# Inclined plane with pulley and spring

• lorenz0
In summary: So the force on the mass is ## \vec{T_1}=T_1 \hat r=T \cos \alpha \hat r ## and the force on the pulley is ## \vec{T_2}=T_2 \hat r=T \sin \alpha \hat r ##.Very good. :) Let ## T ## be positive: ## T=ks+I_o s''/r^2 ## and ## m_p g \sin{\alpha}-T=m_p s'' ##. I agree with your first expression, but I think the signs are incorrect on the second.Thanks. So, ##\begin{cases}(T-ks)r=I_0 (-\alpha)=-I_
lorenz0
Homework Statement
Consider a pulley which is a ring of radius ##r##.
1) Find the moment of inertia of the pulley with respect to an axis through its center and perpendicular to its plane, knowing that the work needed to make the pulley rotate with angular momentum ##L## around this axis is ##W##.
Now this pulley is placed on a frictionless inclined plane of angle ##\alpha##, and an ideal rope connects a mass ##m_p## to a spring placed vertically to the left of the inclined plane, and with one end attached to the ground.
2) Find the length of the spring at equilibrium.
Starting from the equilibrium position we give to mass ##m_p## a speed ##v_i## parallel to the inclined plane and downwards.
If the rope doesn't slide on the pulley and always remains taut, find:
3) Period of the oscillation of the system.
4) The maximum and minimum value of the tension applied to the spring and to mass ##m_p##.
5) The maximum value of the kinetic energy of the system.
Relevant Equations
##\sum_{i}\vec{F}_i=m\vec{a},\ W=\Delta K=K_f-K_i,\ F_{spring}=-ks,\ T=\frac{2\pi}{\omega}.##
1) By the Work-Energy Theorem, ##W=K_f-K_i=\frac{1}{2}I_{0}\omega^2=\frac{L^2}{2I_0}.##
2) By assuming that the initial length of the spring is ##0##, calling its final length ##S## and ##T## the tension in the rope connecting the pulley and mass ##m_p## I have: ##\begin{cases}(kS-T)r=0\\ m_p g\sin(\alpha)-T=0 \end{cases}## which gives ##S=\frac{m_pg\sin (\alpha)}{k}.##
3) mass ##m_p## oscillates around the equilibrium position, so by denoting ##s## the displacement of ##m_p## from the equilibrium position I get ##-ks=m_p s''\Rightarrow s''=-\frac{k}{m_p}s\Rightarrow \omega=\sqrt{\frac{k}{m_p}}\Rightarrow T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m_p}{k}}. ##

Is it justifiable to ignore ##T## and ##-ks## since they "cancel out" each other because of the motion starting from the equilibrium position found before?

Thanks.

#### Attachments

• inclined plane.png
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What about the pulley and its effect on the frequency? They should have given you a value for the mass of the pulley=call it ## m_o ##.

In addition, for part (3), I think the forces on ## m_p ## are more complex than what you have. On ## m_p ## you have the tension from the rope, and the force of gravity, i.e. the component along the incline. You then have the tension in the rope, (that is not the same everywhere), affected by how much the spring is stretched, along with a part that is taken up by the angular acceleration of the pulley.

Suggestion is to write the forces ## F ## on ## m_p ##, so that ## F=m_p s'' ## with tension ## T=-ks-torque/r ##, where ## torque=dL/dt ##, etc.

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What about the pulley and its effect on the frequency? They should have given you a value for the mass of the pulley=call it ## m_o ##.

In addition, for part (3), I think the forces on ## m_p ## are more complex than what you have. On ## m_p ## you have the tension from the rope, and the force of gravity, i.e. the component along the incline. You then have the tension in the rope, (that is not the same everywhere), affected by how much the spring is stretched, along with a part that is taken up by the angular acceleration of the pulley.
From ##\begin{cases}(ks-T)r=I_0 \alpha=\frac{I_0}{r}a\\ T-m_p g\sin(\alpha)=m_p a \end{cases}## I get ##a=-\frac{m_p}{m_p+\frac{I}{r^2}}g\sin(\alpha)+\frac{k}{m_p+\frac{I_0}{r^2}}s## so ##\omega=\sqrt{\frac{k}{m_p+\frac{I_0}{r^2}}}## thus ##T=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{\frac{k}{m_p+\frac{I_0}{r^2}}}}## but I am unsure about whether I have chosen the correct positive and negative signs of the various quantities.

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Very good. :) Let ## T ## be positive: ## T=ks-I_o s''/r^2 ## and ## m_p g \sin{\alpha}-T=m_p s'' ##. I agree with your first expression, but I think the signs are incorrect on the second.
(## s ## is positive to the right, down the incline).

Edit: We later determined the first expression should read ## T=ks+I_o s''/r^2 ##. See posts 8 and 9 below.

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lorenz0
Very good. :) Let ## T ## be positive: ## T=ks-I_o s''/r^2 ## and ## m_p g \sin{\alpha}-T=m_p s'' ##. I agree with your first expression, but I think the signs are incorrect on the second.
Thanks. So, ##\begin{cases}(T-ks)r=I_0 (-\alpha)=-I_0\frac{a}{r}\\ m_p g\sin(\alpha)-T=m_p a\end{cases}\Rightarrow a=\frac{m_p}{m_p-\frac{I_0}{r^2}}g\sin(\alpha)-\frac{k}{m_p-\frac{I_0}{r^2}}s\Rightarrow \omega=\sqrt{\frac{k}{m_p-\frac{I_0}{r^2}}}\Rightarrow T=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{\frac{k}{m_p-\frac{I_0}{r^2}}}}##: is this correct?

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Try your algebra again: I get a plus sign on the ## I_o/r^2 ##.

Try your algebra again: I get a plus sign on the ## I_o/r^2 ##.
I had forgotten the ##-## sign on ##\alpha## in the first equation. With that, ##a=\frac{m_p}{m_p-\frac{I_0}{r^2}}g\sin(\alpha)-\frac{k}{m_p-\frac{I_0}{r^2}}s##

My mistake=I believe the first expression should read ## T=ks+I_o s''/r^2 ##. Then it should get the correct sign on ## I_o ##. Edit: Let me double-check that...

lorenz0
Yes, that is the problem. I think this error resulted because the tension ## T ## in the very right segment of the rope between the pulley and the mass is peculiar in that it pulls the mass ## m_p ## to the left, but it will pull the pulley to the right.

The tension ## T ## is not a vector. From ## T ## the force vector is determined, and it points in different directions for the two items. It points towards the rope in each case.

lorenz0
and be sure and see the last couple of sentences I added to the previous post.

lorenz0
To get the first expression with the correct signs, it helps to look at just the pulley and consider the rope(s) anchored to it with one rope pulling to the left with force ## ks ## and the other pulling to the right with tension ## T ##. We then get ## I_o s''/r^2=T-ks ##, which agrees with post 8.

## What is an inclined plane with pulley and spring?

An inclined plane with pulley and spring is a simple machine that combines the functions of an inclined plane, a pulley, and a spring. It is used to lift objects to a higher level by reducing the amount of force needed to lift the object.

## How does an inclined plane with pulley and spring work?

The inclined plane with pulley and spring works by using the inclined surface to reduce the amount of force needed to lift an object. The pulley then redirects the force to a spring, which stores the energy and releases it to lift the object.

## What are the advantages of using an inclined plane with pulley and spring?

The main advantage of using an inclined plane with pulley and spring is that it reduces the amount of force needed to lift an object. This makes it easier to lift heavy objects and can also reduce the risk of injury. Additionally, the use of a pulley and spring allows for a more controlled and precise lifting motion.

## What are some real-life applications of an inclined plane with pulley and spring?

An inclined plane with pulley and spring can be found in various applications, such as elevators, cranes, and garage doors. It is also commonly used in physical therapy and rehabilitation equipment to assist with exercises and movements.

## What are the limitations of an inclined plane with pulley and spring?

One limitation of an inclined plane with pulley and spring is that it can only lift objects to a certain height. Additionally, the spring may lose its elasticity over time, reducing its effectiveness. It also requires proper maintenance and regular inspections to ensure safe and efficient operation.

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