physics doubt
My attempt is in the second image. I'm getting -1 m/s but it's none of the options. This question is from Newton's Laws of Motion, which we haven't learnt yet, but we have learnt rod, string and wedge constraints but I wasn't able to use those so I tried this. Thanks.
58 Replies
@Gyro Gearloose
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.Its because it asked total velocity at B it will go in x direction too u only found for y direction @Takt
is it B?
Cuz there should be no relative velocity in the direction of x
therefore, sqrt3 must be its velocity in x
and if your calculations are correct, we have a net velocity of 2m/s
C?
teri bhi calculation sahi hi hai dy/dt ki why're we getting different results for vertical velocity ðŸ˜
bruh
x toh sqrt3 hai pakka
bas y ka dekhte
haan y hi toh different aa rha
let me solve smh
$$x^2 + y^2 = l^2$$
$$2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0$$
$$ (x/y)\frac{dx}{dt} +\frac{dy}{dt}=0$$
$$ \sqrt{3} \cdot \sqrt{3} = - \frac{dy}{dt}$$
$$v_{y} = 3$$
haan that's 3
Keshav
right
ok so ig your one's right
but i think i know uska kaise galat hai
can you really say dl/dt=0 tho? yaha l length nahi hai na it's the position vector of point b which is changing wrt time
no wait we are wrong
x/y isn't sqrt3
yea it isn't
x/y is 1/ root 3
see the third and second step from last
tera toh galti se correct answer aa gaya
Lmao
mera method diff
idts dl/dt=0 likh sakte hain since yaha l length nahi hai wo position vector of point b hai
wrt time change hora
wahi soch rha mai bhi
still ho sakta i'm wrong, OP jab answer batayega then we'll know
That's the only mistake that can occur
Warna calculation toh sahi hai
yup, btw adv ka paper dekha?
nope
kaha se milega
official website pe hai bhai #📣│announcements check kr
ohk mil gaya
Hi, the answer is B
well fuck
how am i wrong tho
I doubt the answer key @Takt what's the source?
It's from a test we had yesterday, I can show the solution sheet we received
ig I will ask my teacher about it
Hi guys pls checkout my doubt when you are free "basic gif"
The thing is it can't be
constant when you take x or y as a variable cuz l is depending on other "variables"
Kinda contradictory.
If L is a constant, wouldn't x^2 +y^2 = l^2 hold true the same way x^2 + y^2 = 1 does? As a circle on the graph
L not being dependent on x,y but as a constant like 5, so for a value of x there is a specific value for y
Should I mark as solved?
1 is not a vector quantity na
Aren't x and y also not vectors
Just lengths,
But while differentiating we get dx/dt which describes change in magnitude of x in a particular direction so it's a vector?
isn't l also changing in direction wrt time? 💀
@Takt i pointed out a flaw in your soln, you do in mine
Can someone be honest is this useful in chem
Not sure if this is right but this is how it's in my mind
x^2 + y^2 = l^2
All of these are lengths, not vectors
If we differentiate both sides wrt time, the rod mustn't break, for which length l is a constant, it could be anything like 5m or 3cm
So when we differentiate it, the result would be 0 as if the length changes the rod will break
Thus,
2x(dx/dt) + 2y(dy/dt) = 0
We would get a vector if we had something like dl/dt
just point out a flaw in my calc
I don't think there's in mistake in your's, I think the difference in vertical velocity occurs because of writing L, a constant in terms of other sides of the triangle which changes the differentiation, but that doesn't seem possible either as that value of L in terms of other sides would always yield L if we put in the values
think this way, we can write L as Lcostheta i + Lsintheta j, L ka magnitude will be whole root [(Lsintheta)^2+ (Lcostheta)^2], ab while differentiating wrt time, won't you do d(theta)/dt? theta toh change ho hi rha hai wrt time
But wouldn't L^2 (sin^2theta + cos^2 theta) just be L^2 again
(Pythagorean identity)
haan that was L to begin with we just expressed it in terms of theta XD
idk i might be wrong, let's consult tame impala
What's tame Impala?
@Fermat's Last Theorem
Oh
bro summoned the bigger fish
Ehats the question
Please see the original post
Is this alright or should we ping physics helpers fir se? :D
Use constraint
Is the answer 2root(3)? If it is then I think ik what to do
The rod certainly can’t extend or contract, hence B is always at a fixed distance from A at all times. This means that B moves along a circular path around A and that means the velocity of B is always perpendicular to the direction of the rod
We get a diagram that looks like this
Hence B has a velocity making an angle of 30deg with the horizontal axis (say B's velocity is x)
Since B is currently moving with the box, we see that the horizontal velocity of B and the box must be equal so that B touches the box
(if the horizontal velocity of B is slower than the box then B wouldn't stay in touch,
if the horizontal velocity is greater than the box, that means B is exerting some extra force like a push on the box)
Horizontal velocity of B is x cos 30 deg
that equals to sqrt 3
And hence we find x
Thank you so much!
So the main thing was that since the rod is leaning on the box, it's velocity component in the direction of motion of box is same as the velocity of the box.
Sorry for not closing this thread earlier.
+solved @Comrade Rock Astley
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