A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m= 100 g and length l= 15 cm. Use g=10 m/s2 for the gravitational acceleration, and the ruler has a uniform mass distribution. Note that there is no friction whatsoever in this problem.
The angle is from the vertical.
(a) What is the angular speed of the ruler ω when it is at an angle θ=30∘? (in radians/sec)
m = 0.1 Kg
L = .15 m
moment of inertia of ruler about center = (1/12) m L^2 = 1.88*10^-4 Kg m^2
------------------------------
Method 1
decrease in potential energy = increase in kinetic energy
Height of cg = .075 cos T
height of cg at start = .075 meters
height of cg at 30 deg =.075 cos 30 =.065 meters
loss of PE = m g (.065)
= .1 * 9.81 * .065 = .0638 Joules
x = distance out from wall of cg
y = height of cg
x = .05 sin T
y = .075 cos T
dx/dt = hor velocity = .075 cos T dT/dt
dy/dt = ver velocity =-.075 sin T dT/dt
angular velocity = dT/dt
so at T =30 deg
dx/dt = .065 dT/dt
dy/dt = .0375 dT/dt
total liner velocity = sqrt(dx/dt^2+dy/dt^2)
= dT/dt(.075)
Linear Ke = (1/2) mv^2 = .05(.00563) (dT/dt)^2
= 2.82*10^-4 (dT/dt)^2
Rotational Ke = (1/2) I (dT/dt)^2
= .94 *10^-4 (dT/dt)^2
Total Ke = 3.76*10^-4 (dT/dt)^2 Joules
so
.0638 Joules = 3.76*10^-4 (dT/dt)^2
so
dT/dt = 13 radians/second
Check arithmetic
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 = ah
so
Fx = .1 ah
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
Now solve for dT/dt :)
loss of PE = m g (.065)
= .1 * 9.81 * (.075-.065) = .00981 Joules
then go way down to use that
.00981 Joules = 3.76*10^-4 (dT/dt)^2
so
dT/dt = 5.1 radians/second
Thankyou very much !!
How did you do parts b and c?
for part (c) can theta be found out by equating fx = 0?
Yes
@Damon, could you please explain how you found your linear KE?
I want to do it by myself (and don't even understand the above) , can someone point me at the right chapter/lecture?
I do not see any parts b and c. You only typed part a. I do not have your text or problem sheet or whatever :)
for linear Ke we need (1/2) m v^2
but v = vx i + vy j
so magnitude of v is sqrt (vx^2+vy^2)
Hey Damon,
Part b and c:
(b) What is the force exerted by the wall on the ruler when it is at an angle θ=30∘? Express your answer as the x component Fx and the y component Fy (in Newton)
Fx=
Fy=
(c) At what angle θ0 will the falling ruler lose contact with the wall? (0≤θ0≤90∘; in degrees) [hint: the ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes.]
θ0=
Hi Damon how did you find x=distance out from cg=0.05sinT? I have different numbers in my problem and I did not understand how to find x. Thanks
apply conservation of energy:
U=m*g*hcm
EK= 1/2*I*w^2, I=1/3*m*L^2
Eini= mg(L/2) + 0
Efin= mg(L/2)cos30 + 1/2*I*w^2
solve for Eini=Efin -> w=
any one knows how to answer the other questions?
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 :-).
@fima
Did your conservation of E method worked for part a ?
You used moment of inertia about center of the ruler. But in your calcs you assume that it is rotating about its end. Shouldn't you use instead moment of inertia bout the end, when calculating KE of rotation?
@Damon
You used moment of inertia about center of the ruler. But in your calcs you assume that it is rotating about its end. Shouldn't you use instead moment of inertia bout the end, when calculating KE of rotation?
hey damon, could you solve the full question?
hey Mum, i did as u said but i got only the angular velocity,others are wrong. this is how i got my alpha: Torque=m*g*(l/2)=I_CM*alpha. ie taking torque from origin. am i correct? if yes then what might be the problem.
the equation is Ok, and the w value got green checked. So YES, I'm sure about A.
But anyone knows how to do B and C. And please, try to post the procedure, because we all have different data.
i also tried using I=I_CM+(m*l^2)/4,still my answers were not correct.
hello Damon, i really appreciate that you are sharing your knowledge with us here. please can you solve for part b and c. I followed your procedure. I got my alpha by taking torque at the Origin both using I=I_CM and I=I_CM+I_P. They were all wrong. please do solve for F_x and F_y, thank you
To vivipop: You are missing the sin of angle (theta) in your torque eq. I_CM is (1/3)*M*l*l
I got c) part, the angle is 48.18 for all values
Can someone please tell me the solution for b) part?
has anyone found F_x?
bw did u find f_y?
BW please tell me solution for the 1st problem, my beta is 0.8 and mass is 3kg
to solve the geometrical equations, I need alpha, can I calculate alpha like this:
torque at bottom:
mgl/2*sin(theta)=Ic*alpha, Ic=1/3*3*l^2, though should I add a "-"???
@Phy,
a) Integrate beta*t^2 from 0 to 5... as it is the only force, you shall get someting like: p_f = beta/3 * t_f^3...
b) Its more complicated, need some click on your mind. The block will only start to move when F > Ff... So it won't start to move until beta*t^2 is more then m*g*mu. So you get this: F-Fk = 0, beta*t^2 - m*g*mu=0, t = sqrt(m*g*mu/beta), this will tell you, at what time F its about to surpass Fk, then you need to integrate F=beta*t^2 from that time to 5. In the end, divide it to the mass. And keep this integration fro c).
c) Power = Force * v_f. Force is beta*t_f^2 (you'll only equate for F, not for Ff) and v_f you'll find using the eq above, but from that time to t_f = 4 seconds.
Hope it helps.
just checked and green mark, except for Fy. considering m=0.2, L=0.15, g=10 andDamon eq.
ax = -L/2*sin T w^2 + L/2 cos T alpha
ay = -L/2*cos T w^2 + L/2 cos T alpha
torque=mgL/2 sin T
Icm= 1/3 L^2
alpha=torque.bottom/Icm
with my data alpha=50,w=5.17, ax=2.25, ay=-3.61 so Fx=0.45... and got Fy=2.72 but is not right, I'll be happy if any one can show me were's the error for Fy. :)
ERROR
I used:
ay= -L/2*cos T w^2 - L/2 cos T alpha,
though I'm thinking there might be an error with the signs
@ fima how do you find the torque at the bottom
Didn't get the solution for b) and c) of 1st problem, please help me
@fima hw u got alpha i use your value and i got alpha=40 ! how u got 50??
@Fima
After finding f_x, use that value in the equation Torque=I_cm *alpha equation (containing F_y and F_x), i think you don't need to find a_y.
Also check ur equation for a_y, both are Cos(theta),i think the second one will be sin (theta)
hai anonymous hw @fima got alpha =50 i used his value and got 40 hint plz!
I wrote for mass=200gr L=15cm, and I'm sure about those values, because I got Fx, but I'm having problems with Fy
i tought that if there is no friction the F_y component would be zero, but it's wrong.
Also for the angle I just set the a_x equation equals to zero and find the arctg between two parameters already found on the exercise. I get 61.81° and its also wrong.
After finding f_x, use that value in the equation Torque=I_cm *alpha equation (containing F_y and F_x), i think you don't need to find a_y.
May I ask how to apply to the equation, please?
@Kumar
Trying to "massage" alpha equation, I've got:
3*g*sin(theta)/(2*l)
Wich that for fima values, got alpha = 50, as she said.
Try to evaluate it again.
@KEITANAKO, following your suggestions, I replaced in the torq. eq and got Fy=0.76, though if snyone can confirm that value for MASS=200gr and L=15cm, because I don't waste my last chance
Please tell f_x and f_y for m=150g and l=20 and theta=30 please help !!
@fima: how do you get alpha? if i take your values i do not get alpha=50
Probably 0.76 is wrong.. I have a different equation for Fy but I'm not yet sure if it is correct, but it worked for my values..
Hello guys,
After some massage got an function for f_y, using f_x:
f_y = (2 * I_cm * alpha)/(l*sin(theta)) + f_x * cotangent(theta)
For I_cm used 1/12 * m * l^2, so made it throught the center of mass...
For m=150g and l=20cm I've got 0.958, and an green mark, use those values up here, and check it with care, please don't wast your change by misscalculations. Good luck. (:
what is f_x?
how to calculate alpha ?
thx.. now I have it complete :)
a)
I=1/3*m*L^2
Eini= mg(L/2) + 0
Efin= mg(L/2)cos30 + 1/2*I*w^2
Eini=Efin ->
w=sqrt(3*g (1-cos(theta))/L)
b)
alpha=3*mg/2*sin(theta)/L
ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha
Fx=m*ax
Fy=m*g*cos(theta)-m*w^2*r
c)
cos(theta)=2/3
theta=48.19
but I will not calculate it for you, I have enough for now
What is "r" in the Fy equation? Would it be (L/2)?
bw
Fy=m*g*cos(theta)-m*w^2*r what's r???
i check it r=L/2
yes r=L/2
why is Fx negative?
PLZ WHERE IS PARATHESE I DONT'SEE THIS CLEAR
alpha=3*mg/2*sin(theta)/L
ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha
Fx=m*ax
Fy=m*g*cos(theta)-m*w^2*r
@da: take absolute value
b)
alpha=3*mg/2*sin(theta)/L
ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha
Fx=m*|ax|
Fy=m*g*cos(theta)-m*w^2*L/2
@bw: How did you get alpha from omega in a) ??
tehta alpha=?
Thanks to everyone that contributed to this, now i can rest :)
HAI vivop if u got green tick on this plz help m c
onsider a rocket in space that ejects burned fuel at a speed of vex= 2.0 km/s with respect to the rocket. The rocket burns 11 % of its mass in 250 s (assume the burn rate is constant).
(a) What is the speed v of the rocket after a burn time of 125.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)
v=
incorrect
(b) What is the instantaneous acceleration a of the rocket at time 125.0 s after the start of the engines?(in m/s2)
a=
Thank you bw
Now I'm tired need to sleep.
da can u help m for above question rocket
onsider a rocket in space that ejects burned fuel at a speed of vex= 2.0 km/s with respect to the rocket. The rocket burns 11 % of its mass in 250 s (assume the burn rate is constant).
(a) What is the speed v of the rocket after a burn time of 125.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)
v=
incorrect
(b) What is the instantaneous acceleration a of the rocket at time 125.0 s after the start of the engines?(in m/s2)
a=
much honor
such code
wow
v=113.14
a=0.90
hai #Greco i really appreciate for this rocket question ! i mean it ! thx !
greco can you please solve for my rocket question
Consider a rocket in space that ejects burned fuel at a speed of vex= 1.5 km/s with respect to the rocket. The rocket burns 9 % of its mass in 310 s (assume the burn rate is constant).
(a) What is the speed v of the rocket after a burn time of 155.0 s?