Two ladybugs, Gertrude and Henrietta, are sitting on a record spinning at 33.3 r

Question

Two ladybugs, Gertrude and Henrietta, are sitting on a record spinning at 33.3 rpm, at distances of 10.0 and
5.00 cm from the center of the record, respectively. Gertrude and Henrietta are very svelte ladybugs, each with a mass
of 25.0 milligrams. For each of the following quantities, specify whether Gertrude has the larger value, Henrietta has
the larger value, or their values are equal.

(a) angular speed
(b) tangential speed
(c) centripetal acceleration
(d) angular momentum
(e) rotational kinetic energy

Please provide an explanation! Thank you!

Explanation / Answer

Useful relationships:

atangential = ·r

vtangential = ·r

acentripetal = (vtan)2/r

L = I·

I = m·r2 (true for this situation; not always the right formula, check your book/notes)

KErotational = (1/2)·I·2  

= angular acceleration; = angular velocity; L = angular momentum; I = moment of inertia; r = radius, distance from axis of rotation; m = mass

 

(a) angular speed  Equal... the angular speed is same anywhere on the record, at any distance from the center

 
(b) tangential speed Gertrude's speed is greater: using vtan = ·r, same =33.3 rpm... the larger r, 10 cm, gives the greater tangential speed  10*33.3 > 5*33.3     333>166.5

(c) centripetal acceleration Gertrude's is greater:  the ladybug w/ the greater tangential speed (from previous part) has the greater centripetal acceleration 3332/10 > 166.52/5

(d) angular momentum  Gertrude's is greater: Using L = I·, substitute in I = m·r2, L = (m·r2)·.   Both masses are the same, as well as , so the greater L (ang. momentum) is determined by larger radius, so LGertrude> LHenrietta     (25*102)*33.3 > (25*52)*33.3

(e) rotational kinetic energy  Gertrude's KE is greater: From above KE equation only the radius changes, and again, the larger radius yields the larger calculation.  (1/2)*(25*102)*33.32 > (1/2)*(25*52)*33.32

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