A 2.62kg stone is tied to a thin, light wire wrapped around the outer edge of th
ID: 1262913 • Letter: A
Question
A 2.62kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 9.14kg cylindrical pulley. The inner diameter of the pulley is 51.1 cm, while the outer diameter is 1.23m. The system is released from rest, and there is no friction at the axle of the pully.
a) What is the rotational inertia of the pulley?
I tried using I=(1/2)(m)(rout^2-rin^2) and also I=(1/2)(m)(rout^2+rin^2) but they were both wrong
b) Find the accleration of the stone.
c) Find the tension in the wire.
d) Find the angular acceleration of the pulley.
A 2.62kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 9.14kg cylindrical pulley. The inner diameter of the pulley is 51.1 cm, while the outer diameter is 1.23m. The system is released from rest, and there is no friction at the axle of the pully. a) What is the rotational inertia of the pulley? I tried using I=(1/2)(m)(rout^2-rin^2) and also I=(1/2)(m)(rout^2+rin^2) but they were both wrong b) Find the accleration of the stone. c) Find the tension in the wire. d) Find the angular acceleration of the pulley.Explanation / Answer
I = m * ( r12 + r22) / 2 )
I = 9.14 * (0.25552 + 0.6152) / 2)
I = 2.03 kg.m2
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a = 2.62 * 9.81 / (2.62 + 2.03/0.6152)
a = 3.22 m/s2
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angular accerlation = a/R = 3.22 / 0.615
= 5.24 rad/s2
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T = mg - ma = 2.62 ( 9.81 - 3.22)
= 17.27 N
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