Recent content by physicstime

the initial angular velocity is 0, i knowingly excluded it. ok, i tutor mathematics. while tutoring AP Calc a student posed this problem to me. he is in AP Physics. is there NO way to solve this in a manner a high school student would follow?

ok, so here's what i tried. using W = KEf - KEi = 1/2 I (omega - final)^2 and W = ∫τdθ I can solve for the final angular velocity when the door has closed. I'm just not sure how to get the time it took for it to close from there.

no it won't be constant and it goes to 0 as the door closes.

because the force of the car is on the hinge I'm breaking the force into a component normal to the door and one parallel to the door.

i don't know where to apply the torque. because the car is applying a force at the hinge i tried getting the tangential component there and used the the center of mass as the pivot point. but i can't find a force that makes the center of mass want to go to the right towards the car.

1. A car door is left open at an angle theta with the side of the car. The door is uniform of mass M and length L. The car accelerates with linear acceleration a. How long does it take the door to close. 2. theta = .5(angular acceleration)t^2 angular acceleration = tangential...

but doesn't the tension of the rope on the other side contribute just as much as the man therefore doubling the total force upwards?

Thanks for the response. But doesn't the platform have an equal and opposite normal force?

Can someone explain why the force P with which the man must pull on the rope to achieve an acceleration a m/s2 IS NOT (m+M)(a+g)/2 and is instead (m+M)(a+g). M+m is the combined mass of man and platform. Why does 2T-(M+m)g=(M+m)a not work here?