- #1
sachin
- 60
- 7
- Homework Statement
- while solving a problem on rotational dynamics for ap physics, m finding in the question given below,
I am unable to predict the form of movement of the bottom point as on the whole the ladder is not showing any particular rotational motion,
the question is A uniform ladder of length L is supported by a person holding end A, so the ladder makes an angle θ with the horizontal. There is no friction between the ladder and the floor. When the ladder is dropped it falls down because of gravity. Which of the following correctly describes the movement of end B as the ladder falls?
(A) It stays at rest.
(B) It moves a distance Lsinθ to the right.
(C) It moves a distance 𝐿/2 (1 − cosθ) to the right.
(D) It moves a distance 𝐿/2 (1 − sinθ) to the left.
(E) It moves a distance Lsinθ to the left.
this is question no 35 in the below link https://www.slideshare.net/slideshow/apphyscmrotationaldynamicsmultiplechoiceslides20170116-1pptx/260308461
the answer given is option (C) i.e It moves a distance 𝐿/2 (1 − cosθ) to the right.
I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere, will request if any solution can be given, thanks.
- Relevant Equations
- torque about the point of contact of the ladder on the ground = mg x l/2
I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere
