8.01xers Can anybody help with the falling ruler. I submission left(
Hey, look there is a limit to the number of times that I am going to respond to the same 8.01 question.
If you are taking 8.01 you are capable of reading, at least we could in 1955. Scroll and stop spamming.
except that no one is helping on the last two parts.
Look at what Mum posted below my first response:
Classical Mechanics Physics - Urgent help please - Mum, Monday, December 9, 2013 at 2:52am
I don't want you to cheat so I am giving some hints - not full solution. As for part b - Damon gave you everything in his geometrical approach. Look at the equations for forces - you need x and y (Damon calls it x and z) component of acceleration a of the center of mass. You have the equations for them given by Damon in his geometrical appoach. In these equations you have the angle - you know it, then you have the angular velocity - you just calculated it, at last you have there angular acceleratin - get that either from the torque equation (taking pivot point as origin gets rid of the contact force). Then you can solve. Or if you have a nice equation for omega from part a), take a derivative to get the acceleration.
As for part c): F_x=0 therefore a_x=0 - play with the equations for a_x and a_y and omega from part a).
Sorry for my grammar :-).
Note, when you use the pivot point at the floor for torques be sure to use (1/3)m L^2 instead of (1/12)m L^2 as I did for the origin at the CG.
but how do you find alpha?
alpha is d^2T/dt^2
quoting Mum:
".....Then you can solve. Or if you have a nice equation for omega from part a), take a derivative to get the acceleration. ....."
this is what i have.
dT/dt=sqrt(3g*(1-cos(30))/L)
dt^2/dt=sqrt(3)*g*sin(30)/2L*sqrt(10*(1-cos(30))/L)
however i do not know how to continue.
I know you gave some equations but they are not yielding answers, if you could pls help out?
F = m A
Fz(.075)sin T -Fx(.075)cos T = 1.88*10^-4 alpha
.981 - Fz = .1 az
Fx = .1 ax
ax =
-.075 sin T (dT/dt)^2 + .075 cos T alpha
az =
-.075 cosT (dT/dt)^2 -.075 sin T alpha
how did you get -0.075?
I am taking moments about the cg.
The length of the rod is .15
so half of it is .075
==========================
Harder way
---------------------------
Force up from floor = Fz
Force down from gravity = m g = .1*9.81 = .981 Newtons
so
.981 - Fz = .1 az
where az is acceleration of CG down
Force horizontal out from wall = Fx
acceleration of cg out from wall = ax
so
Fx = .1 ax
Torques about cg = I alpha
Fz(.075)sin T -Fx(.075)cos T = 1.88*10^-4 alpha
Now geometry, relate angular acceleration to linear acceleration components at cg
x = .075 sin T
dx/dt = .075 cos T (dT/dt)
ax = d^2x/dt^2 =-.075 sin T (dT/dt)^2 + .075 cos T alpha
{{note - because alpha = ang accel = d^2T/dt^2}}
y = .075 cos T
dy/dt = -.075 sin T dT/dt
az = d^2y/dt^2 = -.075 cosT (dT/dt)^2 -.075 sin T alpha