Suppose that a person closes a car door. The person's fingers are in the door opening, that is located 0.50 m from the hinges. The rotational inertia of the car door is 5.0 kg⋅m2. The door closes in 0.1 s and the rotational speed of the door is 1.4 rad/s just before it closes

Determine the average force that the car door exerts on the person's fingers

To determine the average force that the car door exerts on the person's fingers, we can use the principle of conservation of angular momentum.

The initial angular momentum of the car door can be calculated using the equation:

Linitial = Iωinitial

where Linitial is the initial angular momentum, I is the rotational inertia of the car door, and ωinitial is the initial rotational speed.

Substituting the given values:

Linitial = 5.0 kg⋅m^2 * 1.4 rad/s
Linitial = 7.0 kg⋅m^2/s

According to the conservation of angular momentum, the final angular momentum should be equal to the initial angular momentum. The final angular momentum can be calculated as:

Lfinal = Iωfinal

where Lfinal is the final angular momentum, and ωfinal is the final rotational speed.

Since the door closes, the final rotational speed is 0 rad/s. Therefore:

7.0 kg⋅m^2/s = 5.0 kg⋅m^2 * 0 rad/s
7.0 kg⋅m^2/s = 0

This equation shows that there is a discrepancy and violation of the conservation of angular momentum. Therefore, there must be an external torque acting on the door.

The average force exerted by the car door on the person's fingers can be calculated using the equation:

torque = force * distance

where torque is the external torque acting on the door, force is the average force exerted on the person's fingers, and distance is the distance from the hinges to the position where the fingers are trapped.

Substituting the given values:

torque = force * 0.50 m

The external torque acting on the door is responsible for changing its angular momentum. It can be calculated using the equation:

torque = ΔL / Δt

where ΔL is the change in angular momentum and Δt is the time over which the change occurs.

Substituting the values:

ΔL = Lfinal - Linitial
ΔL = 0 - 7.0 kg⋅m^2/s
ΔL = -7.0 kg⋅m^2/s

Δt = 0.1 s

torque = (-7.0 kg⋅m^2/s) / (0.1 s)
torque = -70 N⋅m

Now we can rearrange the torque equation to solve for the average force:

force = torque / distance

force = (-70 N⋅m) / (0.50 m)
force = -140 N

The negative sign indicates that the force is exerted in the opposite direction, towards the person's fingers. Therefore, the average force that the car door exerts on the person's fingers is 140 N in the opposite direction.

To determine the average force that the car door exerts on the person's fingers, we can use the principle of torque, which relates the force applied to an object to its rotational motion. The formula for torque is:

Torque = Inertia × Angular Acceleration

In this case, the angular acceleration can be calculated using the formula:

Angular Acceleration = (Final Angular Speed - Initial Angular Speed) / Time

Given that the final angular speed is 1.4 rad/s, the initial angular speed is 0 rad/s (since the door starts from rest), and the time is 0.1 s, we can calculate the angular acceleration.

Angular Acceleration = (1.4 rad/s - 0 rad/s) / 0.1 s

Next, we can substitute the rotational inertia and the calculated angular acceleration into the torque formula to find the torque exerted by the door.

Torque = 5.0 kg·m^2 × Angular Acceleration

Finally, the torque exerted by the door can be used to calculate the average force on the person's fingers. Since the distance from the door hinges to the point of contact on the person's fingers is given as 0.50 m, we can use the formula:

Torque = Force × Distance

Rearranging the formula, we can solve for the force:

Force = Torque / Distance

Substitute the calculated torque and distance values to find the average force exerted by the door on the person's fingers.

yep, it will hurt.

torque*time=I*w
force*distane*time=I*w
you have distance, time, I, and w
solve for force.
I get on the order of 140 N