2. A uniform long, thin rod has a mass of M = 20.0 kg and a length of L = 1.50 m
ID: 2029464 • Letter: 2
Question
2. A uniform long, thin rod has a mass of M = 20.0 kg and a length of L = 1.50 m. It is pinned so that it can pivot in the horizontal plane about an axis that is one-third of the way from the end of the rod, as shown in the figure below. Three forces, all with a magnitude of F = 75.0 N, act on the rod. Force F~ 1 points at an angle of = 30.0 relative to the rod and acts at a distance 1 3 L from the other end of the rod. Forces F~ 2 and F~ 3 act perpendicular to the rod, with F~ 2 acting at the center of the rod and F~ 3 acting at the end of the rod. What is the angular acceleration of the rod?
Ignore question 1.
PIIYS 298 Introductory Mechanics, Ileat, and Sound Spring 2018 Homework 7 Due: March 27th at 11:59 pm 1. Two masses (rn,-8.00 kg and 7722 = 10.0 kg) are connected by a massless string which runs across a frictionless pulley with a mass of 12.0 kg and a radius of 5.00 cm. The heavier mass is also connected to a spring, which is attached to the ground and has a spring constant of k = 500 N/ui. Both masses are initially placed at the same height. When the masses are released, m2 begins to fall and compresses the spring. What is the speed of both blocks when the heavier one has fallen a distance of 3.50 crn? L2 2. A unifg o has a mass of M 20.0 kg and a length of I1.50 . is pined so that it can pivot in the horizontal plane about an axis that is one third of the way from the end of the rod as shown in the figure below. Three forces, al with a magnitude of F 75.0 N, act onte ro. Force F1 points at an angle of 30.0° relative to the rod and acts at a distance L from the other end of the rod. Forces F and F3 act perpendicular to the rod, with F2 acting at the center of the rod and F3 acting at the end of the rod. What is the angular acceleration of the rod? F1Explanation / Answer
moment of inertia of rod about the pivot I = (1/12)*M*L^2 + M*(L/2 - L/3)^2
moment of inertia of rod about the pivot I = (1/12)*M*L^2 + M*(L/6)^2
moment of inertia of rod about the pivot I = (1/12)*M*L^2 + (1/36)*M*L^2
moment of inertia of rod about the pivot I = (1/9)*M*L^2
torque = F*l
l = lever arm = perpendicular distance from pivot to the line of action of force
net torque T = +F1*L/3*sintheta - F1*(L/2-L/3) - F3*L/3
F1 = F2 = F3 = F = 75
net torque T = F*L/3*sin30 - F*L/6 - F*L/3
net torque T = FL/3
from newtons second law net torque = I*alpha
FL/3 = (1/9)*ML^2*alpha
alpha = 3F/(ML) = 3*75/(20*1.5) = 7.5 rad/s^2
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