E-StoreHome » Ask & Discuss » Physics. » Mechanics « Back to Discussion
Mechanics
23 Mar 2008 00:49:51 IST
0 People liked this
10
723
Comments (10)
BALGANESH
Blazing goIITian
Joined: 14 Nov 2007
Posts: 309
23 Mar 2008 01:35:22 IST
Like
1 people liked this
by bhuvankar
we get a range of ws due to the fact that frictn can act both in the upwards and the downwords dirns..........
first let us assume that fn is acting in the upwards dirn...........then we have for vertical eqbm...
Ncos@ + KNsin$=mg.......(K=coeff of fn)
from here we have N=mg/(ksin$+cos$).............1)
then for horizontal eqbm.........we have,(seeing in pseudo frame we apply an outward centrifugal force.......)
kNcos$= Nsin$ + mw2 Rsin$..........
substituting for N from 1) we get the expn that.......
g(sin$-Kcos$)/(ksin$+cos$) = w2 Rsin$
on simplifying which we get the expn for w as..........
sqrt(g(sin$-Kcos$)/((Rsin$)(ksin$+cos$)))...............
similarly on reversing the dirn of frn.ie in dwnwrd dirn.......we will get our result as sqrt(g(sin$+Kcos$)/((Rsin$)(ksin$+cos$))).................
Reply
= v^2/rg = 
= 0.58 = 1/root3 23 Mar 2008 01:46:29 IST
Like
1 people liked this
by amangupta
25. at highest point the normal acc. is g, it can be considered a circle with centirpetal acc=g
thus v^2/r=g
or r=V^2/g and v at highest point is gcosx.
thus answer is (vcos theta)^2/g.
and 27(c) the normal reaction is mv^2/r
and frictional force is the only force acting in tangentail direction thus acc=uN/m
=uv^2/r and since it is opposing motion, the negative sign is added
thus v^2/r=g
or r=V^2/g and v at highest point is gcosx.
thus answer is (vcos theta)^2/g.
and 27(c) the normal reaction is mv^2/r
and frictional force is the only force acting in tangentail direction thus acc=uN/m
=uv^2/r and since it is opposing motion, the negative sign is added
23 Mar 2008 01:50:03 IST
Like
1 people liked this
So we have the inst. acc. as -
v2 /r
And we all know chain rule !!
So we can write V=dS/dt=dS/dV*dV/dt ---------------------------(1)
by madmax , refer earlier post
v2 /r And we all know chain rule !!
So we can write V=dS/dt=dS/dV*dV/dt ---------------------------(1)
by madmax , refer earlier post
Rearranging we get the eqn they have given which is dv/dt=v(dV/dS)
I'll just continue with eqn (1)
Take dv on the other side ,substitute dV/dt=
v2/r and rearrange the eqn to get
(1/V)dV=(-/r)dS
The question is for one revolution so integrate S from 0 to 2
r and dV from Vo to V........where Vo is init vel
I'll just continue with eqn (1)
Take dv on the other side ,substitute dV/dt=
v2/r and rearrange the eqn to get(1/V)dV=(-
/r)dSThe question is for one revolution so integrate S from 0 to 2
r and dV from Vo to V........where Vo is init vel
23 Mar 2008 12:25:10 IST
Like
1 people liked this
26) Q is taken as theeta {not a method 2 solve the problem
easier typing)
easier typing) let v be the velocity at the desired point
horizontal component remains unchanged------->v cos Q/2 = u cos Q
v= u cos Q/ cos Q/2..(1)
radial acceleration is the component of acc perpendicular to velocity
acceleration radial = g cos Q/2
v^2 /r = g cos Q/2....................................(2)
substituting v from .......(1) in (2)
r =( u^2 cos^2 Q) / g cos^3 Q/2
Free Sign Up!
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
Topics
9395
20091
3672
2186
7617
2827
2638
Mathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
1553 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011