Simple Mechanics, relative motion example

AI Thread Summary
The discussion focuses on calculating the angular velocity and acceleration of a bar (OA) at t=1s, given its motion on a sliding prism. The prism's horizontal position is defined by the equation S(t)=4*t*t+5*t+1[cm], leading to a calculated horizontal velocity Ve(1s)=13cm/s. The angular velocity is derived as ω=Va/OA, resulting in a value of 1.3s-1, which differs from the example answer of 1.0s-1. For angular acceleration, the calculations yield Ae(1s)=8cm/s² and an angular acceleration α=0.176s-2, again differing from the example's 0.19s-1. The discussion emphasizes the importance of vector resolution in analyzing the motion of point A relative to the prism.
twowheelsbg
Messages
25
Reaction score
0

Homework Statement


Might be the sketch most useful first:
ex.3.3.jpg

The prism is sliding horizontally from left to right under condition : S(t)=4*t*t+5*t+1[cm]
and the bar OA is rotating, point A is sliding over the prism's slope and point O is the immobilized center of bar's rotation, OA=10cm.

What would be the angular velocity and acceleration of the bar at the moment t=1s,
if at that moment the bar is declined at β=60deg from the vertical?

Homework Equations


Va=ωxOA
Aa=αxOA+ωxVa

Idea is the point A is once transferred by the prism's horizontal motion to the left ( Ve and Ae ),
and twice transferred over the prism's slope in relative motion ( Vr and Ar ).
It's absolute motion ( Va and Aa ) comes also from the circular path with radius OA.

Va=Ve+Vr
Aa=Ae+Ar

The Attempt at a Solution


Ve(t)=dS(t)/dt, so Ve(1s)=13cm/s
point A is following a circle so Va is 60deg above Ox ( τ direction ),
otherwise Vr is parallel to the prism's slope ( η direction ).
This way vector solving of Va=Ve+Vr results in Va=13cm/s, Vr=13cm/s.
Va is perpendicular to OA so ω=Va/OA=1.3s-1
( this is my first problem, examples booklet answer is 1.0s-1 )

Ae(t)=dVe(t)/dt, so Ae(1s)=8cm/s.s ,
Ar is parallel to the prism's slope ( η direction ),
Aa has accelerating part αxOA ( τ direction ) and centripetal part ωxωxOA ( η direction ).
Solving Aa=Ae+Ar in vectors results in α=0.176s-2
( this is my second problem, examples booklet answer is 0.19s-1 )

I could attach my Mathcad sheet if needed.
(My results for Aa=17cm/s.s , Ar=11.5cm/s.s )
 
Last edited:
Physics news on Phys.org
twowheelsbg said:

Homework Statement


Idea is the point A is once transferred by the prism's horizontal motion to the RIGHT ( Ve and Ae ),
and twice transferred over the prism's slope in relative motion ( Vr and Ar ).
It's absolute motion ( Va and Aa ) comes also from the circular path with radius OA.

Later edit, sorry for the typinng bug.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top