Tension, Spring and Moments of Inertia

1. Feb 20, 2017

rahularvind06

1. The problem statement, all variables and given/known data
A given pulley has a radius of 20 cm and moment of inertia 0.2 kgm2. The string going over it is attached to a vertical spring of spring constant 50 N/m on one end and a 1 kg mass on the other end. The system is released from rest with the spring at natural length. Find the speed of the block when it has descended 10 cm.
2. Relevant equations
Γ = r x F
Γnet=Iα
F=ma

3. The attempt at a solution
i let T2 be the tension in the rope pulling the spring upward and T1 be the tension pulling the mass upward.
then mg - T1 = ma ----- (1)
and T2 = kx

then for the pulley,
T1R -T2R =Iα
T1R -T2R = Ia/R
then T1 -T2 = 5a (substituting known values)
then T1 - kx = 5a ---- (2)

eliminating T1 between (1) and (2)
mg - kx - 5a = ma
substituting for m,k and x = 10 cm,
a = 5/6 m/s2
v=√2as
= 5/2√15

the correct answer, however is 0.5.

2. Feb 20, 2017

rahularvind06

EDIT: Also, i have seen valid solutions using energy conservation but, no solutions with tensions..

3. Feb 20, 2017

kuruman

Under what circumstances is this the correct relation between speed and acceleration?

4. Feb 20, 2017

rahularvind06

v²=u²+2as (constant acceleration)
Since motion starts from rest, u=0.

5. Feb 20, 2017

kuruman

Is this equation always applicable or only under a certain assumption?

6. Feb 22, 2017

rahularvind06

isnt this applicable whenever acceleration is constant? or am i missing something...

7. Feb 22, 2017

Staff: Mentor

Yes. And, apparently yes.

Is acceleration constant here? What happens as the spring stretches?

8. Feb 22, 2017

rahularvind06

I get it now, so as the string stretches, the restoring force increases and so does the deceleration from the spring.

9. Feb 22, 2017

rahularvind06

How would you go about calculating the velocity in this case?
I know dV=a dt but i dont know how a varies with time...

10. Feb 22, 2017

Staff: Mentor

You'll end up having to write and solve the differential equation for the motion. Considerably more effort than just using conservation of energy.

11. Feb 22, 2017

kuruman

Exactly my advice. However, there is a shortcut. You know that the motion will be harmonic oscillations obeying the SHM equation
$$\frac{d^2x}{dt^2}=-\frac{k}{m_{eff}}~x=-\omega^2~x$$
A few points to consider if you choose to go this way are,
1. The hanging mass will execute harmonic oscillations about its equilibrium point; you need to find where that is.
2. You need to find the amplitude of oscillations about the equilibrium point.
3. The "$m_{eff}$" in the denominator is the "effective" mass of the oscillator that takes into account the moment of inertia of the pulley. In other words, it is the mass you need to have to get the same frequency of oscillations $\omega$ in the case of a massless pulley.
Once you do all that, you can write down x(t) and find what you need. Still, considerably more effort compared with using energy conservation because you end up deriving expressions that you don't have to derive.

12. Feb 22, 2017

rahularvind06

got it now, thanks everybody for the help!