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Rotational Mechanics Problem (Rotating solid wooden disk)

  • Thread starter savva
  • Start date
  • #1
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I have the working for part d) but I don't understand certain parts, so my problems lie with part d), f) and g)

Homework Statement


3. A solid wooden disk is suspended as shown: it has mass 7.70 kg and radius
0.219 m . It is rotated from rest about its vertical axis, reaching an angular velocity of 3.69 rad/s in 15.1 s.

http://i41.tinypic.com/2u9mgzs.jpg

Calculate
(a) the average angular acceleration
(b) the angular momentum.
(c) the rotational kinetic energy

A 0.17 kg metal block is gently placed on the disk 0.140 m from the rotational axis.
(d) Show that the new angular velocity is 3.62 rad/s
(e) Calculate the new rotational kinetic energy.

The wooden cylinder is very slowly accelerated, and when the new angular velocity is 4.02 rad/s the block is just about to slide.

Calculate
(f) the frictional force on the block.
(g) the coefficient of static friction μs between the block and the disk.

Homework Equations


a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t
I(Disk) = MR^2/2
L=I[itex]\omega[/itex]
Krot = 1/2I[itex]\omega[/itex]^2
W2 = I/I2 x [itex]\omega[/itex]

The Attempt at a Solution



(a) a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t = 3.69/15.1 = 0.244 rad/s^2

(b)
I(Disk) = MR^2/2 = (7.70x0.219^2)/2 = 0.185 kg m^2
L=I[itex]\omega[/itex] = 0.185x3.69 = 0.681 kg m^2 s^-1

(c) Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.185)(3.69)^2 = 1.257J

(d)
I know L1 = L2, can we prove that this is the case if we didn't have given the value of angular velocity?
[itex]\tau[/itex]NET = 0 (Not sure how I can prove this)

I2 = I1 + MR^2 (Inertia of block)
I2 = 0.185 + 0.17(0.14)^2 = 0.188 kg m^2

[itex]\omega[/itex]2 = (I1/I2)x[itex]\omega[/itex] = (0.185/0.188)x3.69 = 3.62 rad/s

(e)
Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.188)(3.62)^2 = 1.23J

(f) & (g) no idea how to work out, can anyone please give me a hand?
 

Answers and Replies

  • #2
5
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Although the wooden disk is rotating, the block's motion is circular, that means there must be a force (centripetal force) that keeps it going in circles. In this particular example, it's the friction.

Fμ = v2 / R
 
Last edited:
  • #3
39
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I tried using the formula given a few different ways, but can't seem to get the correct answer:

so, Fμ = v^2 / R

using v=ωr = 4.02x0.14 = 0.5628 ms^-1

We have R = 0.14m from previous part

Subbing that in we get

Fμ = (0.5628)^2/0.14 = 2.26N

The answer for this part is 0.385N

Why am I not getting this answer? :frown:
 
  • #4
5
0
I tried using the formula given a few different ways, but can't seem to get the correct answer:

so, Fμ = v^2 / R

using v=ωr = 4.02x0.14 = 0.5628 ms^-1

We have R = 0.14m from previous part

Subbing that in we get

Fμ = (0.5628)^2/0.14 = 2.26N

The answer for this part is 0.385N

Why am I not getting this answer? :frown:
Sorry my fault..

It's fμ = m v2/r

I forgot to mention the mass. :tongue: Newton's law is F = ma after all, and the acceleration in this one is the centripetal acceleration that keeps the block moving in circles a = v^2/r


Now that you know the friction, you should be able to calculate the coefficient of fiction
 
  • #5
39
0
Yep, I got both answers correct now. I'll post up my working tomorrow. Thanks alot for your help!
 

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