The movie \"Frames of Reference” showed two experiments where a ball of mass m w

Question

The movie "Frames of Reference” showed two experiments where a ball of mass m was dropped from a height h on a cart that was (a) moving at a constant horizontal velocity v. = (vg,0), and (b) accelerating with a constant acceleration a, = (a,0), both relative to the laboratory reference frame (we're using coordinate systems in both the cart and laboratory where x is horizontal (positive in the direction of the cart's motion) and y is vertical (positive upward). The trajectory of the ball is filmed in both frames – the constant velocity experiments start around 5:50 (min:sec) into the movie, and the accelerated cart experiments at about 13:35. This exercise asks you to determine the trajectory of the ball (i.e. its height vs its horizontal displacement), as seen in each reference frame for both constant velocity and acceleration. (a) Constant velocity (i.e. inertial) reference frames: i. Lab (inertial) frame: Denote the instantaneous height and horizontal position of the ball in this frame by y(1and x,(), respectively. Noting that the only force acting on the ball is the gravitational attraction of the earth F, = (0, -mg), use Newton's 21° Law to find the trajectory y, (x,) shown in the film. ii. Cart (inertial) frame: Denoting the instantaneous height and horizontal position o the ball in the cart frame by yo(1) and xo(t), and using Newton's 2nd Law, repeat the analysis above to find the trajectory yo(x) shown in the film. Over --->

Explanation / Answer

(a) The constant velocity reference frames

i. Lab ( inertial frame)

height y(t)

horizontal position = x(t)

F = -mg j ( where i, j and k are unit vectors)

from the second law of motion

x'(t) = 0 ( as the ball is dropped vertically, ahnce no horixzontal velocity)

hence

x(t) = xo

similiarly

y"(t) = -g

y'(t) = -gt + C1

y(t) = -gt^2/2 + C1t + C2

at t = 0

y(0) = h = C2

y'(0) = 0

C1 = 0

hence

y(t) = -gt^2/2 + h

ii) Cart (inertial frame)

x'(t) = -v i ( the cart moves to the right, hence the ball moves to the left wrt the cart)

hence, x(t) = -vt + C1

let x(0) = xo

So, C1 = xo

then x(t) = -vt + xo

and

y(t) = -gt^2/2 + h

(b) The non inertial reference frame:

i) As the lab is still in rest and the ball is dropped in the labs frame

x(t) = xo

y(t) = -gt^2/2 + h

ii) In the cart's frame of reference

x"(t) = -a

hence, x(t) = -at^2/2 + C1t + C2

at t = 0, x(0) = xo

C2 = xo

at t= 0, x'(t) = 0

C1 = 0

So, x(t) = -at^2/2 + xo

and

y(t) = -gt^2/2 + h

now, xc = xL - xCl

hence

-at^2/2 + xo = xo - xCl

xCl = at^2/2

angle with the vertical = theta

tan(theta) = y/x = (h - gt^2/2)/(xo - at^2/2)

theta = arctan((h - gt^2/2)/(xo - at^2/2))

(c) For the person closed inside the box

Fc = m*ac

where ac = -(ai + gj) where, i and j are unit vectors along x and y axes respectively

Now, for the inertial frame

Fi = m*g = -m*g j

By using tranformation

Fi = Fc + m*xCl

= Fc + m*ai

hence

-mg j = -m*a i - m*g *j + m*a i

which is true, hence this transformation is necessary for the newtons second law to twork in non inertial frames.

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