How Does a Pendulum Reveal Bus Acceleration in Different Reference Frames?

AI Thread Summary
A passenger on a bus accelerating parallel to the Earth's surface can determine the bus's acceleration using a pendulum setup. In the Earth's reference frame, the pendulum's angle relates to the bus's acceleration with the equation a=gtanA, where A is the angle from vertical. However, in the bus's noninertial frame, the pendulum appears stationary, leading to a misleading result of A=0 due to the absence of real forces acting on it. This discrepancy arises from fictitious forces present in noninertial frames, such as the centrifugal force. Accurate acceleration measurements can still be made using other methods, but the pendulum approach is only valid in inertial frames.
StephenPrivitera
Messages
360
Reaction score
0
Suppose we are on a windowless bus that travels parallel to the Earth's surface at a constant acceleration with respect to the Earth's surface. How can a passenger in the bus determine the acceleration of the bus?
We decided we could attach a mass to a string and attach the string to the ceiling. From inside the bus, it would look like this:
________________
...\
...\
...\
...\
...O

Ignore those dots. Theyre the only way I can space things over.
There are two forces acting on this mass: gravity and the tension of the string. From the Earth's referece frame, the mass is accelerating with the same acceleration as the bus, ie, horizontally. So the only force the causes the mass to accelerate is TsinA=ma where T is the tension is the string, A is the angle measured wrt the line perpendicular to the ceiling, and a is the acceleration of the bus (and ball). Also, TcosA-mg=0 since there is no acceleration in this direction. Dividing each by each yields a=gtanA.
From the reference frame of the bus, the mass is stationary. TsinA=0 and TcosA=mg. Dividing each by each yields 0=tanA or A=0. Does this spurious result come from the fact that the reference frames are noninertial with repsect to each other? How come the result is valid for one frame but not the other?
 
Physics news on Phys.org
Yes, with respect to the bus, you cannot say that TsinA=0 because the bus is an accelerating frame of reference. You must include this acceleration in any calculations that you make.

The principle of invariance only applies to inertial (non accelerating) frames of reference.

Claude.
 


Yes, the spurious result comes from the fact that the reference frames are noninertial with respect to each other. In a noninertial reference frame, fictitious forces, such as the centrifugal force or the Coriolis force, may appear to act on objects. These forces are not real, but rather a result of the frame's acceleration. In this case, the fictitious force is the "centrifugal" force that appears to act on the mass attached to the string.

In the Earth's reference frame, the mass is accelerating horizontally due to the acceleration of the bus. This acceleration is counteracted by the tension in the string, resulting in the equation a=gtanA. However, in the reference frame of the bus, the mass is stationary and there is no acceleration. Therefore, the tension in the string is equal to the weight of the mass, resulting in the equation A=0. This difference in results is due to the presence of the fictitious force in the noninertial reference frame.

It is important to note that the acceleration of the bus can still be determined by the passenger using other methods, such as measuring the change in speed or observing the motion of objects outside the bus. However, the method described using the mass attached to a string will only yield accurate results in an inertial reference frame. In a noninertial reference frame, the results will be affected by the presence of fictitious forces.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top