The chain is released from rest with a length b of overhanging links just sufficient to initiate motion. The coefficients of static and kinetic friction between the links and the horizontal surface have essentially the same value, μ. Determine the velocity v of the chain when the last link leaves the edge. Neglect any friction at the corner.
{gL(1+u)}^(1/2)
To determine the velocity of the chain (v) when the last link leaves the edge, we can analyze the forces acting on the chain.
When the chain is just about to move, the static frictional force (fs) between the chain and the surface provides the driving force. Once the chain starts moving, the kinetic frictional force (fk) acts on the chain.
Let's break down the analysis into two parts:
1. Initiation of motion: In this stage, the static frictional force (fs) provides the driving force to initiate motion. The force required to initiate motion is given by:
fs = μ * (weight of the chain)
The weight of the chain can be calculated using the mass of the chain (m) and acceleration due to gravity (g):
Weight = m * g
2. Once motion starts: After the chain starts moving, the kinetic frictional force (fk) acts on it. The force of kinetic friction is given by:
fk = μ * (weight of the chain)
At this stage, the driving force (fs) is no longer required to keep the chain moving, and it is equal to the force of kinetic friction (fk):
fs = fk
Now, we can write equations to determine the velocity of the chain (v) when the last link leaves the edge. Let v1 be the velocity just before the last link leaves the edge, and v2 be the velocity when the last link leaves the edge. We assume that the length of each link is negligible compared to 'b'.
Using the equations of motion:
v1^2 = 2 * a * s
where a is the acceleration and s is the distance traveled.
For the initiation of motion:
fs * b = m * a
fs = μ * (m * g)
Substituting fs = μ * (m * g) and solving for acceleration (a):
a = μ * g
Using this value of acceleration, we can calculate the distance traveled (s) during initiation of motion:
s = (b - 0.5 * v1^2) / a
For the stage after motion starts, we assume uniform acceleration. Using the equation of motion:
s = (v2^2 - v1^2) / (2 * a)
Substituting the values of acceleration (a) and distance traveled (s) obtained above, we can solve for v2:
(v2^2 - v1^2) / (2 * μ * g) = (b - 0.5 * v1^2) / (μ * g)
Simplifying and solving for v2:
v2^2 - v1^2 = 2 * (b - 0.5 * v1^2)
v2^2 = 2b - v1^2
v2 = √(2b - v1^2)
Therefore, the velocity (v) of the chain when the last link leaves the edge is given by √(2b - v1^2).
To determine the velocity of the chain when the last link leaves the edge, we can use the principle of conservation of energy.
The energy of the system initially is in the form of potential energy due to the overhanging links. As the chain falls, it converts this potential energy into kinetic energy.
Here are the steps to solve the problem:
1. Identify the potential energy and kinetic energy involved:
- Potential energy: The potential energy is due to the overhanging links and is given by the formula PE = mgh, where m is the mass of the chain, g is the acceleration due to gravity, and h is the height of the overhang.
- Kinetic energy: The kinetic energy is given by the formula KE = (1/2)mv^2, where m is the mass of the chain, and v is the velocity of the chain.
2. Equate potential energy to kinetic energy:
Set the initial potential energy (PE) equal to the final kinetic energy (KE) because energy is conserved: PE = KE.
mgh = (1/2)mv^2
Simplify the equation by canceling out the mass (m) on both sides:
gh = (1/2)v^2
3. Solve for velocity (v):
Divide both sides by (1/2) to isolate v^2:
v^2 = 2gh
Take the square root of both sides to solve for v:
v = sqrt(2gh)
where sqrt denotes square root.
4. Substitute the value of h:
The problem states that the length b of the overhanging links is just sufficient to initiate motion. Therefore, the height of the overhang is equal to b, h = b.
v = sqrt(2gb)
Substitute the given values for g (acceleration due to gravity) and b (length of overhang) to find the velocity v of the chain.
Note: Since the problem neglects any friction at the corner, the velocity calculated will be the ideal velocity neglecting any losses due to friction. In reality, friction may affect the velocity of the chain.