Question

The rod is massless.
At t = 0 rod is pointing up.
It is of length R. rod is rotating without friction
around the lower end which is hinged to rotation axis.
The mass M is attached at the top end of the rod.
The mass is pushed at t=0 with initial velocity .
g is gravitational acceleration.
All these parameters are given and known.

Answer 1

WHEN WILL THE MASS REACH THE LOWEST POINT ?

Answer 2

@mukushla @UnkleRhaukus @experimentex @estudier

Answer 3

@ParthKohli @radar @RaphaelFilgueiras

Answer 4

@sauravshakya

Answer 5

@Hero @apoorvk @AccessDenied

Answer 6

find the angular velocity

Answer 7

Well it is not constant and it is not explicit in energy - the linear v is there

Answer 8

@ganeshie8 @ajprincess

Answer 9

@ganeshie8 @ajprincess

Answer 10

@ParthKohli @mathslover

Answer 11

\[0.5mv^2 + mgh = E_0 = constant\]

Answer 12

CAN ANYONE SOLVE IT AROUND HERE ?

Answer 13

@Omniscience ?

Answer 14

HINTS ?

Answer 15

Please solve it someone

Answer 16

At least some SIGNIFICAN part of the way forward ?

Answer 17

@radar you can help ?

Answer 18

Hmm I guess you will just have to find the general term for velocity at any instant, and integrate the term in 'dt' - since energy conservation and angular velocity fundae won't work here.

Answer 19

let me try this out...

Answer 20

You ARE 20% forward

Answer 21

Hi @ajprincess

Answer 22

help

Answer 23

Ok , the prize for the sover will be 25 minute-worth field integral from EM that I am ready to do in return !

Answer 24

OK the beauty of this situation is that it is VERY common and simple setting. Buuut very unusual answer

Answer 25

@demitris @Directrix can you help ?

Answer 26

Hi @sauravshakya

Answer 27

@demitris @Directrix can you help ?

Answer 28

Hi @Mikael ..

Answer 29

I AM VERY SORRY I STARTED WITHOUT YOU.
Complete please

Answer 30

Wait a minute.... Let me see what I can do.

Answer 31

@Directrix ?

Answer 32

@asnaseer Please help

Answer 33

Hello @Callisto - try to help please

Answer 34

Well imagine my sour expression when a nice timid student asks me this question in the middle of a nice sunny day ....

Answer 35

Why no rotational energy term?

Answer 36

Yea , you can use that. Although this is not the only way. Just do it please

Answer 37

This requires physics too

Answer 38

why not to use law of conservation of energy?

Answer 39

Calculus and mechanics are essentially one united universal soul

Answer 40

Good @mathslover USE it and do the soln. please

Answer 41

potential energy + kinetic energy = constant.

I cn help after 1 n half hours

Answer 42

Actually I have already pointed that in a hint above. But go forward - when is the bottom reached ?

Answer 43

@ganeshie8 @UnkleRhaukus help pls

Answer 44

\[\frac{ t ^{2} }{2 } + \frac{ V _{o}}{ R } t =\int\limits_{ }^{ }\int\limits_{ }^{ } \frac {-R }{ g \cos \theta } \]
0..pi

Answer 45

A) What is the integration variable ?
B) Why (suddenly ) double int ????
c) What is the reason you write this so categorically ?

Answer 46

A) g, of course
B) equation of motion
C) integral is tough, not bothering

Answer 47

theta is the angle the rod makes with the vertical 0...pi

Answer 48

g CANNOT CHANGE - it is CONSTANT !

Answer 49

it's theta, why would you ask?

Answer 50

would it be g or R?

Answer 51

Please Show the FULL argument. Double integral seems, on the face of it, out of place in this problem

Answer 52

PLEASE SOLVE @mukushla @TuringTest

Answer 53

argument from torque = -mgRcos(theta)
I=mR^2
d^2/ dt^2 (theta) =- gRcos(theta)

Answer 54

\[E_0 = 2mgR +0.5 mv^2\]

Answer 55

Giving the solution soon in 3-4 minutes

Answer 56

E = CONSTANT

Answer 57

h for you is like the y axis ?

Answer 58

yes

Answer 59

OK here goes the solution

Answer 60

I have concluded this problem may be above my pay-grade.

Answer 61

\[t(h) = \int\limits_{h=2R}^{h=0} \frac{dl}{v(h)} dh\]

Answer 62

Now \[dl = \frac{dh}{\cos \theta}\]

Answer 63

Forgot to say - thta is the angle with the horizon of the rod.The relation dh to dl
is a geometric fact that follows from\[h = Sin \theta ====> dh = dl * Cos \theta\]

Answer 64

sorry h = R*Sin theta

Answer 65

Thus gathering all the above observations until now we obtain the following integral

Answer 66

\[\int\limits_{h=2R}^{h=0} dh \frac{1}{\sqrt{1-\frac{h^2}{R^2}}*\sqrt{\frac{2E_0}{m} -2gh}}\]

Answer 67

Here the 1-st square root in the denominator is simply 1/Cos theta

Answer 68

The second square root is the velocity as a function of h - of course derived from the law of conservation of energy stated above

Answer 69

can you please explain how h = rsintheta
means dh = dl costheta

Answer 70

Now by rescaling the distance by R and taking one convenient constant (from the denominator roots) out of the integral we simplify it all to

Answer 71

I did this