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# Physics Problem!

Find the relation b/w v1 and v2 in terms of v1,v2, and theta

Note by Ajitesh Mishra
4 years ago

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Let $$L$$ be the length of the string and $$H$$ be the height difference between the pulleys.

Let $$x$$ be the length of the horizontal piece of string and $$y$$ be the length of the vertical piece of string.

The slanted piece of string has length $$L-x-y = \sqrt{x^2+H^2}$$. Thus, $$(L-x-y)^2 = x^2+H^2$$.

Differentiation yields: $$2(L-x-y)\left(-\dfrac{dx}{dt}-\dfrac{dy}{dt}\right) = 2x \dfrac{dx}{dt}$$.

By definition, $$\dfrac{dx}{dt} = -v_2$$ and $$\dfrac{dy}{dt} = v_1$$. Using trigonometry $$x = H\tan \theta$$ and $$L-x-y = H\sec \theta$$.

Therefore, $$2H\sec\theta \cdot (v_2 - v_1) = 2H\tan\theta \cdot (-v_2)$$. Solving yields $$v_1 = v_2(1+\sin \theta)$$. · 3 years, 12 months ago

Awesome Solution Dude...!!! · 3 years, 12 months ago

Wow! Good job Jimmy! · 3 years, 12 months ago

using the virtual work method..-T(1+sin(thetha))x2 + Tx1=0. Differentiating this constraint equation we get the relation v2(1+sin(thetha))=v1. (x2 is wedge's displacement and x1 is the block's displacement) · 3 years, 12 months ago

Another way to derive the relation is by using the fact that work done by tension is zero (by newton's third law). We get $$\sum$$T.s=0 . Differentiating we get, $$\sum$$T.v=0 · 3 years, 12 months ago

constraint relations · 3 years, 12 months ago

xdx=ldL sinθ=dL/dx sinθ+1=(dL+dx)/dx sinθ+1=V1/V2 · 3 years, 12 months ago

concept behind the problem is use of constraint motion · 3 years, 12 months ago

v1/(1+sinθ)=v2 I think so!!! · 3 years, 12 months ago

$$v_2(1+\sin\theta)=v_1\sin\theta$$ · 3 years, 12 months ago

is the velocity v1 with respect to wedge or with respect to ground? · 3 years, 12 months ago

won't it also be in terms of the masses of the two objects? (determining their acceleration from the tension in the thread and gravity) · 3 years, 12 months ago

nope ..masses aren't needed.. · 3 years, 12 months ago

Does $$\theta$$ stay constant? My guess is no, but I can't tell for certain from the diagram. · 3 years, 12 months ago

Yes , $$\theta$$ will definitely not remain the same . $$\theta$$ would gradually decrease . · 3 years, 12 months ago

use the concept that length on the thread remains constant. :) · 3 years, 12 months ago

i think $$v_1$$=$$v_2$$$$cos\theta$$ Not pretty sure but still..i think.. · 3 years, 12 months ago

no .. it isn't the right ans · 3 years, 12 months ago

My guess... it's "v1 = v2 + v2*sin(theta)". Is that right? · 3 years, 12 months ago

is it a guess or something else?? · 3 years, 12 months ago

More than a guess. I have used logic but might be wrong. · 3 years, 12 months ago