- #1

decentfellow

- 130

- 1

## Homework Statement

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.

## Homework Equations

## 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