a) Write an expression for the ring\'s internal energy at point 1, assuming the
ID: 2296457 • Letter: A
Question
a) Write an expression for the ring's internal energy at point 1, assuming the gravitational potential at point 3 is 0.
b) If the ring rolls (without slipping) to point 2, what is the ring's energy in terms of of h2 and v2?
c) Given h2 = 32 m, what is the velocity of the ring at point 2 in m/s?
d) What is the the ring's rotational velocity in rad/s?
e) After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s?
Explanation / Answer
I = Moment of inertia of ring = MR^2 = 8 * 0.08^2 = 0.05 kg m^2
A) E1 = internal energy at point 1 = potentail energy at 1 = Mg *h1= 8*9.8* 83 =6507.2 J
B) let V2 = linear velocity at point 2
W2 = angular velocity at point 2 = V2/R
K2 = Kintetic energy at point 2
= rotational energy + linear kinetic energy
= I*(W2)^2/2 + MV2^2/2
=(MR^2 )* (V2/R)^2/2 + 4V2^2
= M V2^2 /2 + 4V2^2
= 8V2^2
P.E = potentail energy at 2 = mgh2
E2 = total energy at 2
= K2 +P.E
= 8 V2^2 +mgh2
C) h2 = 32
P.E=8*9.8*32 =2508.8 J
E2 = total energy at 2
= K2 +P.E
=8V2^2 +2508.8 J
from conservation of energy :: E2 =E1
8 V2^2 +2508.8 = 6507.2
V2 = 22.35 m/s
d) W2 = V2/R = 22.35/0.08 =279.45 rad/s
e) K3 = kinetic energy at point 3
=I*W2^2/2 + mV3^2/2
= 1952.3 + 4V3^2
P.E = potentail energy at 3 =0
E3 = total energy at 3 =1952.3 + 4V3^2
From conseravtion of energy ::: E3 = E1
1952.3 + 4V3^2 = 6507.2
V3 = 33.74 m/s
V3 = velocity at point 3 = 33.74 m/s
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