The Journey of the Solitary Rod
A
uniform
rod of mass
M
and length
L
is released from the position as shown. A
slight
disturbance causes the rod to start its journey to the floor.
When the rod makes an angle θ with the horizontal , find the angular speed of the rod (in s r a d ), to the nearest integer .
Details and Assumptions:
∙
The floor is smooth and frictionless.
∙
M
=
1
0
K
g
∙
L
=
2
m
∙
g
=
9
.
8
m
/
s
2
∙
θ
=
3
0
o
The answer is 3.
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5 solutions
I about one end of the rod is ml^3/3 not ml^3)/12.
We need two things conservation of energy and the fact that the horizontal velocity of the centre of mass is zero. There is no horizontal force on the rod. We can get a relation between the angular velocity of the rod about the axis passing through the centre of mass and the velocity of centre of mass of the rod at any time. I made a mistake. Moment of inertial I took a wrong value. Science is unforgiving. Even though we know the concept we are wrong unless we get the correct solution.
We can simply apply Energy conservation principle In general for rod problem as- Change in PE= Change in KE mgl/2 - mg(l/2) \sin \theta = (Iw^{2})/2 , where I is moment of Inertia about one end of rod which is (mgl^{2})/3 and w is angular velocity and angle from horizontal. The answer will be \sqrt{15} =2.7386..
A uniform rod of mass M and length L is released from the vertical
position. A slight disturbance causes the rod to start its journey to the
floor. When the rod makes an angle θ with the horizontal, find the
angular speed of the rod (in rad/s), to the nearest integer.
Details and Assumptions:
1.The floor is smooth and frictionless.
-
M = 10Kg;
-
L = 2m;
-
g = 9.8m/s^2
-
θ = 30°
Solution:
The physical system is conservative, so we can apply the law of
conservation of energy:
Eo = Ep + Ec; (1)
Eo = MgL/2, (initial energy, all potential);
Ep = Mg(L/2)sinθ, (potential energy);
Ec = (1/2)Jcω^2 + (1/2)Mv(C)^2; (kinetic energy), where:
Jc := inertia momentum of the rod respect to the center of mass C,
C is the midpoint between A, B the extremal point of the rod
(A at the bottom);
Jc = ML^2/12; for a uniform rod;
v(C) := velocity of the center of mass;
the equation for Ec is note as Konig Theorem;
Let (O, x, y, z) a system of cartesian coordinates with:
y // initial position of the rod;
x ⊥ y in O at the bottom of the rod on the plane x, z that is the floor;
A ≡ O, at time t = 0; A ≡ O' at time t; O'(x', y, 0);
the position vector of C is:
C - O = C - O' + O' - O;
C - O = [(L/2)cosθ + x']ix + (L/2)sinθiy;
v(C) = (d/dt)(C - O) = (L/2)cosθ(dθ/dt)iy, because:
Fw = - Mg(iy); weight force is applied on C,
(Fw)x = 0; so v(C)x = constant = 0;
x(C) = (L/2)cosθ + x' = constant, =>
x(C) = x(O) = 0;
v(O')x = dx'/dt = - (L/2)sinθ(dθ/dt) = - (L/2)ωsinθ;
v(O')y = 0;
dθ/dt = ω;
v(C)y = (L/2)ωcosθ; v(C)x = 0;
substituting in (1), we obtain:
Mg(L/2) = Mg(L/2)sinθ + (1/2)Jcω^2 + (1/2)Mv(C)^2;
Mg(L/2) = Mg(L/2)sinθ + (1/2)(ML^2/12)ω^2 +
- (1/2)(ML^2/4)(ω^2)(cosθ)^2;
Solving, we obtain:
ω = √{(12g/L)(1 - sinθ)/(1 + 3(cosθ)^2)};
substituting the values, we obtain:
ω = √(12 x 9.8/13) = 3.007682471..
ω ≈ 3 rad/s (ANSWER).
The main challenge is to find a relation between velocity of center of mass(which is vertically down at all instants) and the angular velocity of the rod.
When the rod makes an angle theta with the horizontal, the center of mass has moved down by a distance of
0 . 5 L × ( 1 − sin θ )
Differentiate this with respect to time to get,
v=- 0 . 5 L × ω × cos θ
Now, just use conservation of energy on the rod+earth system to obtain
ω =3
Since the only forces acting on the rod are vertical, the center of mass must move in a straight line.
Ans is coming to be 3.6 something.. then nearest integer should be 4
This is the first time I have used LaTeX all I can see on my screen is the code and not the actual equation. If you are also seeing only the code, I am sorry! I dont know if I need to have some font or something else in my computer for this to work, Please advise.
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Do not forget to enclose what you type in brackets as follows:
\ (....\ )
without the spaces. Hope that helps : )
What is the exact answer you got?
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3.0076824 Which is approximately 3.
I am not sure about my solution so tell me if i am wrong. I used newton's laws of uniform acceleration to calculate the linear velocity of the top of the rod. When the rod is at 3 0 ∘ the top of the rod will have moved a displacement equals the chord of the arc. So S = 2 ∗ 2 ∗ sin 3 0 = 2 m . And V 2 = u 2 + 2 g s , u = 0 then, V = 2 ∗ 9 . 8 ∗ 2 = 6 . 2 6 m . / s e c . And ω = r V So ω = 3 . 1 3 r a d . / s e c . = 3 r a d . / s e c . to the nearest integer.
Not a complete solution, however, some hints:
1) Find the position of the instantaneous axis of rotation:
The bottom most point has a horizontal velocity, while the centre of mass has a vertical velocity. It is easy to see that when the angle with the horizontal is θ , the point about which these velocities are perpendicular, is located at the coordinates ( − 0 . 5 l cos θ , 0 . 5 l sin θ ) , taking the origin as the point where the centre of mass will finally fall. This point lies on the instantaneous axis of rotation (perpendicular to the plane of the motion of the rod)
2) About this axis, the motion is purely rotational. Conserve energy and get the answer. 3
mgh=1/2 Iw^2 or, mgl/2(1-sinthita)= 1/2(ml^2/12 +ml^2/4 cos^2 thita)w^2 solving this we get, w=root(12g(1-sinthita)/l(1+3cos^2thita)). w=3.007 which is nearly equal to 3. rad per sec. Here we have to consider the moment of inertia obt ICOR!