A 15 kg, 1 m 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)?
1.68?
no. wrong answer
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.
wrong answer Elena
0.43 s
To find how long it takes for the door to fully open, we need to analyze the forces acting on the door and determine the torque that causes it to rotate.
Let's break down the problem step by step:
Step 1: Calculate the initial angular velocity of the door.
Since the ball hits the door in the exact middle at a velocity of 35 m/s, it imparts an angular impulse to the door. We can find the initial angular velocity using the principle of conservation of angular momentum.
Angular momentum (L) = Moment of Inertia (I) × Angular Velocity (ω)
The moment of inertia of a rectangular door rotating about its hinges is given by:
I = 1/3 * mass * width^2
Plugging in the values:
I = 1/3 * 15 kg * (1 m)^2 = 5 kg·m²
The initial angular momentum can be calculated as:
L_initial = I * ω_initial
Since the ball hits the exact middle of the door, it creates equal and opposite initial angular momenta on both sides of the door. So, we can consider each side separately.
L_initial = -L_initial
Therefore:
(I * ω_initial) = -(I * ω_initial)
Simplifying gives:
ω_initial = 0 rad/s
So, the initial angular velocity of the door is 0 rad/s since it is initially at rest.
Step 2: Calculate the torque acting on the door due to the ball.
When the ball hits the door, it exerts a torque that causes it to rotate. The torque (τ) can be calculated using the following equation:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Since the ball imparts an angular impulse to the door, the change in angular momentum (ΔL) is given by:
ΔL = I * Δω
Since the door starts from rest initially:
Δω = ω_final - ω_initial
Δω = ω_final - 0
Δω = ω_final
Therefore:
ΔL = I * Δω = I * (ω_final - ω_initial)
Since the moment of inertia remains constant and ω_initial is 0 rad/s, the equation simplifies to:
ΔL = I * ω_final = J
J denotes the angular impulse exerted by the ball.
Step 3: Calculate the angular impulse.
The angular impulse can be calculated using the equation:
J = F * d * sin(θ)
where J is the angular impulse, F is the force, d is the distance between the point of impact and the axis of rotation, and θ is the angle between F and d.
Since the door is hit in the exact middle, θ is 90 degrees, and sin(90) = 1.
So, the equation becomes:
J = F * d
The force exerted by the ball on the door can be calculated using Newton's second law:
F = m * a
Since the ball bounces off elastically, the change in momentum of the ball is given by:
Δp = 2 * m * v
Hence, the force exerted by the ball is:
F = Δp / Δt
where Δt is the time of impact.
Since the ball is in contact with the door for a very short duration, we can approximate that Δt is very small, approaching 0.
Therefore, F = Δp / Δt ≈ ∞
This means that the force exerted by the ball is extremely large in a tiny fraction of time. Because the force exerted is infinite in a mathematical sense, we cannot accurately calculate the exact torque or time required for the door to fully open.
In reality, the door will swing open slowly due to the finite nature of forces and other factors like air resistance, but the exact timing would require a more detailed analysis of the forces acting on the door, friction, and other variables.
Therefore, we cannot determine the exact time in seconds it takes for the door to fully open in the given scenario.