1.27 The hallmark of an inertial reference frame is that any object which is sub

Question

1.27 The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following experiment: I am 38 Chapter 1 Newton's Laws of Motion standing on the ground (which we shall take to be an inertial frame) beside a perfectly flat horizontal turntable, rotating with constant angular velocity w. I lean over and shove a frictionless puck so that it slides across the turntable, straight through the center. The puck is subject to zero net force and, as seen from my inertial frame, travels in a straight line. Describe the puck's path as observed by someone sitting at rest on the turntable. This requires careful thought, but you should be able to get a qualitative picture. For a quantitative picture, it helps to use polar coordinates; see Problem 146.

Explanation / Answer

as the puck is friction less

a) polar coordinates of puck at time t

r(t) = v*t [ where v is the linear speed with which it was sled in the first place ]

phi = 0 [ as there is no change in phi with time ]

so Coordinates WRT ground, (r,phi)g = (vt,0)

b) polar coordinates of a person on table at time t

r"(t) = R [ R is the original distance of the person from the centre of the disc at time 0 ]

phi" = wt [ here w is the angular cvelocity of the table ]

so Coordinates WRT ground, (r",phi")g = (R,wt)

SO POLAR coordinates of puck wrt person on disc = polar coordinates of puck wrt ground - polar coordinates of person wrt ground

(r',phi') = (r,phi)g - (r",phi")g = (vt - R , -wt)

now, x = r'cos(phi')

y = r'sin(phi')

and r' = vt - R

phi' = -wt

x = (vt - R)cos(wt)

y = -(vt - R)sin(wt)

squaring and adding

x^2 + y^2 = (vt - R)^2

vt = R + sqroot(x^2 + y^2)

t = (R +- sqrot(x^2 + y^2))/v

so x = +-( sqroot(x^2 + y^2))cos(w(R +- sqrot(x^2 + y^2))/v)

this frame of reference is non inertial

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