A 0.20 m uniform bar has a mass of 0.75 kg and is released from rest in the vertical position, as the drawing shows. The spring is initially unstrained and has a spring constant of 25.0 N/m. Find the tangential speed with which end A strikes the horizontal surface.
To find the tangential speed with which end A strikes the horizontal surface, we need to use the principle of conservation of mechanical energy. The initial potential energy at the top is converted into kinetic energy at the bottom.
1. The initial potential energy can be calculated using the formula: PE = mgh, where m is the mass of the bar, g is the acceleration due to gravity, and h is the height of the bar.
- m = 0.75 kg (given)
- g = 9.8 m/s² (acceleration due to gravity)
- h = 0.20 m (height of the bar)
PE = (0.75 kg) * (9.8 m/s²) * (0.20 m) = 1.47 J
2. The final kinetic energy can be calculated using the formula: KE = (1/2)mv², where m is the mass of the bar, and v is the tangential speed.
KE = (1/2) * (0.75 kg) * v² = 0.375v²
3. The potential energy at the bottom is zero since the spring is initially unstrained.
4. According to the principle of conservation of mechanical energy, the initial potential energy is equal to the final kinetic energy.
PE = KE
1.47 J = 0.375v²
5. Solve for v²:
v² = (1.47 J) / (0.375)
v² = 3.92 m²/s²
6. Take the square root of both sides to find v:
v = √(3.92 m²/s²)
v ≈ 1.98 m/s
So, the tangential speed with which end A strikes the horizontal surface is approximately 1.98 m/s.
To find the tangential speed with which end A strikes the horizontal surface, we need to apply the principles of conservation of energy.
First, let's consider the potential energy of the system. When the bar is released from rest in the vertical position, it has gravitational potential energy, which will be converted into kinetic energy as it falls.
The gravitational potential energy (U_g) of the bar can be calculated using the formula:
U_g = m * g * h
Where:
m = mass of the bar
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height from which the bar is released
Given:
m = 0.75 kg
h = length of the bar = 0.20 m
U_g = 0.75 kg * 9.8 m/s^2 * 0.20 m
U_g = 1.47 J
Now, since the spring is initially unstrained, all of the gravitational potential energy will be converted into the kinetic energy of the bar (ignoring any loss due to friction).
The kinetic energy (K) of the bar can be calculated using the formula:
K = (1/2) * I * ω^2
Where:
I = moment of inertia of the bar about the pivot point
ω = angular velocity of the bar
For a uniform bar rotating about an axis through its center of mass perpendicular to its length, the moment of inertia (I) can be calculated using the formula:
I = (1/3) * m * L^2
Where:
L = length of the bar
Given:
m = 0.75 kg
L = 0.20 m
I = (1/3) * 0.75 kg * (0.20 m)^2
I = 0.02 kg·m^2
Now, we can equate the gravitational potential energy (U_g) to the kinetic energy (K) and solve for the angular velocity (ω):
1.47 J = (1/2) * 0.02 kg·m^2 * ω^2
2.94 J = 0.02 kg·m^2 * ω^2
ω^2 = (2.94 J) / (0.02 kg·m^2)
ω^2 = 147 rad^2/s^2
ω ≈ √(147 rad^2/s^2)
ω ≈ 12.12 rad/s
Now, to find the tangential speed (v) with which end A strikes the horizontal surface, we can use the formula:
v = ω * r
Where:
r = distance from the pivot point to end A.
Given:
r = 0.20 m
v = 12.12 rad/s * 0.20 m
v ≈ 2.42 m/s
Therefore, the tangential speed with which end A strikes the horizontal surface is approximately 2.42 m/s.