Velocity of a body in a pulley problem involving a spring fixed to the ground

Question

So the picture shows the whole situation. My question is, how do you find the velocity of the object after it had descended 10 cm (0.1 m) ? Pulley's radius 0.2 m, moment of inertia 0.2 kg/m

I have a problem because the pulley is supposed to have friction if it's rotating,right? Because the torques on either side shouldn't be equal. My teacher used $$ \frac {Iw^2}{2} + \frac {mv^2}{2} + \frac {kx^2}{2} = mgh$$ Where $$ v=rw $$ and $$h=0.1m $$ as given. So the answer came as 0.5 m/s. My problem is, if there is friction why aren't we taking the energy loss due to that into account?

$mgh $ would have to be spent on that as well. Plus where do the individual tensions on each side come into play? I understand that $mgh $ kind of accounts for those tensions, but not exactly how.

If mgh is doing work on the pulley, does that include work done by/against tension as well as the energy loss due to friction?

Plus, tension on one side is stretching the spring, right? So why shouldn't we consider that separately?

If somebody could clarify I'd be grateful. Thanks so much. No working out needed, I just want to know exactly what happens to the energy

Answer 1

Firstly, for clarity add an $x$-axis.

You correctly, in the absence of friction, wrote the Total Energy (Hamiltonian) equation as:

$$\frac {Iw^2}{2} + \frac {mv^2}{2} + \frac {kx^2}{2} = mgh$$

With the new $x$-axis:

$$\frac {Iw^2}{2} + \frac {mv^2}{2} + \frac {kx^2}{2} = mgx$$

You also correctly stated that $v=\frac{dx}{dt}=r\omega$, so that with substitution we get:

$$\big(\frac{I}{2r^2}+\frac{m}{2}\big)v^2+\frac{kx^2}{2}=mgx$$

For simplicity, set:

$$\alpha=\big(\frac{I}{2r^2}+\frac{m}{2}\big)$$

So:

$$\alpha v^2+\frac{kx^2}{2}=mgx$$

Now derive both sides to $t$, which gives us the Newtonian equation of motion:

$$2\alpha v\frac{dv}{dt}+kx\frac{dx}{dt}=mg\frac{dx}{dt}$$

$$2\alpha va+kxv=mgv$$

$v$ cancels out, so:

$$2\alpha a+kx=mg$$

Or:

$$2\alpha \frac{d^2x}{dt^2}+kx-mg=0$$

Which has the general solution:

$$x(t)=\frac{mg}{k}+c_1\sin(\sqrt{\frac{k}{2\alpha}}t)+c_2\cos(\sqrt{\frac{k}{2\alpha}}t)$$

Use initial conditions (e.g. $t=0$, $v=0$, $x=x_0$) to determine $c_1$ and $c_2$:

$$x(t)=\frac{mg}{k}+\big(x_0-\frac{mg}{k}\big)\cos\big(\sqrt{\frac{k}{2\alpha}}t\big)$$

So the mass enters into a simple harmonic oscillation.

Answer 2

If there is friction why aren't we taking the energy loss due to that into account?

The energy loss due to friction has been considered by the term of $\frac 12 I\omega^2$. In fact, some portion of reduction of potential energy of the object $m$ is used (wasted) to rotate the disc by friction.

I think you can understand the role of the friction on the object $m$ motion better if we check the question in other (difficult) way.

At first, let’s draw free body diagram for object $m$:

$$mg-T=ma\;\tag 1$$ Then, the string:

The magnitude of the force that is exerted on the string by the spring is equal to $kx$. Because the string is inextensible. As the string is massless ($I_\textrm {string}=0$), sum of torques that are exerted on it about any arbitrary point must be zero. $$\Sigma M_\mathrm O=I_\textrm {string}\alpha=0 \; \Longrightarrow \; Tr-kxr-\int_{\mathrm A-\mathrm B}r\mathrm dF_f=0$$ $$\Longrightarrow \; \int_{\mathrm A-\mathrm B}r\mathrm dF_f=(T-kx)r\;\tag 2$$ Now, the disc:

All forces except $\color{red}{\mathrm dF_f}$s cross from the point $O$. $$\Sigma M_\mathrm O=\int_{\mathrm A-\mathrm B}r\mathrm dF_f=I\alpha\;\tag 3$$ As you are seeing, the friction force appeared in equations $2$ and $3$.

According to non-slipping condition, we have: $$a=r\alpha\;\tag 4$$ Now, we can calculate the acceleration $a$. $$a=\large{\frac 1{\large{1+\large{\frac I{mr^2}}}}}\left(g-\frac km x\right)\;\tag 5$$ If we replace known amounts, then: $$a=\frac 53-\frac{25}3 x$$ Finally we can find speed $v$ by using the formula $vdv=adx$. $$v^2=\frac{10}3x-\frac{25}3x^2$$ $$x=0.1 \; \Longrightarrow \; v=0.5 \;\textrm{m/s}$$ If there was no friction between string and disc, then equation $2$ was changing to $T-kx=0$ and equations $3$ and $4$ weren’t be apparent (because if there is no friction, then the disc won’t rotate). Under this condition (no friction), the acceleration $a$ will be: $$a=g-\frac km x\;\tag 6$$ Hence, we can obviously see the efficacy of the friction on the motion of the object $m$ by comparing equations $5$ and $6$.

We can also see this efficacy by considering your teacher’s equation. If there was no friction between string and disc, then that equation was changing to $\frac 12 mv^2+\frac 12 kx^2=mgx$.

Answer 3

You have a few questions here. They mostly concern the energy loss due to friction. In this problem there is no energy loss because of friction.

Friction is present between the rope and the pulley. This is static friction. It makes the pulley rotate. Energy is lost due to friction (dissipated as heat) only when there is relative motion between the surfaces in contact - ie kinetic friction. In this problem we assume the rope does not slip against the pulley - there is no relative motion. So there is no work done by the rope against the force of friction, and therefore no loss of energy as the rope makes the pulley rotate.

(There might also be friction between the pulley and the axle on which it rotates. Work is done - and energy lost - in that case, because there is relative motion between the surfaces in contact - the pulley slides over the axle. But the question assumes the energy loss in this case is negligible.)

I think you are concerned that there must be energy lost because the tension is different on both sides of the pulley. These tensions are different, but this does not mean that energy is lost because the rope is doing work against friction. No work is done against friction, because the rope does not slip. The tension in the rope is doing useful work accelerating the pulley, giving it kinetic energy. This energy is not lost. (See my answer to "Tension in a string with pulley and two objects at opposite ends" https://physics.stackexchange.com/q/264798.)

The tensions in the rope and spring (and the torque on the pulley) are brought into the equations only if you are trying to solve the problem by applying F=ma. The alternative method (used by your teacher) is to apply Conservation of Energy. This method ignores the forces and looks instead at the effect they have in storing and transferring energy - ie stretching the spring or raising the mass or making the mass and pulley move faster/slower.

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Back to your 1st question :

how do you find the velocity of the object after it had descended 10 cm?

The answer is : use the equation for conservation of energy which your teacher wrote. The RHS tells you how much gravitational PE the system loses when the mass descends by $h$. That is equal to the LHS, which is the elastic PE gained by the spring plus the KE gained by the pulley and the mass. Because the rope is inelastic, the extension of the spring is $x=h$. Because the rope does not slip against the pulley then $v=r\omega$. Substitute in the equation to find v, the velocity of the 1kg mass.

There is no need to solve the equation of motion to find x(t).