Rotational Dynamics and string Tension

  • Thread starter Juniper7
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  • #1
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Homework Statement



A light string is wrapped around a solid cylinder and a 300g mass hangs from the free end of the string. when released, the mass falls a distance of 54cm in 3.0s.
a) Calculate the tension in the string.
b) Calculate the mass of the cylinder.

Homework Equations



y=y0 + v0t + 1/2at^2
I=mr^2
α=a/r
T-mg=ma
F=ma
τ=Iα
τ=rF

The Attempt at a Solution



a) y=y0 + v0t + 1/2at^2
0.54m = (0.5)a(3s^2)
a=0.12m/s^2

T-mg=ma
T-(0.3kg)(9.8m/s^2) = (0.3kg)(0.12m/s^2)
T=2.979N

b) τ=Iα
τ=mr^2α
Fr=mr^2α
F=mrα

I'm not sure if thats right for a), I feel like I'm missing something.... :frown:
I'm very confused for b). I have been playing around with the equations, as you can see above, but I keep going in circles and I feel like I don't have enough information. Thanks in advanced for any help!!
 

Answers and Replies

  • #2
SammyS
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Homework Statement



A light string is wrapped around a solid cylinder and a 300g mass hangs from the free end of the string. when released, the mass falls a distance of 54cm in 3.0s.
a) Calculate the tension in the string.
b) Calculate the mass of the cylinder.

Homework Equations



y=y0 + v0t + 1/2at^2
I=mr^2
α=a/r
T-mg=ma
F=ma
τ=Iα
τ=rF

The Attempt at a Solution



a) y=y0 + v0t + 1/2at^2
0.54m = (0.5)a(3s^2)
a=0.12m/s^2

T-mg=ma
T-(0.3kg)(9.8m/s^2) = (0.3kg)(0.12m/s^2)
T=2.979N

b) τ=Iα
τ=mr^2α
Fr=mr^2α
F=mrα

I'm not sure if that's right for a), I feel like I'm missing something.... :frown:
I'm very confused for b). I have been playing around with the equations, as you can see above, but I keep going in circles and I feel like I don't have enough information. Thanks in advanced for any help!!
Hello Juniper7. Welcome to PF !

First of all, there is at least one missing detail. Is the cylinder free to rotate on its axis ?

The form of the moment of inertial you have, I = mr2 is for a hollow cylinder, not a solid one.

You use both F and T, somewhat interchangeably.


Draw 2 free body diagrams; one for the hanging mass, and one for the cylinder?

Doing these things should help take the guesswork out of what equations to employ.
 
  • #3
19
0
Oh, ok. I think I have it now. The cylinder is freely rotating. Thanks!
 

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