# Is energy conserved in this spring-pulley-block question?

1 vote
93 views I believe that energy isn't conserved in this question as the pulley has mass and tension on both sides would be different (due to friction?).

So my attempt is:

$T_2 = kx$
$mg- T_1 = ma$

$(T_1-T_2)R = I \alpha$ , $\alpha = \frac aR$

Where the symbols carry their usual meaning.

On solving for a, and using $v^2 = 2ah$, I get $v = 0.408 {m s^{-1}}$, but answer is 0.5 ms^-1.

I checked the solution and they had done it simply using $mgh = 1/2 mv^2 + 1/2 kx^2$ to get $v= 0.5 ms^{-1}$ as the right answer.

Please let me know whats wrong with my approach.

asked May 26, 2018
edited May 27, 2018

## 1 Answer

1 vote

Best answer

Both approaches are correct, but the energy method is quicker in this case.

Mechanical energy is conserved because there is no sliding (kinetic) friction. There is static friction between the string and the pulley, but this friction force does not dissipate energy because there is no relative motion (sliding) between the string and the pulley. The distance through which the point of application of the force moves is the same as the distance which the pulley moves. So all of the work done by the string is recovered by the pulley. There is no lost work in between.

The statement of energy conservation which you have given does not include the rotational KE of the pulley, which is $\frac12 I \omega^2$. However, I expect this was a typing error.

The error in your solution is your use of $v^2=2ah$. This equation applies only for constant acceleration. It does not apply here because in SHM acceleration $a$ is not constant. Like displacement and velocity it is sinusoidal with time.

To find displacement using your method :

$(T_1-T_2)R=I\alpha=I\frac{\ddot x}{R}$
$T_1-T_2=\frac{I}{R^2}\ddot x$

$T_2=kx$
$mg-T_1=m\ddot x$
$T_1-T_2=mg-m\ddot x-kx=\frac{I}{R^2}\ddot x$

$(m+\frac{I}{R^2})\ddot x+kx=mg$
$(m+\frac{I}{R^2})\frac{d^2}{dt^2}(\frac{mg}{k}-x)+k(\frac{mg}{k}-x)=0$
which is in the form for SHM :
$\ddot y+\omega^2 y=0$
with $\omega^2=\frac{k}{m+I/R^2}$ and $y=\frac{mg}{k}-x$.

Try a solution $y=\frac{mg}{k}-x=A\cos\omega t$.
At $t=0$ we have $x=0$ therefore $A=\frac{mg}{k}$.
Note that $\dot y=-\dot x=0$ as required.

So we have
$x=\frac{mg}{k}(1-\cos\omega t)$
$\dot x=\omega\frac{mg}{k}\sin\omega t$
Solve the 1st eqn for $\cos\omega t$ using $x=10cm$ and other values as in the question.
Then apply $\sin^2x+\cos^2x=1$ to find $\sin\omega t$.
Finally substitute into the 2nd eqn to find $\dot x$ when $x=10cm$.

This is considerably longer than the energy method.

answered May 26, 2018 by (28,746 points)
selected May 28, 2018