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Rotation and Translation

  • Thread starter hayowazzup
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  • #1
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[SOLVED] Rotation and Translation

Homework Statement


A uniform cylinder of mass M=21.3 kg and radius R = 18.1 cm rests on a horizontal table-top. The cylinder can rotate about a frictionless axle that has a light handle connected to it. A light string passes over a disk shaped pulley of mass mp=5.2 kg and radius r = 8.51 cm and connects a block of mass m = 11.3 kg to the cylinder handle as shown in the diagram. Upon release, the cylinder rolls without slipping on the table, the string moves without slipping over the pulley surface and the block descends with acceleration a = 2.42 m/s²

(a)What is the tension in the string immediately above the block?
(b)Determine the tension in the string directly connected to the light handle?

for the first one i got 83.4, but i dunno how to work out the second question


Homework Equations


I=rw?
T= (9.8-2.42)11.3


The Attempt at a Solution


..
 

Answers and Replies

  • #2
Doc Al
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Start by analzying the forces acting on the cylinder. The big difference here is that you have to worry about rotation as well as translation, so you'll be applying Newton's 2nd law twice. Hint: There are two horizontal forces acting on the cylinder. What are they?
 
  • #3
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oh, i guess one of them should be the friction of the cylinder, but i have no idea how to get it from its radius and mass
 
  • #4
Doc Al
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oh, i guess one of them should be the friction of the cylinder, but i have no idea how to get it from its radius and mass
Yes, friction and string tension are the two horizontal forces acting on the cylinder. You'll solve for the friction (if you need to find it, which you don't). For now, just label it "F" and write Newton's 2nd law for rotation and translation.

Be sure to draw yourself a simple free body diagram of the cylinder.
 
  • #5
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...........N
F<-- ( ) T-->
.........(mg)

Torq= I(a/r)
F=ma
hmm...I still don't get this
 
Last edited:
  • #6
Doc Al
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Torq= Iw
That should be:
[tex]\tau = I \alpha[/tex]
F=ma
OK. Now write these two equations as they apply to this problem, in terms of friction' and tension.

How does the angular acceleration relate to the translational acceleration?
 
  • #7
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a= radius * angular acceleration
2.42m/s^2 = 0.181m* ang acceleration
angl acc = 13.37rads^2


I = (1/2)*21.3kg*0.181^2
I= 0.3489

torque = I * angular acceleration
torque = 0.3489 * 13.37 = 4.66
 
Last edited:
  • #8
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Torque = I * angular acc
Torque = F*r
F=ma

Force 1= ma = 21.3k*2.42= 51.546
Force 2= Torque/ r = ((1/2)*M*r^2)*(a/r)/ r = 4.66/0.181 = 25.74
Ftotal= Force 1 + force 2= 51.546+25.74

can you tell me where it goes wrong?
by comparing it to the one in the book, they are close if i round it off to 2s.f.
 
Last edited:
  • #9
Doc Al
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What you call "Force 1" is really the net force. How does this relate to the friction' and tension?

Rework this a bit. There are two forces acting on the cylinder: Tension and friction. Label them T and F. Express all your equations in terms of those forces.

What's the net force? What torque does each force produce?
 
  • #10
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so the Net force= Tension - friction
Tension = Netforce + friction ?
 
  • #11
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Net force= Tension - friction = mass*acceleration= 21.3*2.42= 51.546
Friction= Torque/ r = ((1/2)*M*r^2)*(a/r)/ r = 4.66/0.181 = 25.74
Tension= Net force + Friction= 51.546+25.74
 
Last edited:
  • #12
Doc Al
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That's much clearer.
Net force= Tension - friction = mass*acceleration= 21.3*2.42= 51.546
OK.
Friction= Torque/ r = ((1/2)*M*r^2)*(a/r)/ r = 4.66/0.181 = 25.74
Good. Realize that F = 1/2 M r^2*(a/r^2) = 1/2 Ma = 1/2 (51.546)
Tension= Net force + Friction= 51.546+25.74
Looks good to me.
 

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