Solving for Time Gain in Pendulum Clock Covered by Lead Layer

[karnten07]

Homework Statement

A pendulum clock in the centre of a large room is observed to keep correct time. How many seconds per year will the clock gain if the floor is covered by a 1m thick layer of lead of density 11350kgm^-3 ?

Newtons gravitational constant is G = 6.67 x10^-11 Nm^2 kg^-2

Homework Equations

The Attempt at a Solution

Im not sure if its relevant to use but i thought i could use the equation for a pendulum
T = pi(l/g)^1/2

Any ideas?

 

[Kurdt]

It seems this question wants you to work out the new acceleration due to gravity when the lead is added.

Is that the question in full as you've typed it?

 

[karnten07]

It seems this question wants you to work out the new acceleration due to gravity when the lead is added.

Is that the question in full as you've typed it?

It is the second part to a question but i didnt think the first part related. The firt part is as follows:

Show that the gravitational field due to a horizontal uniform thin disc (thickness d, radius R and density p ) at a distance h vertically above the centre of the disc has magnitude

2piGpd (1 - (h/(R^2 + h^2)^1/2 )

I was able to do this. But in the second part i don't see that i can use this as it is for a disc and i don't know what shape the room is or how big it is. Can anyone spot something I've missed here?

p.s thanks for the speedy response

 

[Kurdt]

Well since the first part is talking about discs, I'd say you would have to make some assumptions about the room in the second part. Namely that the room is circular and that its radius is a lot larger than the pendulum's height from the disc. If you state the assumptions in the solution then that should satisfy whoever is marking it. You can then work out a correction to g, and apply it in the pendulum period formula from the first post.

 

[karnten07]

Well since the first part is talking about discs, I'd say you would have to make some assumptions about the room in the second part. Namely that the room is circular and that its radius is a lot larger than the pendulum's height from the disc. If you state the assumptions in the solution then that should satisfy whoever is marking it. You can then work out a correction to g, and apply it in the pendulum period formula from the first post.

excellent idea, thanks