Maximum tension in thread connecting loads

[Bling Fizikst]

Homework Statement

see below

Relevant Equations

see below

Let's say the center of mass of the ring (wheel) moved a distance $x$, tension in the thread connecting wheel and $A$ be $T$ and tension in the thread connecting $A,B$ be $T'$ :
Due to no slip condition , $a=R\alpha$ where $R$ is the assumed radius of wheel
Since $P$ is IAOR , hence we can balance torque about it :
$$\tau_P= (T-kx)R=2mR^2\alpha \implies T-kx = 2ma$$
Force eqn on wheel : $$T-f-kx=ma$$
Force eqn on loads $m,3m$:
$$T'-T=ma$$ $$3mg-T'=3ma$$

Simplifying them gives :
$$3mg-kx=6ma$$ --(1) $$T'=3ma+kx$$
From (1) , we can say $$x=A\cos\omega t$$ where $\omega=\sqrt{\frac{k}{6m}}$
Using this : $$T'\leq \frac{kA}{2}$$
How do i find $A$? I am not sure i know the boundary conditions for it

 

[haruspex]

Force eqn on wheel : $$T-f-kx=ma$$

Since that introduces f, it won’t give you any more info than you have from your torque equation.

Force eqn on loads $m,3m$:
$$T'-T=ma$$

You've left something out.

From (1) , we can say $$x=A\cos\omega t$$

What is the value of x at equilibrium?

 

[Bling Fizikst]

Since that introduces f, it won’t give you any more info than you have from your torque equation.

You've left something out.

What is the value of x at equilibrium?

I made two blunders . Here is my re attempt :
The force eqn on $m$ : $$T'+mg-T=ma$$
Using the other equations , we get :
$$T'=3ma+kx-mg$$ $$ 6ma=-kx+4mg$$

The differential eqn can be written as :
$$6m\ddot{x}= -k\left(x-\frac{4mg}{k}\right)$$ $$\equiv 6m\ddot{u}=-ku$$ where $u=x-\frac{4mg}{k}$ and $\omega=\sqrt{\frac{k}{6m}}$

Hence , $$u=A\cos\omega t \implies x=A\cos\omega t +\frac{4mg}{k}$$
We know that intially the wheel was at rest . Hence , $x(0)=0$
Using this , we find : $$A=-\frac{4mg}{k}$$
Plugging $x, \ddot{x}=a$ in the eqn for $T'$ and simplifying:
$$T'= -2mg\cos\omega t +3mg \leq 5mg\implies T'\leq 5mg$$

 

[haruspex]

Hence , $$u=A\cos\omega t \implies x=A\cos\omega t +\frac{4mg}{k}$$

The general form has a sine term as well …

We know that initially the wheel was at rest .

… which eliminates the sine term, but does not lead to …

Hence , $x(0)=0$

… which comes from your definition of where x is being measured from.

$$T'= -2mg\cos\omega t +3mg \leq 5mg\implies T'\leq 5mg$$

Two things:

• It is not quite enough to deduce $T'\leq 5mg$. You should note that this value is achieved.
• You should also check that $T'$ is never negative, otherwise SHM is not sustained, possibly leading to a higher tension at some point.

 

[Bling Fizikst]

The general form has a sine term as well …

… which eliminates the sine term, but does not lead to …

… which comes from your definition of where x is being measured from.

Two things:

• It is not quite enough to deduce $T'\leq 5mg$. You should note that this value is achieved.
• You should also check that $T'$ is never negative, otherwise SHM is not sustained, possibly leading to a higher tension at some point.

But $T'$ is atleast $mg$ (equality at $\cos \omega t=1$) . So , $T'$ is always non negative if i am not missing out on something . I checked the answer key , it says $5mg$ too .

 

[haruspex]

But $T'$ is atleast $mg$ (equality at $\cos \omega t=1$) . So , $T'$ is always non negative if i am not missing out on something . I checked the answer key , it says $5mg$ too .

Quite so, but did you check that before I mentioned it?

 

[Bling Fizikst]

Quite so, but did you check that before I mentioned it?

Yes i did check once before . So , maybe i am biased at this point .

 

[haruspex]

Yes i did check once before .

Good. It could have been a trick question.

 

[kuruman]

Here is an alternative solution using energy conservation. I think it is a simpler approach.

Maximum acceleration occurs at maximum displacement of the spring where all kinetic energy terms are zero. From mechanical energy conservation we have, from the release point to maximum displacement, $$0=\Delta U_{\text{spring}}+\Delta U_{\text{grav.}}=\frac{1}{2}k~x_{\text{max}}^2-(m+3m)g~x_{\text{max}} \implies x_{\text{max}}=\frac{8mg}{k}.$$ At any other point $x$ we have $$\begin{align} & \Delta U_{\text{elastic}}+\Delta K_{wheel}+\Delta K_{\text{masses}}+\Delta U_{\text{gravity}}=0 \nonumber \\
& \left(\frac{1}{2}k~x^2\right)+\left(\frac{1}{2}mv^2+ \frac{1}{2}mR^2\omega^2\right)+\left(\frac{1}{2}(4m)v^2\right)+\left(-4mgx\right)=0.\nonumber
\end{align}$$ For rolling without slipping $\omega=v/R$ and the equation simplifies to $$\frac{1}{2}k~x^2+3mv^2-4mgx=0\implies \left(\frac{1}{2}v^2\right)=\frac{2}{3}gx-\frac{1}{12}\frac{kx^2}{m}.$$ We can now find the acceleration $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{d}{dx}\left(\frac{1}{2}v^2\right)=\frac{2}{3}g-\frac{k}{6m}x.$$Thus, the initial value of the acceleration is $a_0=\frac{2}{3}g.$ It is positive, i.e. in the same direction as the displacement $x.$ The wheel accelerates to the right and the hanging masses accelerate down. As the spring is extended, the acceleration decreases until it reaches zero at the equilibrium position. Past that it turns negative, which means "to the left" for the wheel and "up" for the hanging masses.

The maximum acceleration is $$a_{\text{max}}=\frac{2}{3}g-\frac{k}{6m}x_{\text{max}}=\frac{2}{3}g-\frac{k}{6m}\frac{8mg}{k}=-\frac{2}{3}g$$ and is directed "up". Taking "down" as positive, we write Newton's second law for mass B as $$-T+3mg=(3m)\left(- \frac{2}{3}\right)g$$ and solve it to find the required tension.

 

[haruspex]

Here is an alternative solution using energy conservation. I think it is a simpler approach.

Maximum acceleration occurs at maximum displacement of the spring where all kinetic energy terms are zero. From mechanical energy conservation we have, from the release point to maximum displacement, $$0=\Delta U_{\text{spring}}+\Delta U_{\text{grav.}}=\frac{1}{2}k~x_{\text{max}}^2-(m+3m)g~x_{\text{max}} \implies x_{\text{max}}=\frac{8mg}{k}.$$ At any other point $x$ we have $$\begin{align} & \Delta U_{\text{elastic}}+\Delta K_{wheel}+\Delta K_{\text{masses}}+\Delta U_{\text{gravity}}=0 \nonumber \\
& \left(\frac{1}{2}k~x^2\right)+\left(\frac{1}{2}mv^2+ \frac{1}{2}mR^2\omega^2\right)+\left(\frac{1}{2}(4m)v^2\right)+\left(-4mgx\right)=0.\nonumber
\end{align}$$ For rolling without slipping $\omega=v/R$ and the equation simplifies to $$\frac{1}{2}k~x^2+3mv^2-4mgx=0\implies \left(\frac{1}{2}v^2\right)=\frac{2}{3}gx-\frac{1}{12}\frac{kx^2}{m}.$$ We can now find the acceleration $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{d}{dx}\left(\frac{1}{2}v^2\right)=\frac{2}{3}g-\frac{k}{6m}x.$$Thus, the initial value of the acceleration is $a_0=\frac{2}{3}g.$ It is positive, i.e. in the same direction as the displacement $x.$ The wheel accelerates to the right and the hanging masses accelerate down. As the spring is extended, the acceleration decreases until it reaches zero at the equilibrium position. Past that it turns negative, which means "to the left" for the wheel and "up" for the hanging masses.

The maximum acceleration is $$a_{\text{max}}=\frac{2}{3}g-\frac{k}{6m}x_{\text{max}}=\frac{2}{3}g-\frac{k}{6m}\frac{8mg}{k}=-\frac{2}{3}g$$ and is directed "up". Taking "down" as positive, we write Newton's second law for mass B as $$-T+3mg=(3m)\left(- \frac{2}{3}\right)g$$ and solve it to find the required tension.

Yes, using energy is neater. It does have the disadvantage that it does not show the tension never goes negative (and of course, if it were to then that would be followed by an unlimited tension when going taut again).

 

[kuruman]

It does have the disadvantage that it does not show the tension never goes negative (and of course, if it were to then that would be followed by an unlimited tension when going taut again).

If I understand you correctly, negative tension means that the rope connecting the hanging masses to the wheel goes slack. For this to happen, the center of the wheel must have acceleration $a \geq g$ directed to the right in which case the hanging masses will be in free fall. However, the maximum acceleration to the right is the initial acceleration $a_0=\frac{2}{3}g.$ If the system were released from rest with the spring already compressed, then the story would be different.

 

[haruspex]

If I understand you correctly, negative tension means that the rope connecting the hanging masses to the wheel goes slack.

No, the question concerns the rope connecting the hanging blocks.

 

[kuruman]

No, the question concerns the rope connecting the hanging blocks.

Yes, I agree, but isn't it sufficient to show that the upper rope connecting the upper mass to the wheel does not go slack? The "going slack" condition for a hanging mass undergoing oscillations is $\omega^2 x\geq g$ and depends on the frequency of oscillations but not on the mass. We have seen that the maximum acceleration here is $\frac{2}{3}g$. Thus, the "going slack" condition is not satisfied at any point during an oscillation.

 

[haruspex]

Yes, I agree, but isn't it sufficient to show that the upper rope connecting the upper mass to the wheel does not go slack? The "going slack" condition for a hanging mass undergoing oscillations is $\omega^2 x\geq g$ and depends on the frequency of oscillations but not on the mass. We have seen that the maximum acceleration here is $\frac{2}{3}g$. Thus, the "going slack" condition is not satisfied at any point during an oscillation.

Yes, that works, thanks.