions on all the modules in my degree (I've done over 30, ugh!)

hmm … I think I'd call that panoptic :wink:
Although I still can't figure out the normal way of getting there!

Do F = ma and τ = Iω.

Do a forum search for "rolling" to find some other examples to work through. :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
Replies
13
Views
2K
  • · Replies 17 ·
Replies
17
Views
874
Replies
12
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
3
Views
1K