parts b and c (7%) Problem 7: A rod of mass M= 107 g and length L 48 cm can rota
ID: 1631620 • Letter: P
Question
parts b and c
Explanation / Answer
moment of inertia of rod, I1 = M L^2 / 3
= (0.107)(0.48^2) / 3
= 8.218 x 10^-3 kg m^2
moment of inertia of ball = m D^2 = (0.013) (0.48/2)^2
= 0.749 x 10^-3 kg m^2
(A) moment of inertia of rod-ball system,
I = I1 + I2
= 8.97 x 10^-3 kg m^2
(b) Applying angular momentum conservation,
m v D cos(theta) = I w
0.013 x 7 x (0.48/2) x cos35 = (8.97 x 10^-3) w
w = 2 rad/s
(c) after the collision,
KE = I w^2 /2
= (8.97 x 10^-3) (2^2) / 2
= 17.9 x 10^-3 J
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