1. A wheel of radius 82 cm, mass 6 kg, and moment of inertia 2.0172 kg m2 is mounted on a frictionless, horizontal axle as shown. A light cord wrapped around the wheel supports an object of mass 1.5 kg. Find the tension T in the cord. The acceleration
due to gravity is 9.8 m/s2. Answer in units of N.
I have answered this one, and it is: 9.81.
2. What is the angular acceleration of the wheel? Answer in units of rad/s ^2.
I have not answered this one, and I'm really confused.
Physics Question?
2017-02-07 4:13 am
回答 (2)
2017-02-07 11:27 am
✔ 最佳答案
2. torque = T * r = I * α9.81N * 0.82m = 2.0172kg·m² * α
α = 4.0 rad/s² ◄
Hope this helps!
2017-02-07 6:31 am
Let’s use the following equation to determine the acceleration and tension.
Torque = T * r = T * 0.82
Torque = I * α = 2.0172 * α
T * 0.82 = 2.0172 * α
T = 2.46 * α
α = a ÷ r = a ÷ 0.82
T = 2.46 * a ÷ 0.82
T = 3 * a
Weight = 1.5 * 9.8 = 14.7 N
Net force = 14.7 – T
14.7 – T = 1.5 * a
T = 14.7 – 1.5 * a
3 * a = 14.7 – 1.5 * a
4.5 * a = 14.7
a = 14.7 ÷ 4.5
This is approximately 2.27 m/s^2
T = 3 * 14.7 ÷ 4.5 = 9.8 N
To determine the angular acceleration of the wheel, divide the acceleration by the radius of the wheel.
α = (14.7 ÷ 4.5) ÷ 0.82
This is approximately 4 rad/s.
The only force on the wheel and object is the object’s weight.
14.7 = (1.5 + m) * 14.7 ÷ 4.5
66.15 = 22.05 + m * 14.7
m * 14.7 = 44.1
m = 3 kg
In this type of problem, the relative mass of the wheel is one half of its actual mass.
I = 2.0172
Let’s divide this number by r^2
2.0172 ÷ 0.82^2 = 3 kg
In this type of problem, the radius of the wheel does not affect its angular acceleration. If you add 3 kg to 1.5 kg, the answer is the 4.5kg that we used to determine the acceleration of the 1.5 kg object. This simplifies is type of problem. The radius of the wheel cancels out the equations.
Torque = T * r = T * 0.82
Torque = I * α = 2.0172 * α
T * 0.82 = 2.0172 * α
T = 2.46 * α
α = a ÷ r = a ÷ 0.82
T = 2.46 * a ÷ 0.82
T = 3 * a
Weight = 1.5 * 9.8 = 14.7 N
Net force = 14.7 – T
14.7 – T = 1.5 * a
T = 14.7 – 1.5 * a
3 * a = 14.7 – 1.5 * a
4.5 * a = 14.7
a = 14.7 ÷ 4.5
This is approximately 2.27 m/s^2
T = 3 * 14.7 ÷ 4.5 = 9.8 N
To determine the angular acceleration of the wheel, divide the acceleration by the radius of the wheel.
α = (14.7 ÷ 4.5) ÷ 0.82
This is approximately 4 rad/s.
The only force on the wheel and object is the object’s weight.
14.7 = (1.5 + m) * 14.7 ÷ 4.5
66.15 = 22.05 + m * 14.7
m * 14.7 = 44.1
m = 3 kg
In this type of problem, the relative mass of the wheel is one half of its actual mass.
I = 2.0172
Let’s divide this number by r^2
2.0172 ÷ 0.82^2 = 3 kg
In this type of problem, the radius of the wheel does not affect its angular acceleration. If you add 3 kg to 1.5 kg, the answer is the 4.5kg that we used to determine the acceleration of the 1.5 kg object. This simplifies is type of problem. The radius of the wheel cancels out the equations.
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