A solid wood door, 90.0 cm wide by 2.00 m tall has a mass of 35.0 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20.0 m/s and hits the door 60.0 cm from the hinged side. The ball rebounds with a speed of 16.0 m/s along the same line. What is the angular speed of the door just after the collision with the ball?
Well, well, well, we have quite the physics problem here! And guess what? I have just the right answer for you, with a touch of humor of course.
To find the angular speed of the door after the collision, we need to take a spin into the world of rotations. You see, after the collision, the door starts to rotate like a graceful gymnast on a pommel horse. But instead of pommels, we have hinges!
To solve this, we can use the principle of conservation of angular momentum. Before the collision, the door is at rest, so its initial angular momentum is zilch. But after the collision, it starts to spin, and we need to find its final angular momentum.
The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed. Now, the moment of inertia for a rigid door rotating around one side is I = 1/3 ML², where M is the mass of the door and L is the length of the door.
Plugging in the numbers, we have I = (1/3) x 35.0 kg x (2.00 m)². Let me grab my handy-dandy calculator... *beep beep beep*
Calculating... calculating... and voila! I get I = 93.33 kg·m².
Now, we can rearrange the formula to find ω = L/I. Since we know the final angular momentum is a result of the ball's collision, and the ball bounces with a speed of 16.0 m/s, we can use that to find the final angular momentum. You know, because the ball had a momentum, and then it rebounds... like a heart after a breakup.
So, L final = mvr, where m is the mass of the ball, v is its velocity, and r is the distance from the ball's trajectory to the hinge. Plugging in the numbers, we get L final = (0.5 kg) x (16.0 m/s) x (0.60 m). More calculator action...
And I end up with L final = 4.80 kg·m²/s. Look at that!
Finally, we can find the angular speed by dividing the angular momentum by the moment of inertia. So, ω = L final / I. Let me do the final calculation...
Drum roll, please! *ba dum tss*
After some mathematical magic, I get ω = 4.80 kg·m²/s / 93.33 kg·m². And the result is approximately 0.0515 radians per second. Ta-da!
So, the angular speed of the door just after the collision with the ball is around 0.0515 radians per second. Isn't it delightful to dive into the world of physics and rotations?
To find the angular speed of the door just after the collision with the ball, we need to use the principle of conservation of angular momentum.
The angular momentum formula is given by:
L = Iω
Where:
L is the angular momentum
I is the moment of inertia
ω is the angular speed
1. Calculate the moment of inertia (I) of the door.
The moment of inertia for a rectangular object (door) rotating about an axis perpendicular to its plane through its center is given by:
I = (1/12) * m * (h^2 + w^2)
Where:
m is the mass of the door
h is the height of the door
w is the width of the door
Substituting the given values:
m = 35.0 kg
h = 2.00 m
w = 90.0 cm = 0.90 m
I = (1/12) * 35.0 kg * (2.00^2 + 0.90^2)
2. Calculate the initial angular momentum (L_initial) of the door-ball system before the collision.
The initial angular momentum is given by the product of the moment of inertia and the initial angular speed of the door, which is zero since the door is at rest.
L_initial = I * 0
3. Calculate the final angular momentum (L_final) of the door-ball system after the collision.
The final angular momentum is given by the product of the moment of inertia and the final angular speed of the door.
L_final = I * ω
4. Apply the conservation of angular momentum.
According to the conservation of angular momentum, the initial angular momentum is equal to the final angular momentum:
L_initial = L_final
Since L_initial is zero:
0 = I * ω
5. Solve for the angular speed (ω) of the door just after the collision.
ω = 0
Therefore, the angular speed of the door just after the collision with the ball is zero.
To find the angular speed of the door after the collision with the ball, we can use the principle of conservation of angular momentum.
Angular momentum is defined as the product of moment of inertia and angular velocity. In this case, the moment of inertia of the door is given by:
I = (1/3) * m * r^2
Where:
- m is the mass of the door
- r is the perpendicular distance from the axis of rotation to the point of impact
Given:
- m = 35.0 kg (mass of the door)
- r = 60.0 cm = 0.60 m (distance of impact from the hinged side)
Substituting these values, we can find the moment of inertia of the door.
I = (1/3) * 35.0 kg * (0.60 m)^2
I = 4.20 kg * m^2
Now, let's consider the angular momentum before and after the collision. The angular momentum before the collision is zero since the door is at rest. The angular momentum after the collision is the angular momentum of the door plus the angular momentum of the ball.
Angular momentum before collision = 0
Angular momentum after collision = I * ω
Where:
- ω is the angular speed of the door after the collision
The angular momentum of the ball can be calculated using its linear momentum. The linear momentum of the ball before the collision is:
p = m_ball * v_ball
Where:
- m_ball is the mass of the ball
- v_ball is the velocity of the ball before the collision
Given:
- m_ball = 500 g = 0.500 kg (mass of the ball)
- v_ball = 20.0 m/s (speed of the ball before the collision)
Substituting these values, we can find the linear momentum (p) of the ball.
p = 0.500 kg * 20.0 m/s
p = 10.0 kg * m/s
The angular momentum of the ball is then given by:
L_ball = r * p
L_ball = 0.60 m * 10.0 kg * m/s
L_ball = 6.00 kg * m^2/s
Now, we can equate the angular momentum before and after the collision to find the angular speed of the door.
0 = I * ω + L_ball
Substituting the values we calculated:
0 = 4.20 kg * m^2 * ω + 6.00 kg * m^2/s
To solve for ω, we rearrange the equation:
ω = - (6.00 kg * m^2/s) / (4.20 kg * m^2)
ω = -1.43 rad/s
The negative sign indicates that the angular speed is in the opposite direction to the initial motion.
Therefore, the angular speed of the door just after the collision with the ball is approximately -1.43 rad/s.