What is the tension in the string immediately above the block?

In summary, a uniform cylinder with a mass of 21.3 kg and a radius of 18.1 cm is connected to a block of mass 11.3 kg through a light string that passes over a disk shaped pulley with a mass of 5.2 kg and a radius of 8.51 cm. The cylinder is able to rotate about a frictionless axle and upon release, it rolls without slipping on a horizontal table while the string moves without slipping over the pulley and the block descends with an acceleration of 2.42 m/s². The tension in the string above the block is 83.4 N and the tension directly connected to the light handle is 77.3 N. The forces acting
  • #1
hayowazzup
64
0
[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 don't know how to work out the second question


Homework Equations


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


The Attempt at a Solution


..
 
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  • #2
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
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
hayowazzup said:
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
...N
F<-- ( ) T-->
...(mg)

Torq= I(a/r)
F=ma
hmm...I still don't get this
 
Last edited:
  • #6
hayowazzup said:
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
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
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
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
so the Net force= Tension - friction
Tension = Netforce + friction ?
 
  • #11
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
That's much clearer.
hayowazzup said:
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.
 

1. What is the difference between rotation and translation?

Rotation is a type of transformation that involves rotating an object around a fixed point, whereas translation involves moving an object from one location to another without changing its orientation.

2. How are rotation and translation used in real-world applications?

Rotation and translation are used in various fields such as robotics, computer graphics, and engineering to manipulate and control the movement of objects.

3. Can rotation and translation be combined to create more complex transformations?

Yes, rotation and translation can be combined together to create more complex transformations known as rotational translation or rigid body motion. These transformations are commonly used in 3D animation and modeling.

4. What are the mathematical representations of rotation and translation?

Rotation can be represented using angles, matrices, or quaternions, while translation can be represented using vectors.

5. How do rotation and translation affect the shape and size of an object?

Rotation and translation do not affect the shape and size of an object, as they only change its position and orientation in space.

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