Its a tunnel in a planet

Consider a planet of radius $r$ having density $\lambda$.A tunnel is dig inside it it at a distance $\frac{r}{2}$ from its centre as shown. An object of mass m is left in the tunnel at the surface at $t=0$

Find the time taken by the object to reach the mid of the tunnel and the normal reaction applied by wall of the tunnel on the object.

Note by Tanishq Varshney
5 years, 3 months ago

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Hint, it doesn't matter how far the tunnel is from the center of the Earth.

Staff - 5 years, 3 months ago

- 5 years, 3 months ago

How much are you getting in AIITS-8.

@Tanish

- 5 years, 3 months ago

The answer key webpage is showing error so i couldn't match my answers

- 5 years, 3 months ago

I totally messed up with the paper.

- 5 years, 3 months ago

- 5 years, 3 months ago

Use dimensional analysis(during exam).

The actual way is to use restoring force and stuff, with it you will get that the motion of the object in the tunnel is S.H.M.

- 5 years, 3 months ago

I want that stuff @Raghav Vaidyanathan

- 5 years, 3 months ago

Hi Raghav

I need your view on this problem

- 5 years, 2 months ago

First of all, thanks for mentioning me @Tanishq Varshney !! Is the answer for time period = $\frac { \pi }{ 2 } \sqrt { \frac { 3 }{ 4\pi \lambda G } }$ ???

- 5 years, 3 months ago

I have same guess.

- 5 years, 2 months ago

Can u provide a hint or some expression to get the answer

- 5 years, 2 months ago

Since the motion inside is SHM , we know that force at amplitude is $F=A \omega^2$ here A= √3r/2 and you know what F is. You will get \omega there. And the rest is piece of cake.

- 5 years, 2 months ago

See, first of all, i found the restoring force, in this case, a component of the gravitational force between the Earth and the object.

Tunnel In a Planet

Let's consider that the object starts from the centre and moves towards the surface. Consider a very small displacement $'x'$ from the centre such that the restoring force remains constant. Now, the motion of the body through this $'x'$ is an SHM which i will prove later. According to shell theorem, the Earth around this motion won't apply any force. So, the force is applied by a small sphere with radius $R=\sqrt { \frac { { r }^{ 2 } }{ 2 } +{ x }^{ 2 } }$ which is equal to $\frac { Gm\frac { 4 }{ 3 } \pi \lambda { R }^{ 3 } }{ { R }^{ 2 } }$. But the restoring force will be the vertical component of this force = $\frac { G(\frac { 4 }{ 3 } \pi { R }^{ 3 }\lambda )m }{ { R }^{ 2 } } \frac { x }{ R }$ = $G\frac { 4 }{ 3 } \pi \lambda m(x)$

Clearly this force is directly proportional to $x$ and so, the motion is a simple harmonic one. So, the spring factor in the case is =$G\frac { 4 }{ 3 } \pi \lambda m$ and the time period is given by = $2\pi \sqrt { \frac { Mass\quad factor }{ Spring\quad factor } }$ = $2\pi \sqrt { \frac { m }{ \frac { 4 }{ 3 } \pi \lambda Gm } } =2\pi \sqrt { \frac { 3 }{ 4\pi \lambda G } }$

But since, you have asked about the time to move from extreme to mean, it is equal to $Time period/4$ = $\frac { \pi }{ 2 } \sqrt { \frac { 3 }{ 4\pi \lambda G } }$

And the normal reaction (I guess :P) will be given by the other component of the attractive force = $\frac { Gm\frac { 4 }{ 3 } \pi \lambda { R }^{ 3 } }{ { R }^{ 2 } } (\frac { r/2 }{ R } )\quad =\quad \frac { 2 }{ 3 } Gm\pi \lambda r$

So, how did i do?? Lol:D

- 5 years, 2 months ago

Thats it.

- 5 years, 2 months ago

Yayy! lol

- 5 years, 2 months ago

Can I have your views over here

- 5 years, 2 months ago