A point mass of 0.100 kg (size greatly exaggerated in the diagram) is attached t
ID: 2244631 • Letter: A
Question
A point mass of 0.100 kg (size greatly exaggerated in the diagram) is attached to the end of a meter stick of mass 0.200 kg. The whole system is in the horizontal plane. Three forces, making constant angles with the stick, are applied to it. The 1.00 N force is applied to the middle of the meter stick at a 37o angle with respect to the stick. I = (1/3) ML2 for the meter stick.
a. What is the total moment of inertia?
b. What is the torques due to the: i. 3.00 N Force
ii. 2.00 N Force
iii. 1.00 N Force
c. What is the total torque?
d. Find the angular acceleration.
e. If the mass were not a point mass but rather a small sphere of radius 0.050 m with center 1.050 m from the axis, what would be the new angular acceleration?
Please help!
A point mass of 0.100 kg (size greatly exaggerated in the diagram) is attached to the end of a meter stick of mass 0.200 kg. The whole system is in the horizontal plane. Three forces, making constant angles with the stick, are applied to it. The 1.00 N force is applied to the middle of the meter stick at a 37o angle with respect to the stick. I = (1/3) ML2 for the meter stick. What is the total moment of inertia? What is the torques due to the: i. 3.00 N Force 2.00 N Force 1.00 N Force What is the total torque? Find the angular acceleration. If the mass were not a point mass but rather a small sphere of radius 0.050 m with center 1.050 m from the axis, what would be the new angular acceleration?Explanation / Answer
a) I = 0.2*1^2/3 + 0.1*1^2
= 0.1667 kg.m^2
b)
i) T1 = 0.5*1*sin(37) = 0.3009 N.m
ii)T2 = 0
iii) T3 = 3*1*sin(90) = 3 N.m
c) T(applied) = T3 - T1 = 2.699 N.m
torque due grvity
Tg = 0.5*0.2*9.8 + 1*0.1*9.8
Tg = 1.96 N.m
T(total = 2.699 + 1.96 = 4.569 N.m
d)
alfa = T/I = 4.569/0.1667 = 27.95 rad/s^2
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