Tension, Spring and Moments of Inertia

[rahularvind06]

Homework Statement

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.

Homework Equations

Γ = r x F
Γ_net=Iα
F=ma

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.
any conceptual errors or oversights in my answer? please help..

 

[rahularvind06]

Homework Statement

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.

Homework Equations

Γ = r x F
Γ_net=Iα
F=ma

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.
any conceptual errors or oversights in my answer? please help..

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

 

[kuruman]

v=√2as

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

 

[rahularvind06]

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

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

 

[kuruman]

v²=u²+2as

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

 

[rahularvind06]

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

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

 

[gneill]

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

Yes. And, apparently yes.

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

 

[rahularvind06]

Yes. And, apparently yes. [emoji2]

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

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

 

[rahularvind06]

Yes. And, apparently yes. [emoji2]

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

How would you go about calculating the velocity in this case?
I know dV=a dt but i don't know how a varies with time...

 

[gneill]

How would you go about calculating the velocity in this case?
I know dV=a dt but i don't know how a varies with time...

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

 

[kuruman]

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

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.

 

[rahularvind06]

got it now, thanks everybody for the help!