NAME SHOW ALL WORK! 11. (15) Three objects of uniform density-a solid sphere, a

Question

NAME SHOW ALL WORK! 11. (15) Three objects of uniform density-a solid sphere, a solid cylinder, and a hollow cylinder-are placed at the top of an incline (see figure below). They are all released from rest at the same elevation and roll without slipping. (a) Which object reaches the bottom first? b) Why? (Please write 2-3 sentences. Make sure your reasoning makes sense and feel free to include equations. PLEASE WRITE NEATLY!!!) c) If the disk has a mass of 2 kg and a radius of 0.5 m and is released from an elevation of 0.3 m, what is its center of mass velocity when it reaches the ground?

Explanation / Answer

moment of inertia of solid sphere Iss = 2/5 mr2 = 0.4 mr2

moment of inertia of solid cylinder Isc = 0.5 mr2

moment of inertia of hollow cylinder Ihc = mr2

as the solid sphere has less moment of inertia , it reaches first

hence the answer is solid sphere

b) from conservation of energy PE at h = energy at bottom = 0.5 mv2 + 0.5 Iw2

for solid sphere mgh = 0.5mv2 + 0.5 x 0.4 mr2 x v2 / r2 since w=v/r

gh = 0.7v2  

i.e. v2 = gh/0.7 = 1.43 gh ................... (1)

for solid cylinder mgh = 0.5 mr2 + 0.5 x 0.5 mr2 x v2/r2 = 0.75 mv2

hence v2 = gh/0.75 = 1.33 gh ................... ( 2 )

for hollow cylinder mgh = 0.5 mv2 + 0.5 x mr2 x v2/r2 = mv2

hence v2 = gh ............................. (3)

from 1 , 2 , 3 solid sphere has highest velocity when they reaches the bottom

hence solid sphere reaches first.

c) mass m = 2 kg radius r = 0.5m h = 0.3m

moment of inertia I = 0.5mr2 = 0.5 x 2 x 0.5 x 0.5 = 0.25 kg-m2

from conservation of energy

mgh = 0.5 mv2 + 0.5 I w2

= 0.5mv2 + 0.5 x 0.5 mr2 v2/r2

= 0.75 mv2

hence v2 = gh / 0.75 = 9.81 x 0.3 / 0.75

= 3.924

v = sq.root of 3.924 = 1.98 m/s

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