Period of simple pendulum on an inclined moving platform

[songoku]

Homework Statement

A person, holding a simple pendulum, stands on a moving platform inclined at 30 degrees to horizontal. When the platform is at rest, the period of the pendulum is 2 s. When the platform moves with constant acceleration $2 m/s^2$, the period will be
a) 2.36 s
b) 2 s
c) 1.64 s
d) 2.11 s
e) 1.89 s

Relevant Equations

$T=2\pi \sqrt{\frac{L}{\text{apparent gravity}}}$

When the platform moves with constant acceleration, the equation of Newton's 2nd law of motion is

Forward force - W sin 30^o = m.a
Forward force = m (a + g sin 30^o) ⇒ apparent gravity = a + g sin 30^oFinal period of pendulum = $\sqrt{\frac{g}{a+g \sin 30^{0}}} \times 2 = 2.38 s$

Is this correct? Thanks

 

[DrClaude]

What changes between being at rest and moving? (Hint: it is not the direction of the force of gravity.)

Also, does the problem mention in which direction the platform is accelerating?

 

[Steve4Physics]

Homework Statement:: A person, holding a simple pendulum, stands on a moving platform inclined at 30 degrees to horizontal. When the platform is at rest, the period of the pendulum is 2 s. When the platform moves with constant acceleration $2 m/s^2$, the period will be
a) 2.36 s
b) 2 s
c) 1.64 s
d) 2.11 s
e) 1.89 s
Relevant Equations:: $T=2\pi \sqrt{\frac{L}{\text{apparent gravity}}}$

When the platform moves with constant acceleration, the equation of Newton's 2nd law of motion is

Forward force - W sin 30^o = m.a
Forward force = m (a + g sin 30^o) ⇒ apparent gravity = a + g sin 30^o

Final period of pendulum = $\sqrt{\frac{g}{a+g \sin 30^{0}}} \times 2 = 2.38 s$

A few thoughts…

Given $T=2\pi \sqrt{\frac{L}{\text{apparent gravity}}}$, will an increase in ‘apparent gravity’ make $T$ larger or smaller?

What direction is ‘forward’ when you use the term ‘forward force’?
Is the acceleration uphill or downhill? (Does it matter?)

Also note:
- the values supplied in the question are given to only 1 significant figure; this is inconsistent with most of the answers;
- the (unrounded) answer will depend on what (unspecified) value of g you have used.

 

[songoku]

What changes between being at rest and moving? (Hint: it is not the direction of the force of gravity.)

The direction of tension?

Also, does the problem mention in which direction the platform is accelerating?

Ah my bad. Yes, the question states that the acceleration is uphill.

Given $T=2\pi \sqrt{\frac{L}{\text{apparent gravity}}}$, will an increase in ‘apparent gravity’ make $T$ larger or smaller?

Smaller

What direction is ‘forward’ when you use the term ‘forward force’?
Is the acceleration uphill or downhill? (Does it matter?)

Uphill and although I am not sure whether the direction matters (because I still don't know the correct way to solve this problem), my guess is yes, the direction matters (taking the case of pendulum in lift accelerates upwards or downwards).

I think I have difficulty analyzing the FBD.

- the (unrounded) answer will depend on what (unspecified) value of g you have used.

I use g = 9.81 ms^-2This imy FBD:

Angle α must be small so the motion of the pendulum can be simple harmonic.

When trying to find apparent gravity of the pendulum, which axis should I use?

The second axis is where the x-axis is parallel to the inclined platform.

I tried to use the third one and ended up with W sin α = m.a (as the restoring force) and get the same period as the case where the platform is stationary (which I don't think making any sense).

 

[PeroK]

Smaller

That narrows down the answers.

Uphill and although I am not sure whether the direction matters (because I still don't know the correct way to solve this problem), my guess is yes, the direction matters (taking the case of pendulum in lift accelerates upwards or downwards).

I think I have difficulty analyzing the FBD.

Have you studies non-inertial refrence frames?

I tried to use the third one and ended up with W sin α = m.a (as the restoring force) and get the same period as the case where the platform is stationary (which I don't think making any sense).

It doesn't make sense.

 

[PeroK]

I noticed that you posted a similar homework problem here:

https://www.physicsforums.com/threa...-system-and-a-pendulum-inside-a-lift.1014679/

You need to learn how to work using a non-inertial reference frame.

 

[Steve4Physics]

@songoku, here’s one way to tackle these sorts of problems but it may not be the method you are expected to use.

Imagine the pendulum is in your mobile laboratory.

If the laboratory accelerates at 2m/s² (uphill, 30º to horizontal), then inside the lab’ the effects of the acceleration are identical to having a new gravitational field of 2m/s² acting the opposite direction (downhill, 30º to horizontal). (For example, as if a massive alien spacecraft had just materialised.)

The apparent gravitational field in your lab’ is the vector sum of the earth’s field (g=10m/s² vertically downwards) and the new 2m/s² (downhill, 30º to horizontal) field.

So you need to do this vector sum.

If g’ is the magnitude of the vector sum, then you know the new period is $T’=2\pi \sqrt{\frac{L}{\text{g’}}}$.

Additional Notes.
The same approach works just as well in other situations (e.g. in simple elevator problems). It is based on what is called the ‘equivalence principle’, which basically says that you can’t distinguish between the affects of a gravitational field and the effects of acceleration. Look it up.

 

[jbriggs444]

Uphill and although I am not sure whether the direction matters

Then the first move is to decide on the direction and magnitude of the apparent gravity relative to the uphill-accelerating platform.

The approach that I would take (to unclutter my mind and focus on the details that matter) would be to consider an object dropped from rest in the platform's frame of reference.

It will experience a downward acceleration from gravity. Superimposed on (added to) this will be the rearward apparent acceleration based on the platform's forward acceleration. Take the vector sum of the two to find the resultant acceleration.

Edit: One thing to be wary of is how to express the two accelerations -- what angles to use for each. Make sure you use the same coordinate system for both. Otherwise you'll be adding apples with oranges.

 

[songoku]

That narrows down the answers.

But I don't know whether the apparent gravity is smaller or larger

Have you studies non-inertial refrence frames?

I haven't

@songoku, here’s one way to tackle these sorts of problems but it may not be the method you are expected to use.

Imagine the pendulum is in your mobile laboratory.

If the laboratory accelerates at 2m/s² (uphill, 30º to horizontal), then inside the lab’ the effects of the acceleration are identical to having a new gravitational field of 2m/s² acting the opposite direction (downhill, 30º to horizontal). (For example, as if a massive alien spacecraft had just materialised.)

The apparent gravitational field in your lab’ is the vector sum of the earth’s field (g=10m/s² vertically downwards) and the new 2m/s² (downhill, 30º to horizontal) field.

So you need to do this vector sum.

If g’ is the magnitude of the vector sum, then you know the new period is $T’=2\pi \sqrt{\frac{L}{\text{g’}}}$.

Additional Notes.
The same approach works just as well in other situations (e.g. in simple elevator problems). It is based on what is called the ‘equivalence principle’, which basically says that you can’t distinguish between the affects of a gravitational field and the effects of acceleration. Look it up.

Then the first move is to decide on the direction and magnitude of the apparent gravity relative to the uphill-accelerating platform.

The approach that I would take (to unclutter my mind and focus on the details that matter) would be to consider an object dropped from rest in the platform's frame of reference.

It will experience a downward acceleration from gravity. Superimposed on (added to) this will be the rearward apparent acceleration based on the platform's forward acceleration. Take the vector sum of the two to find the resultant acceleration.

Edit: One thing to be wary of is how to express the two accelerations -- what angles to use for each. Make sure you use the same coordinate system for both. Otherwise you'll be adding apples with oranges.

I am using this coordinate system:

Is the vector sum of the acceleration = g + a sin 30^o?

If I want to solve this problem using ground's frame of reference, what is the correct approach?

Thanks

 

[PeroK]

I am using this coordinate system:
View attachment 300685

That diagram doesn't tell me very much.

Is the vector sum of the acceleration = g + a sin 30^o?

Acceleration is a vector. A vector has components.

If I want to solve this problem using ground's frame of reference, what is the correct approach?

Go back to first principles, set up the differential equations and solve them!

That's why you need to learn about what to do in a non-inertial frame of reference.

 

[songoku]

Thank you very much for the help and explanation DrClaude, Steve4Physics, PeroK, jbriggs444

 

[kuruman]

If I want to solve this problem using ground's frame of reference, what is the correct approach?

It is the usual FBD approach. Note that
(a) There are two forces acting on the mass, tension $T$ and gravity $mg$. In the steady state (no oscillations) the pendulum string is at fixed angle $\phi$ relative to the vertical. There is no a priori relation between this angle and the angle of the incline.
(b) The acceleration of the mass (in the steady state) is $a$ m/s², say down the incline.
Use a horizontal-vertical coordinate system, write down Newton's second law in each direction and find an expression for the magnitude of the tension.

How is the tension related to $g_{\text{eff}}$ in the steady state?

 

[songoku]

It is the usual FBD approach. Note that
(a) There are two forces acting on the mass, tension $T$ and gravity $mg$. In the steady state (no oscillations) the pendulum string is at fixed angle $\phi$ relative to the vertical. There is no a priori relation between this angle and the angle of the incline.
(b) The acceleration of the mass (in the steady state) is $a$ m/s², say down the incline.
Use a horizontal-vertical coordinate system, write down Newton's second law in each direction and find an expression for the magnitude of the tension.

Sorry I am taking the acceleration to be up the incline (to match the original question)

Equation of Newton second law:
(i) T cos ∅ - W = m . a sin θ
(ii) T sin ∅ = m . a cos θ

Is that correct?

How is the tension related to $g_{\text{eff}}$ in the steady state?

I am not sure. But I think I am expecting something like:

T = m (...) where (...) is the $g_{\text{eff}}$

Thanks

 

[kuruman]

Sorry I am taking the acceleration to be up the incline (to match the original question)
View attachment 300735Equation of Newton second law:
(i) T cos ∅ - W = m . a sin θ
(ii) T sin ∅ = m . a cos θ

Is that correct?I am not sure. But I think I am expecting something like:

T = m (...) where (...) is the $g_{\text{eff}}$

Thanks

That is exactly what I had in mind. Can you find an expression for $T$ using the equations that you wrote down and complete the solution?

 

[songoku]

That is exactly what I had in mind. Can you find an expression for $T$ using the equations that you wrote down and complete the solution?

I tried but...

First attempt:
T cos ∅ - W = m . a sin θ → then using small angle approximation cos ∅ ≈ 1

T = m (g + a sin θ) → g_eff = g + a sin θ (which is not correct because from non-inertial frame of reference I got $g_{\text{eff}}=\sqrt{a^{2} + g^{2} + ag}$Second attempt:
(a)
T sin ∅ = m . a cos θ → using small angle approximation sin ∅ ≈∅

T . ∅ = m . a cos θ

∅ = (m . a cos θ) / T

(b)
T cos ∅ - W = m . a sin θ ----> then using small angle approximation cos ∅ ≈ $1-\frac{1}{2} \phi^{2}$

$$T \left(1-\frac{1}{2} \phi^{2} \right)=mg + m.a \sin \theta$$
$$T \left(1-\frac{1}{2} \frac{m^2 a^2 \cos^{2} \theta}{T^2}\right)=mg + m.a \sin \theta$$
$$T-\frac{m^2 a^2 \cos^{2} \theta}{2T}=mg + m.a \sin \theta$$
$$2T^2-2(mg + m.a \sin \theta)T-m^2 a^2 \cos^{2}=0$$
$$T=\frac{2(mg+m.a \sin \theta) \pm \sqrt{4m^2 g^2+8m^2 g a \sin \theta + 4m^2 a^2 \sin^{2} \theta+8m^2 a^2 \cos^{2} \theta}{}}{4}$$

Is this even correct?

Thanks

 

[kuruman]

Suppose you wrote
$T \cos\phi=\dots$
$T\sin\phi=\dots$
then squared and then added the two equations $\dots$

 

[songoku]

Suppose you wrote
$T \cos\phi=\dots$
$T\sin\phi=\dots$
then squared and then added the two equations $\dots$

I got the answer.

I want to ask about the coordinate system used. Is there a certain reason to use horizontal - vertical coordinate system (other than maybe it is easier)? Can I use another coordinate system, maybe like the x-axis is parallel to the inclined platform?

Thanks

 

[jbriggs444]

I want to ask about the coordinate system used. Is there a certain reason to use horizontal - vertical coordinate system (other than maybe it is easier)? Can I use another coordinate system, maybe like the x-axis is parallel to the inclined platform?

As a pretty general rule, you can always use any coordinate system you like. They'll all deliver the same answer in the end. Some may make the calculations easier and some may make them more difficult.

We can adopt an inertial coordinate system in which the x-axis is aligned with the upward slope. [So that we are all visualizing the same thing, let us say we have rails sloping diagonally up and to the right at 30 degrees above the horizontal. We have a platform with a pendulum device that is accelerating up the slope].

So we switch to this coordinate system. Gravity is now down and 30 degrees to the left of the negative y axis. The pendulum apparatus is accelerating directly in the positive x direction. We write $\sum F=ma$ and ask about force required to overcome gravity and achieve the known acceleration.

It is the same vector sum that one would have computed in a frame aligned with the ground. It's just that the vectors you are adding have been rotated 30 degrees now because we're in a coordinate system that is rotated by 30 degrees. The magnitude of the vector sum is the same either way.

[In fact, they are the same vectors, just reported in coordinate form using a different basis]

 

[Steve4Physics]

I want to ask about the coordinate system

If you use the approach described in Post #7 you don’t need any coordinate system!

In the platform’s frame of reference, the magnitude of apparent acceleration due to gravity ($g_{app}$) can be calculated (see Post #7) by simple vector addition (using the cosine rule):

Since the period is inversely proportional to $\sqrt g$, the new period is: $\text {old period} \times \sqrt{\frac {g}{g_{app}}}$.

Edit - minor changes for readability.

 

[kuruman]

I got the answer.

I want to ask about the coordinate system used. Is there a certain reason to use horizontal - vertical coordinate system (other than maybe it is easier)? Can I use another coordinate system, maybe like the x-axis is parallel to the inclined platform?

Thanks

First of all, the magnitude of the tension $T$ is a scalar quantity and, as such, it is independent of the choice of coordinate axes. That said, the next consideration is choosing a system that makes the algebra simpler.
If one chooses the horizontal-vertical axes (my choice),

• the tension is $\vec T=T\sin\phi~\hat x +T\cos\phi~\hat y$
• the weight is $\vec W=0~\hat x -mg~\hat y$
• the acceleration is $\vec a=a\cos\theta~\hat x +a\sin\theta~\hat y$

If one chooses the parallel-perpendicular-to-incline axes,

• the tension is $\vec T=T\sin\varphi~\hat x +T\cos\varphi~\hat y$
• the weight is $\vec W=-mg\sin\theta~\hat x -mg\cos\theta~\hat y$
• the acceleration is $\vec a=a~\hat x +0~\hat y$

I note that $\varphi$ in the second set of equations is measured from the direction perpendicular to the incline and that is why I used a different symbol. The two sets of equations are comparable and the algebra for finding $T$ is equally difficult. My personal preference is the first set because it references the tip angle of the string to a vertical line which stays the same regardless of the incline angle.

 

[songoku]

Thank you very much for all the explanation and help kuruman, jbriggs444, Steve4Physics.