One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.48 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed v0, so that the rod begins to rotate upward about the pivot. What must be the value of v0, such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?
I'm going to take a shot and guess they're trying to equate angular KE with gravitational PE.
SO:
½ I ω^2 = mgh
Where you need to look up I for a rod spun about one end, ω = v/r, and h = r.
Pretty sure the m’s will cancel
To solve this problem, we can use the principle of conservation of angular momentum.
The angular momentum of an object is given by the equation:
L = Iω
Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity
In this case, the initial angular momentum is zero because the rod is not moving initially. When the rod reaches the straight-up orientation, its angular momentum will also be zero. Therefore, the initial and final angular momenta are equal.
To find the initial angular velocity, we can use the equation:
L = Iω
Since the moment of inertia of a uniform rod rotating about its end is given by:
I = (1/3) * m * L^2
Where:
m = Mass of the rod
L = Length of the rod
We can rearrange the equation to solve for initial angular velocity:
ω = L / (I)
Substituting the moment of inertia of the rod into the equation:
ω = L / ((1/3) * m * L^2)
We can cancel out the length of the rod:
ω = 1 / ((1/3) * m * L)
Now, we know that angular velocity is related to linear velocity by the equation:
ω = v / r
Where:
v = Linear velocity
r = Distance from the pivot point to the point where the linear velocity is applied (in this case, the length of the rod)
Substituting the equation for angular velocity into the equation for linear velocity:
v = ω * r
v = (1 / ((1/3) * m * L)) * L
Simplifying the equation:
v = 3 * m * L
Now, we need to find the mass of the rod. Since it is uniform, we can use the equation:
m = ρ * V
Where:
ρ = Density of the rod
V = Volume of the rod
The volume of the rod can be calculated as:
V = A * L
Where:
A = Cross-sectional area of the rod
Since the rod is thin, we can assume it has a negligible cross-sectional area. Therefore, the cross-sectional area is very close to zero, and the volume is also very close to zero. Thus, the mass of the rod is approximately zero.
Substituting the mass into the equation for linear velocity:
v = 3 * 0 * L
Since the mass of the rod is zero, the linear velocity required for the rod to come to a momentary halt in a straight-up orientation is also zero.
Therefore, v0 = 0.
To solve this problem, we can use the principle of conservation of angular momentum. Angular momentum is defined as the product of moment of inertia and angular velocity. In this case, the moment of inertia of the rod can be calculated using the formula for the moment of inertia of a uniform rod rotating about one end:
I = (1/3) * m * L^2,
where I is the moment of inertia, m is the mass of the rod, and L is the length of the rod.
Since the rod is hanging vertically downward, the initial angular momentum is zero because it is not rotating initially. When the rod comes to a momentary halt in a straight-up orientation, the final angular momentum is also zero. This means that the change in angular momentum is equal to zero.
The change in angular momentum can be calculated using the formula:
ΔL = L_final - L_initial
Since L_initial is zero, the equation simplifies to:
ΔL = L_final
The final angular momentum can be calculated as the product of the final moment of inertia and the final angular velocity (which we are trying to find):
ΔL = I * ω_final
Substituting the expression for the moment of inertia and rearranging the equation, we get:
ΔL = (1/3) * m * L^2 * ω_final
Since ΔL = 0, the equation becomes:
0 = (1/3) * m * L^2 * ω_final
Simplifying the equation further, we get:
0 = ω_final
This implies that the final angular velocity must be zero. Therefore, to find the value of v0 such that the rod comes to a momentary halt in a straight-up orientation, we need to find the linear speed v0 that corresponds to a final angular velocity of zero.
The linear speed v0 corresponds to the initial angular velocity through the formula:
v0 = ω_initial * L,
where ω_initial is the initial angular velocity.
Since the rod begins to rotate upward, the initial angular velocity is positive. Combining the equation for v0 and the fact the final angular velocity is zero, we get:
v0 = 0 * L
v0 = 0.
Therefore, the value of v0 must be zero for the rod to come to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation.