IIT JEE , AIEEE Forum - Physics , Mechanics

chimanshu_007

Friction is required to roll a disc in a particular direction cos to rotate a disc , we require torque wich can be provided by tangential force......

as force of friction is the only tangential force in this case...so its required

force of friction after perfect rolling begins will be zero

siddharthsaxena

i've got a simliar one ................

Q. Would days become longer or shorter in near future.???

Ans - Due to global warming, ice at the pole, will melt into water, and thus the moment of inertia (I) also increases..

But by conservation of angular momentum,

I₁W₁=I₂W₂

therefore W, the angular velcity decreases, and days will become longer.............

Hope u find it helpful.................

tka

here is a question on rotational mechanics

there is a cylinder on a smooth inclined plane, at an angle 'w' wrt horizontal. mass of cylinder = m, radius of cylinder = r. it is initially at a height h on the incline, calculate its translational velocity at the end of the incline.

you would be thinkin, the potential energy of the cylinder [ mg (r+h) ] would be converted into rotational and translational kinetic energy at the end of the incline.

NO. it will be converted only into translational k.e. for rotational motion, there must be a torque. there is no question of torque when the inclined plane is smooth

so the answer is, v = (2gh + 2gr)^ (1/2)

chimanshu_007

For a rolling body of same mass and radius (vaise this is a common 1)

V_sphere > V_disc > V_shell > V_ring

and acceleration also follows the same trend (naturally yaar)

but the time taken will be in order

t_sphere < t_disc < t_shell < t_ring

--> If a particle is moving on a circular path with an angular speed omega about the centre of the circle ,

its angular speed about a point on the opposite side of diameter passing thru it is (omega/2)

tka

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sorry for the bad diagram

Q. a ball is projected at an angle @ to the horizontal, between two walls 'd' distant from each other, with a velocity v0. the height of the walls is to be assumed to extend to infinity. assuming that the ball collides elastically with the walls, find the total time it takes for the ball to come down.

A.

the time that the ball takes to come down does not depend on horizontal component of velocity, only on the vertical component of it.

upon collision, the walls only may change the horizontal component of velocity. so actually this is nothing but a simple projectile motion sum;

t = 2usin@/g

siddharthsaxena

Take one more dude................

Q.Even a sphere rolling on a FRICTIONLESS ground slows down, why???????????

Ans - Because there is a slight deformation in the contact surface (and hence it is no more a point). this shifts the centre of mass and hence the normal force.

The torque due to this normal force opposes the rolling and sphere stops .

Hope it helps

tka

hcv question: why does a fat man fall over while bending?

A. for a fat man, the center of mass is above his waist. so if he bends, it is horizontally displaced as compared to his feet.

the normal reaction does not shift its line of action. the man's weight, due to changed position of center of mass, acts along another line.

generally, the normal reaction and weight act along same line so he doesn't fall over. however if they act along different lines, this sets up a torque that makes him fall frontwards

chimanshu_007

SOME IMPORTANT RELATIONS....

1st the value of vel at bottom

for -:

a circular ring or cyl.shell , vel at bottom = (gs sin@)^1/2

a circular disc or solid cyl , vel at bottom = (1.33 gs sin@)^1/2

solid sphere , vel at bottom = [ (10/7) gs sin@ ] ^1/2

spherical shell , vel at bottom = [ (6/5) gs cos@ ]^1/2

2nd acc at bottom

for

circular ring or cyl shell , its = (1/2) gsin@

circular disc or solid cyl , its = (2/3) gsin@

solid sphere , its = (5/7) gsin@

spherical shell , its (3.5) gsin@

3rd time taken to reach at bottom....

for

circular ring or cyl shell , its { (4s) / gsin@ }^1/2

circular disc or solid cyl , its { (3s) / gsin@ }^1/2

solid sphere , its { (14s) / 5gsin@ }^1/2

spherical shell , its { (10s) / 3gsin@ }^1/2

hope this helps.....

waterdemon

Here are some more important things to "Remember" :

1)When a body is projected vertically upwards with Velocity
having the magnitude equal to the magnitude of Acceleration
due to Gravity then time taken to reach max Height is 1sec
and max height reached is g/2.

2)The Ratio of Distances travelled in 1st,2nd,3rd second
of freely falling body is 1 : 3 : 5.

3)The ratio of maxim. height reached by different bodies
projected upwards with Velocities U₁,U₂,U₃ is equal to
U₁² : U₂² : U₃² .

4)For a body projected upwards,the distance covered by
the body in the last second of its Upward journey remains
4.9m irrespective of Velocity of Projection.

Hope you find it useful.
Cheers!!!!!!!!!!!!!!!

anandghegde

Tension in the string

Guess whats the tension in the string????

anandghegde

Ans: 49 N

kislay

@siddarth
Due to global warming, ice at the pole, will melt into water
and will flow into sea . thus mass from the axis of rotation will get away from it,therefore moment of inertia sud increase
I=MR^{2
BECAUSE MASS IS CONSTANT BUT R WILL INCREASE???}
and hance the day and night sud be longer????

PLS crct if i am wrong,...........

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