Physics Pulley with Inertia?
2017-02-06 5:01 am
An Atwoods machine is formed by passing a string over a solid disc pulley having mass 50 grams and radius 2.5 cm. The masses on each end of the string are m1 = 200g and m2 = 300g. Assume friction is negligible on the axle of the pulley and the string doesn't slip. Find the angular acceleration of the pulley and the linear acceleration of m1.
回答 (2)
2017-02-06 5:23 am
✔ 最佳答案
I = ½M*r² = 15.625E-6 kg∙m²a = (m2 - m1)*g/[m1+m2+I/r²] = 1.867 m/s² (m1 & m2)
α = a/r = 74.67 rad/sec²
2017-02-06 7:14 am
Let’s determine the weights.
w1 = 0.2 * 9.8 = 1.96 N
w2 = 0.3 * 9.8 = 2.94 N
The net force on the system is equal to the difference of these two weights.
Net force = 2.94 – 1.96 = 0.98 N
In this type of problem, the relative mass of a solid disc pulley is one half of its actual mass.
Total mass = 0.2 + 0.3 + 0.025 = 0.525 kg
0.98 = 0.525 * a
a = 0.98 ÷ 0.525
This is approximately 1.8667 m/s^2. To determine the angular acceleration of the pulley, divide this number by the radius of the pulley.
α = (0.98 ÷ 0.525) ÷ 0.025 = 74⅔ rad/s^2
w1 = 0.2 * 9.8 = 1.96 N
w2 = 0.3 * 9.8 = 2.94 N
The net force on the system is equal to the difference of these two weights.
Net force = 2.94 – 1.96 = 0.98 N
In this type of problem, the relative mass of a solid disc pulley is one half of its actual mass.
Total mass = 0.2 + 0.3 + 0.025 = 0.525 kg
0.98 = 0.525 * a
a = 0.98 ÷ 0.525
This is approximately 1.8667 m/s^2. To determine the angular acceleration of the pulley, divide this number by the radius of the pulley.
α = (0.98 ÷ 0.525) ÷ 0.025 = 74⅔ rad/s^2
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