What is the relationship between friction and velocity?
2 answers:
- Friction is typically unrelated to speed. When the surfaces start to melt or otherwise change because of heat build up or inability to follow the surface then things will change.
- Friction is related to velocity in that it acts to reduce velocity. In other words it causes moving objects to decelerate.
- The friction force may also act as one of the forces in a moment that is causing motion. If the motion is rotary, or following any path, then the friction force may be simply a part of the system of forces that influences motion.
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Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb
Answer:
The final components of velocity of the composite object is 3.33 m/s.
Explanation:
Given;
mass of the first object, m₁ = 3.35 kg
initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction
mass of the second object, m₂ = 1.88 kg
initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction
initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s
initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s
The resultant velocity of the two objects is given by;
R² = 16.415² + 5.8656²
R² = 303.858
R = √303.858
R = 17.432 kgm/s
Apply the principle of conservation of linear momentum for inelastic collision;
total initial momentum before = total final momentum after collision
P₁(x) + P₂(y) = Pf
R = Pf
R = v(m₁ + m₂)
17.432 = v(m₁ + m₂)
where;
v is the final components of velocity of the composite object
Therefore, the final components of velocity of the composite object is 3.33 m/s.
Answer:
(a) Time t = 16.46 sec
(b) Time t =13.466 sec
(c) Deceleration =
Explanation:
(a) As the train starts from rest its initial velocity u = 0 m/sec
Acceleration
Final speed v = 80 km/hr
From first equation of motion v =u+at
So
(b) Now initial speed u = 22.22 m/sec
As finally train comes to rest so final speed v=0 m/sec
Deceleration
So
(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec
Final velocity v = 0 m/sec
Time t = 8.30 sec
So acceleration is given by
As acceleration is negative so it is a deceleration
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
