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
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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
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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=