Accelerating Vehicles

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. In the lab, a trolley is held on a smooth ramp, with a small bob hanging from its roof on a string. You let go, and the trolley rolls freely down the ramp. While it rolls, will the string still hang straight down, as it did at rest? Hmm. Most students say yes. Hold on to your answer; we will settle it in Example two. Today: bobs and pendulums in accelerating vehicles. Questions on these do come up in JEE, so let's begin.

0:35Bob in a car: which way, and how tight?

So, look at the left half of the figure: a car on a level road. The thick blue arrow in front of it, a with a little arrow on top, is its acceleration, forwards. Inside, a bob hangs from the roof. The dashed line is the vertical, and theta, the Greek letter, is the string's angle from it. The right half is a wedge, for the wedges lesson. With me? Now, work in the car's frame. It accelerates, so it is not an inertial frame, and we add a pseudo force. On the bob: its weight, m g, the mass m times g, the acceleration due to gravity, the green arrow, straight down, and the pseudo force, m a, the red arrow, backwards, opposite to the acceleration. Okay? Once the bob settles, it is at rest relative to the car. The string can only pull along itself. So, to balance m g and m a together, which way must the string point? Yes, exactly! Along the two of them combined, the effective gravity: g eff vector, the g with an arrow on top and the small letters e f f below. So the string hangs along g eff, down and backwards. That is why it tilts back in the figure. Right? Now the angle. Tan theta is the sideways force over the downward force: m a over m g. The m cancels, so tan theta equals a over g, the line under the car in the figure. And the tilt is always towards the back. Clear? Now the tension in the string; we call it capital T, okay? Many students simply write T equals m g. Is that right? No, no, careful! The string balances both m g and m a, which are at right angles. So T is the square root, that long tick sign, of m g in brackets squared, plus m a in brackets squared. Take m outside: m times the square root of, g squared plus a squared. More than m g, whenever the car accelerates. Got it?

2:32Pendulum in the car

Now, the root in that tension is the size of g eff; without the arrow, g eff just means its size. So g eff equals the square root of, g squared plus a squared, along the tilted string. In the car, the bob feels a stronger gravity, tilted back. With me? Now, let the bob swing. It becomes a pendulum of length l. You already know the period formula for a simple pendulum. So in this car, what will the period be? Just put g eff in place of g! For small swings, the period is two pi times the square root of, l over, root of g squared plus a squared. Pi is the Greek letter, and l is the string's length. Bigger g eff, so a shorter period than in a car at rest. Okay? And careful: it swings about the tilted line, not about the vertical. That line makes an angle of tan inverse of a over g with the vertical. The minus one up on tan means inverse tan: the angle whose tan is a over g. Right?

3:36Vehicle accelerating up an incline

Now take the vehicle onto a slope that makes an angle alpha, this Greek letter, with the horizontal. The vehicle accelerates up the slope; the arrow on top of a says it is a vector. Clear? So, in the vehicle's frame, which way does the pseudo force point? Down the slope, correct! Always opposite to the acceleration. And its size is m times a. Okay? Now the weight. Its part along the slope is m g sine alpha, also down the slope. Add the two, and for each unit of mass, along the slope, we get g sine alpha plus a, downwards. Right? And into the slope, the weight gives m g cos alpha, so g cos alpha for each unit of mass. The pseudo force has no part in this direction. With me? These two are the components of g eff, so, unless a is zero, a bob in this vehicle does not hang vertically. And if the vehicle simply slides freely down a smooth slope? That is our puzzle, Example two. Got it?

4:35Example 1: bob angle and tension in a car at 5 m/s2

Example one. Read it with me: a bob of mass m hangs from the roof of a car accelerating at five metres per second squared; that is the m s to the power minus two. Find the angle of the string with the vertical, and its tension. Pause the video, and try it first. Done? Okay, let's check. Take g as ten metres per second squared. Then tan theta equals a over g: five over ten, that is one half. Simple, right? So theta is tan inverse of one half, about twenty six point six degrees. The wavy equals sign means approximately, and the small circle means degrees. And the string tilts towards the back of the car. Clear? Now the tension. The weight is m times ten, so the tension is ten m newtons, right? No, no, wait! That is the trap we just saw. T equals m times the square root of, ten squared plus five squared. That is m times root one hundred and twenty five. With me? Root one hundred and twenty five is five root five. So T is five root five times m, about eleven point two m newtons, with m in kilograms; the upright N is the newton. More than ten m, as expected. Perfect!

5:53Example 2: bob in a trolley sliding down a smooth incline

Now, remember our puzzle? Here it is. Read it with me: a bob of mass m hangs from the roof of a trolley sliding freely down a smooth incline of angle alpha. Find the direction of the string, and its tension. Pause the video, and try it. Done? Okay, first the big question. Many students say the string hangs straight down, just like at rest. Do you agree? Let's find out. Smooth means no friction, and sliding freely means no engine. So only gravity acts along the slope, and the trolley's acceleration is g sine alpha, down the slope. Okay? Now sit in the trolley's frame. The pseudo force on the bob is m times that acceleration, m g sine alpha, pointing up the slope, opposite to the acceleration. Right? But the weight also has a part along the slope: m g sine alpha, pointing down the slope. Equal and opposite, so they cancel completely! Clear? So only one part of the weight is left: m g cos alpha, perpendicular to the incline, pressing into it. So here, g eff is just g cos alpha. Got it? So the string hangs perpendicular to the incline, not vertically! And the tension is m g cos alpha, less than m g. Puzzle settled: not straight down, but at right angles to the ramp. Perfect!

7:16Quick revision: tilt angle, tension, pendulum in a vehicle

So, quick revision. In a car with acceleration a, the bob tilts back at tan inverse of a over g, and the tension is m times the square root of, g squared plus a squared. Sliding freely down a smooth incline, the string is perpendicular to the incline, and the tension is m g cos alpha. And always: the string hangs along g eff, and a pendulum uses g eff in place of g. Now go and try the practice problems on this page. All the best!