The position of a training helicopter (weight 2.75 ×10^5 N ) in a test is given by r̂=(0.020 m

The position of a training helicopter (weight 2.75 ×10^5 N...

The position of a training helicopter (weight $2.75 \times 10^{5} \mathrm{~N}$ ) in a test is given by $\hat{r}=\left(0.020 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\imath}+$ $(2.2 \mathrm{~m} / \mathrm{s}) t \hat{\jmath}-\left(0.060 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \hat{k} .$ Find the net force on the helicopter at $t=5.0 \mathrm{~s}$

The position of a training helicopter (weight $2.75 \times 10^{5} \mathrm{~N}$ ) in a test is given by $\hat{r}=\left(0.020 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\imath}+$
$(2.2 \mathrm{~m} / \mathrm{s}) t \hat{\jmath}-\left(0.060 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \hat{k} .$ Find the net force on the helicopter at
$t=5.0 \mathrm{~s}$

Key Concepts

Mass Determination from Weight

Mass can be determined from the weight of an object by dividing the weight by the acceleration due to gravity. Since weight is the force due to gravity acting on an object, calculating its mass is necessary when applying Newton's Second Law to relate force and acceleration in mechanics.

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This law is fundamental in mechanics and allows us to relate the acceleration experienced by an object to the forces acting on it.

Kinematics and Vector Differentiation

Kinematics deals with the motion of objects and is often described using vector functions which represent position. By differentiating a position vector with respect to time, one can obtain the velocity vector, and differentiating again yields the acceleration vector. This process is essential for determining how the motion of an object changes over time.

Transcript

00:01 Okay, so we have a calculus problem involving a helicopter, and we, for some reason, very precisely in how it's position vector as a function of time, and they tell us the weight in newton's.

00:13 So let's actually go ahead and convert this into a mass right off the bat, because mass is more useful than weight.

00:20 So we just do that by dividing the weights by our local gravitational acceleration of 9 .81, and that gives us a mass.

00:30 Of 2 .80 times 10 to the fourth kilograms.

00:36 Great, that is a much more useful number.

00:38 And then again, we know the position vector as a function of time very, very precisely.

00:44 So i'm going to suppress the units for elegance, but be rest assured that everything is an si units.

00:54 So we know that it all works out.

00:56 I have 2 .2 t, j hat over here.

01:01 And then, oh, sorry, that's a minus sign.

01:06 0 .0 .0 .0 .0 .t squared k.

01:13 And we basically just do calculus to this in order to find, to take two -time derivatives in order to find the acceleration so that we can then multiply it by the mass that we just solve for in order to get the force.

01:25 So we're going to take one -time derivative.

01:28 So three jumps out in front.

01:30 So this becomes zero -point -06.

01:35 This t just disappears via the chain rule and then this one the two hops out in front great so that's velocity and now we do the same thing one more time we already get the time derivative of velocity which is acceleration so now this two is going to jump out in front the middle term actually disappears because it was a constant and this will also be 0 .120 that t just disappears...

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