# Mechanics question

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I don't know whether it's Maths or Physics question, but we only use 1 Physics formula for that.

The pendulum oscillates with small angle, when it's hung freely its period T = 2pi.rt(l/g) where l is length of the string, g = 9.81 m/s

This pendulum is hung in a car travelling with acceleration a. Find the new period of the pendulum.

The pendulum oscillates with small angle, when it's hung freely its period T = 2pi.rt(l/g) where l is length of the string, g = 9.81 m/s

This pendulum is hung in a car travelling with acceleration a. Find the new period of the pendulum.

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#2

In a non-accelerating frame, the period of a simple pendulum oscillating freely solely under the influence of gravity is: T = 2Pi.Sqrt(l/g), for relatively small angles.

In an accelerating frame, such as when a simple pendulum oscillates in a car with acceleration a, the effective g acting on the pendulum is different.

By considering vectors, we need to find the effective g acting on the pendulum; this is the done by calculating the sum of the individual accelerations acting on the pendulum. There is an acceleration of g, acting vertically downwards and an acceleration of a, acting horizontally forwards.

Hence: g (Effective) = Sqrt (a^2 + g^2).

Hence: New T = 2Pi.Sqrt{l/[g (Effective)]}

--->

Nima

In an accelerating frame, such as when a simple pendulum oscillates in a car with acceleration a, the effective g acting on the pendulum is different.

By considering vectors, we need to find the effective g acting on the pendulum; this is the done by calculating the sum of the individual accelerations acting on the pendulum. There is an acceleration of g, acting vertically downwards and an acceleration of a, acting horizontally forwards.

Hence: g (Effective) = Sqrt (a^2 + g^2).

Hence: New T = 2Pi.Sqrt{l/[g (Effective)]}

--->

**New T = 2Pi.Sqrt{l/[(a^2 + g^2)]}**Nima

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#3

Here are more details on Nima's solution.

Let theta be the angle that the pendulum makes with the vertical, measured so that theta increases in the direction of the car's acceleration.

The transverse acceleration of the bob is l theta'' + a cos(theta). So

l theta'' + a cos(theta) = -g sin(theta)

l theta'' = -sqrt(a^2 + g^2)sin(theta - theta0)

where theta0 = arctan(a/g).

[ sin(theta - theta0)

= sin(theta)sin(theta0) + cos(theta)cos(theta0)

= sin(theta) [a/sqrt(a^2 + g^2)] + cos(theta) [g/sqrt(a^2 + g^2)] ]

Let psi = theta - theta0. Then

l psi'' = -sqrt(a^2 + g^2) sin(psi) . . . . . (*)

For small psi, we can approximate (*) by

l psi'' = -sqrt(a^2 + g^2) psi . . . . . (**)

Solutions to (**) have period 2pi sqrt[l/sqrt(a^2 + g^2)].

Let theta be the angle that the pendulum makes with the vertical, measured so that theta increases in the direction of the car's acceleration.

The transverse acceleration of the bob is l theta'' + a cos(theta). So

l theta'' + a cos(theta) = -g sin(theta)

l theta'' = -sqrt(a^2 + g^2)sin(theta - theta0)

where theta0 = arctan(a/g).

[ sin(theta - theta0)

= sin(theta)sin(theta0) + cos(theta)cos(theta0)

= sin(theta) [a/sqrt(a^2 + g^2)] + cos(theta) [g/sqrt(a^2 + g^2)] ]

Let psi = theta - theta0. Then

l psi'' = -sqrt(a^2 + g^2) sin(psi) . . . . . (*)

For small psi, we can approximate (*) by

l psi'' = -sqrt(a^2 + g^2) psi . . . . . (**)

Solutions to (**) have period 2pi sqrt[l/sqrt(a^2 + g^2)].

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Don't worry Nima, I did rep him

But there's another one much much more difficult than that one.

A shooter put the target in a big circular tray that can rotate with constant angular speed w. The radius of the tray is R. Initialy, the line join him and the target passes through the centre of the tray. (as in the graph)

Find the angle @ that the bullet makes with the horizontal so that it hits the target.

But there's another one much much more difficult than that one.

A shooter put the target in a big circular tray that can rotate with constant angular speed w. The radius of the tray is R. Initialy, the line join him and the target passes through the centre of the tray. (as in the graph)

Find the angle @ that the bullet makes with the horizontal so that it hits the target.

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