Video: Deflection of a Suspended Object due to Acceleration

A plumb-bob hangs from the roof of a railroad car. The car travels around a circular track of radius 228 m at a speed of 63 km/h. At what angle from the vertical does the plumb bob hang while the car travels around the circular track?

04:48

Video Transcript

A plumb bob hangs from the roof of a railroad car. The car travels around a circular track of radius 228 meters at a speed of 63 kilometers per hour. At what angle from the vertical does the plumb bob hang while the car travels around the circular track?

A plumb bob is simply a mass that hangs on a string and it’s used to indicate vertical lines. In this scenario, we have a railroad car moving at a given speed around the circular track of a given radius. The plumb bob we’re told hangs from the ceiling of the rail car that’s in motion.

And due to the rail car’s motion, the plumb bob hangs not vertically downward, but at an angle to that vertical line we can call πœƒ. It’s that angle we want to solve for based on the speed of the rail car and the size of the track it moves on.

To solve for πœƒ, we can start by considering the forces that act on the plumb bob. We know there’s the weight force that acts on it. That’s equal to the mass of the plumb bob times gravity. And there’s another force. That’s the tension force in the wire that holds the plumb bob up. We can call that capital 𝑇.

Looking at these forces, we can consider Newton’s second law of motion, which says that the net force on an object is equal to the object’s mass times its acceleration. If we decide that motion in the positive π‘₯ direction is to the left and motion in the positive 𝑦 direction is upward, then we can apply Newton’s second law of motion both to the vertical and the horizontal directions on our plumb bob.

As we consider the forces acting on the plumb bob in these two dimensions, we want to divide the tension force into its π‘₯- and 𝑦-components. When we draw in the horizontal and vertical components of this force, we see they help to form a right triangle, where the topmost angle is the angle πœƒ we want to solve for.

Now that all the forces in our diagram can be expressed in components in the π‘₯ and 𝑦 directions, let’s apply Newton’s second law of motion to both the vertical and the horizontal forces. In the vertical direction, we can say that 𝑇 times the cosine of πœƒ minus the weight force π‘šπ‘” is equal to zero. There is no acceleration of the plumb bob vertically. So 𝑇 cosine πœƒ is equal to π‘š times 𝑔.

In the horizontal direction, we can say that 𝑇 times the sine of πœƒ is equal to the mass of the plumb bob times its acceleration in that dimension. Considering this acceleration π‘Ž sub π‘₯, here’s where the motion of our rail car comes into play.

Recall that our rail car is moving at a constant speed, but in a circular path. Therefore, it’s experiencing centripetal acceleration towards the centre of this arc. This means we can rewrite π‘Ž sub π‘₯ as π‘Ž sub 𝑐, the centripetal acceleration experienced by the plumb bob.

The centripetal acceleration of an object moving in a circle we recall is equal to its linear speed squared divided by the radius of the circular path it moves in. This means we can rewrite π‘Ž sub 𝑐 as 𝑣 squared over π‘Ÿ, where 𝑣 is the given speed of the rail car and π‘Ÿ is the radius 228 meters.

Now that we have these two force balance equations, we can combine them to help us solve for the angle at which the plumb bob hangs from the vertical. We can take our horizontal force balance equation and divide it by our vertical force balance equation. When we do, we see that the tention force capital 𝑇 cancels out as does the mass of the plumb bob π‘š.

We can further recall that the sine of a given angle divided by the cosine of that same angle equals the tangent of that angle. The left-hand side of our equation, therefore, simplifies to the tangent of πœƒ. And the right-hand side simplifies to 𝑣 squared over π‘Ÿ times 𝑔, where 𝑔 is the acceleration due to gravity; that is, 9.8 metres per second squared.

If we then take the inverse or arc tangent of both sides of this equation, here’s what it simplifies to. πœƒ β€” the angle we want to solve for β€” is equal to the inverse tangent of 𝑣 squared over π‘Ÿ times 𝑔. We’re given 𝑣 and π‘Ÿ and can look up 𝑔. So we’re ready to plug in and solve for πœƒ.

As a final step, before we do, let’s convert our given speed in units of kilometres per hour to a speed in metres per second. To do so, we’ll multiply it by one hour every 3600 seconds times 1000 meters for one kilometre. We see that in this expression the units of hours cancel as do the units of kilometres, leaving us with a speed in metres per second.

We plug in this speed now in metres per second into our equation as well as the radius in metres and 𝑔 in units of metres per second squared. πœƒ we find is 7.8 degrees. That’s the angle off of the vertical at which the plumb bob hangs.

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