# Why is the tangential force not same?

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1. Dec 19, 2016

### decentfellow

1. The problem statement, all variables and given/known data
A chain of length $l\lt \dfrac{\pi r }{2}$ is placed on a smooth hemispherical surface of radius R with one of its ends fixed at the top of the hemisphere. What will be the tangential acceleration of the chain when it starts sliding down.

2. Relevant equations

3. The attempt at a solution

Attempt-1 (Calculus Approach):-
There is an obvious approach by considering the differential mass at an angular position of $\theta$ w.r.t vertical, and then integrating it over all the angular positions at which the chain is present. In the case when the chain just starts to slide the limits of integration will be $0\rightarrow \theta(=\dfrac{l}{r})$

As, the mass is distributed uniformly along the chain, so its linear mass density is given as $\lambda=\dfrac{M}{\ell}=\dfrac{dm}{d\ell}$

Now, the length of the differential mass is given as $\ell=r\cdot d\theta$. Hence, the differential mass of the differential element is given as $dm=\dfrac{M}{\ell}\cdot rd\theta$

Hence, the tangential force acting on the differential element is given as
$$dF_{\text{t}}=(dm)\text{g}\sin\theta=\dfrac{M\text{g}}{\ell}r\sin\theta d\theta$$

On integrating the differential force for all the elements we get,
$$\int{dF_{\text{t}}}=\dfrac{M\text{g}r}{\ell}\int_{0}^{\theta}{\sin\theta}d\theta\\ \implies F=\frac{M\text{g}r}{\ell}\left(1-\cos\theta\right)$$

As, $\theta=\dfrac{\ell}{r}$, so the tangential force comes out to be
$$F_{\text{t}}=\frac{M\text{g}r}{\ell}\left(1-\cos\left(\frac{\ell}{r}\right)\right)$$

As, tangential acceleration is given as $a_{\text{t}}=\dfrac{F_{\text{t}}}{M}=\dfrac{\text{g}r}{\ell}\left(1-\cos\left(\dfrac{\ell}{r}\right)\right)$

Attempt-2 (Center of Mass Approach):-

My second attempt dealt with finding the tangential acceleration of the center of mass, which will be the tangential acceleration of the chain. Now to answer why do I think that finding the tangential acceleration of the center of mass will give the tangential acceleration of the chain. My line of thought was that all the external forces can be assumed to act on the center of mass hence the tangential acceleration of the center of mass will be the same as that of the whole chain.

Now, first we need to locate the center of mass of the chain. What I thought was that there is no need to find the exact location of the center of mass, the angular position of the center of mas suffices to find the angular acceleration of the chain because the distance of the center of mass from the origin as the chain moves on the hemisphere remains constant(as long as the chain is on the hemisphere and has a shape identical to a circular arc). This tells us that the center of mass traces a circular path as the chain slides on the hemisphere. Now, the angular position of the chain will $\dfrac{\theta}{2}(=\dfrac{\ell}{2r})$ due to symmetric distribution of the mass about the angle $\theta/2$.

So, the tangential force acting on the center of mass is given as $$F_{\text{t,COM}}=M\text{g}\sin(\frac{\theta}{2})=M\text{g}\sin\left(\frac{\ell}{2r}\right)$$

Hence, the center of mass has a tangential acceleration
$$a_{\text{t,COM}}=\dfrac{F_{\text{t,COM}}}{M}=\text{g}\sin(\frac{\theta}{2})=\text{g}\sin\left(\frac{\ell}{2r}\right)$$

Why does my second approach give a different answer than the first one? Is there some conceptual misunderstanding in my second approach

2. Dec 19, 2016

### haruspex

That is true of the gravitational forces, but the normal force will be greatest near the top of the chain.

3. Dec 20, 2016

### TSny

For determining the motion of the CM, it is true that you can treat the CM as a point particle of mass M acted on by all of the external forces.

Does the net normal force have a component in the direction of $a_t$ of the CM? (I think this is related to haruspex's comment.)

Also, does the CM have the same value of $a_t$ as the chain, or do they have the same angular acceleration?

Last edited: Dec 20, 2016
4. Dec 20, 2016

### decentfellow

Hmm...after what you and haruspex said I think that I was indeed missing that point but I had some trouble finding the components of Normal reaction to make it easier to find the tangential component of the normal for the center of mass. My work is as follows:-

$$dN_x=dN\sin\theta=dm\text{g}\cos\theta\sin\theta=\frac{M\text{g}r}{\ell}\sin\theta\cos\theta d\theta=\dfrac{M\text{g}r}{2\ell}\sin(2\theta)d\theta\\ \implies \int{dN_x}=\dfrac{M\text{g}r}{2\ell}\int_{0}^{\theta_0}{\sin(2\theta)}d\theta=\dfrac{M\text{g}r}{4\ell}\left(1-\cos2\theta_0\right)$$
$$\theta_0=\dfrac{l}{r}$$
$$dN_y=dN\cos\theta=dm\text{g}\cos^2\theta\\ \implies \int{dN_y}=\dfrac{M\text{g}r}{\ell}\int_{0}^{\theta_0}{\cos^\theta}d\theta=\dfrac{M\text{g}r}{\ell}\left(\theta_0+\dfrac{\sin2\theta_0}{2}\right)$$

But, the resultant of the two components isn't $\displaystyle\int{dN}=\displaystyle\int_{0}^{\theta_0}{dm\text{g}\cos\theta}=\dfrac{Mgr}{\ell}{\sin\theta_0}$

In that case it doesn't make any sense to me to define tangential acceleration for the chain as it is an extended body, angular acceleration would have been more sensible to me. So, what is the significance of tangential acceleration of the chain.

5. Dec 20, 2016

### haruspex

You integrated N as though it is a scalar.

6. Dec 20, 2016

### decentfellow

Didnt I seperate into componenets ? Or are you talking about the other one, i.e. $\displaystyle\int{dN}=\displaystyle\int_{0}^{\theta_0}{dm\text{g}\cos\theta}=\dfrac{Mgr}{\ell}{\sin\theta_0}$

And if you are talking about the last one then my first attempt on integrating the tangential force for all of the chain was also similar, but the answer was correct. So, if it is the last integration of $dN$ that I am doing wrong then is the first attempt correct?

7. Dec 20, 2016

### haruspex

Yes.
Your original attempt was ok because as the tension in one element is transferred to the next one up it gets rotated through dθ, much like a rope over a pulley.

8. Dec 20, 2016

### decentfellow

I am not able to comprehend as to how that affects the components of the normal.

9. Dec 20, 2016

### haruspex

Your original attempt (1) did not consider the normal.
What you did in effect (though you did not set it out this way) was to consider the tensions each end of each element and the weight of each element.
Because the tensions each end of an element are not quite aligned, you should technically consider the net tangential force from the tension as something like $T(\theta+\Delta\theta)- T(\theta)\cos(\Delta\theta)$. But for small Δθ, cos(Δθ) approximates 1. So $T(\theta+d\theta)- T(\theta)+g\sin(\theta)dm=a_tdm$. Integrating, the tension disappears, since it is zero at both ends.

10. Dec 20, 2016

### decentfellow

Your last post had just made sense to me and I was redoing my solution in the first attempt exactly the way you told me right now, but the only thing which was puzzling me were the limits of the integration....you cleared that up for me.

Although in post #6 I was referring to my first attempt of calculating the normal reaction via finding the components of the total normal. After reading that post again i think it was pretty easy to confuse it with the first attempt in my original post (I.e. post #1).

11. Dec 20, 2016

### TSny

I believe these are correct. You know the direction of the acceleration of the CM. So, what is the component in this direction of the total normal force?

Each link has a tangential acceleration. The magnitude of this acceleration is the same for each link, and I think that's what they are asking for.

How does the magnitude of the tangential acceleration of a link compare to the magnitude of tangential acceleration of the CM?