A 15 kg, 1m wide door which has frictionless hinges is closed but unlocked. A 400 g ball hits the exact middle of the door at a velocity of 35 m/s and bounces off elastically, thereby causing the door to slowly swing open. How long in seconds does it take for the door to fully open (rotate 90 degrees)?
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1.68?
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The angular momentum of the door ‘p’= the change in amgular momentum of the ball Δp.
In vector form:
Δp⃗ =p₂⃗-p₁⃗ =mv₂⃗-mv₁⃗,
for magnitudes
Δp =p₂-(-p₁) =mv₂+mv₁= 2mv.
p=Mu =>
Mu=2mv
The speed of the center of the door is
u=2mv/M=2•0.4•35/15 = 1.37 m/s
Its angular speed is
ω=u/r=2u/L =2. •1.87/1 = 3.73 rad/s
t= φ/ ω =π/3.73 = 0.84 s.
please write right answer
To find the time it takes for the door to fully open, we need to consider the torque acting on the door and apply the principles of rotational motion.
The torque equation is given by:
τ = I * α
where:
τ is the torque,
I is the moment of inertia,
α is the angular acceleration.
The moment of inertia for a thin door can be calculated using the formula:
I = (1/3) * m * (L^2)
where:
m is the mass of the door,
L is the length of the door.
In this case, the door is 1 meter wide, so the length is equal to 1 meter.
Given:
mass of the door (m) = 15 kg
length of the door (L) = 1 m
Using the moment of inertia formula, we can find the moment of inertia:
I = (1/3) * 15 kg * (1 m)^2
I = 5 kg * m^2
Now, we need to calculate the torque acting on the door. The torque is caused by the impulse provided by the ball hitting the door.
The torque equation can be rewritten as:
τ = r * F
where:
r is the distance from the axis of rotation to the point where the force is applied,
F is the force.
Since the force acts at the exact middle of the door, the distance (r) would be half the width of the door, which is 0.5 meters.
Given:
distance (r) = 0.5 m
mass of the ball (m_ball) = 0.4 kg
velocity of the ball (v) = 35 m/s
We can calculate the force using the impulse equation:
F * Δt = m_ball * Δv
where:
Δt is the time of impact,
Δv is the change in velocity.
Since the ball hits the door and bounces back elastically, the change in velocity is twice the initial velocity of the ball:
Δv = 2 * v
Plugging in the values, we have:
F * Δt = 0.4 kg * (2 * 35 m/s)
Simplifying,
F * Δt = 28 kg * m/s
Now, we can find the torque (τ) using the torque equation and the calculated force:
τ = r * F
τ = 0.5 m * (28 kg * m/s)
τ = 14 N * m
Since τ = I * α, we can equate the two equations and solve for α:
I * α = 14 N * m
Substituting the value of moment of inertia (I), we have:
5 kg * m^2 * α = 14 N * m
Simplifying,
α = 14 N * m / 5 kg * m^2
α = 2.8 N / kg * m
The angular acceleration (α) is given by:
α = ω / t
where:
ω is the angular velocity,
t is the time.
Since we want to find the time (t) it takes for the door to fully rotate, we rearrange the equation:
t = ω / α
We know that the angular velocity (ω) for a door rotating 90 degrees can be calculated using the formula:
ω = θ / t
where:
θ is the angle in radians,
t is the time.
Since the door rotates 90 degrees, θ = (π/2) radians.
Substituting the value of angular velocity (ω) in the equation for time (t):
t = (π/2) radians / (ω / α)
t = (π/2) radians * (α / ω)
Now we can substitute the values of α and ω:
t = (π/2) radians * (2.8 N / kg * m) / (θ / t)
Simplifying,
t^2 = (π * 2.8 N / kg * m * t) / 2
Multiplying both sides by 2 to get rid of the denominator:
2t^2 = π * 2.8 N / kg * m * t
Dividing both sides by t:
2t = π * 2.8 N / kg * m
Simplifying,
t = (π * 2.8 N / kg * m) / 2
t = π * 2.8 N / (2 * kg * m)
Now, we can substitute the values:
t ≈ 3.50 seconds
Therefore, it takes approximately 3.50 seconds for the door to fully open (rotate 90 degrees) given the given conditions.