E-StoreHome » Ask & Discuss » Physics. » Mechanics « Back to Discussion
Mechanics
5 Mar 2008 12:54:05 IST
0 People liked this
14
748
Can somebody plz answer these H.C verma Questions
None
Circular motion : -
Exercise :- Q 18 , 19
Work energy and power :-
Exercise :-Q 58
Center Of mass :-
Exercise :-Q 6 , 24 , 25 , 52 , 57
Rotational :-
Short Answer :- Q 17
Objective I - Q 2 , 9
Exercise :- Q 44, 45. 49, 57, 68, 69
Gravitation :-
Exercise :- Q 9 , 34,
SHM:-
Objective II :- Q 9 , 12
Exercise :- Q 19, 26 , 31, ,39, 51,
Comments (14)
aNdRoMeDa
Blazing goIITian
Joined: 10 Sep 2007
Posts: 1319
5 Mar 2008 13:02:08 IST
Like
1 people liked this
circular motion
18thr=20m
v=10m/s
us=.4
tan@=v^2/rg
tan2=1/2
Vmax=(gr(sin@+us*cos@)/cos@-us*sin@))^1/2
substitutiin in this equation u'll get d ansewr as 54km/hrf or velocity maximum
for mimnimum
Vmin=(gr(sin@-us*cos@)/cos@+us*sin@)^1/2
substitutin in thsi equation u'll get d ansewr as14.7km/hr
hope u finda useful :)
Reply
5 Mar 2008 13:07:33 IST
Like
1 people liked this
for cemtre of mass 6th
draw d diagram or tz same as fig(9-Q2)
takin d center of disk as origin
Xcom=summation of mixi/summation of mi
=m1x1+m2x2/m1+m2
m1=mass per unit area*area
=sigma*pi*d^2/4......sigma =mass per unit area....(for d disk)
m2=sigma*d^2.....for d squareplate
nw x1=0
x2=d
substittuin this in d equation 4 centre of mass i.e
m1x1+m2x2/m1+m2
u'll get d answer
hope u finda useful :)
7 Mar 2008 08:32:08 IST
Like
1 people liked this
45)taking t=theta
N2 + N1 cos t = mg ------1
N1 sin t = f=
N2 ------2
balancing torque abt the bottommost pt--
N1 cos t * h/tan t + N1 sin t * h =mg *L/2 cos t
or
N1=mgLcos t / 2[h cos^2 t/sin t + sin t]
=mgL cos t sin t / 2h
=N1 sin t/ N2 from 2
=N1 sin t / mg - N1 cos t from 1
=L cos t sin ^2 / [2h - L cos ^2 t sin t]
N2 + N1 cos t = mg ------1
N1 sin t = f=
N2 ------2balancing torque abt the bottommost pt--
N1 cos t * h/tan t + N1 sin t * h =mg *L/2 cos t
or
N1=mgLcos t / 2[h cos^2 t/sin t + sin t]
=mgL cos t sin t / 2h
=N1 sin t/ N2 from 2=N1 sin t / mg - N1 cos t from 1
=L cos t sin ^2 / [2h - L cos ^2 t sin t]
7 Mar 2008 08:46:48 IST
Like
1 people liked this
68)m=0.1 kg
M=0.24 kg
I = moment of inertia of m and M when stuck together
=Ml^2/3 + ml^2=[M+3m]/3 as l=1m
conserving angular momentum before and after collision--
m root(2g) *1 = I w=[M+3m]/3 w
or
w=3m root(2g) / (M + 3m)
=15 root(20) / 17 on putting values
then we can conserve energy and put h=1- cos t to get the answer
M=0.24 kg
I = moment of inertia of m and M when stuck together
=Ml^2/3 + ml^2=[M+3m]/3 as l=1m
conserving angular momentum before and after collision--
m root(2g) *1 = I w=[M+3m]/3 w
or
w=3m root(2g) / (M + 3m)
=15 root(20) / 17 on putting values
then we can conserve energy and put h=1- cos t to get the answer
7 Mar 2008 08:52:01 IST
Like
1 people liked this
short answer 17)
torque of the two weight abt middle pt is not zero
however net torque is zero because it is equal and in opposite directions,
still the arm rotates and finally becomes horizontal because of the inertia of motion of the pans.
Objective--
2)
A would be in the Z plane while B would be in x,y plane
hence angle between A and B would be 90 , thus their dot product would be zero
9)the horizontal component of the force would be equal to centripetal force= mw^2r=mw^2L/2
torque of the two weight abt middle pt is not zero
however net torque is zero because it is equal and in opposite directions,
still the arm rotates and finally becomes horizontal because of the inertia of motion of the pans.
Objective--
2)
A would be in the Z plane while B would be in x,y plane
hence angle between A and B would be 90 , thus their dot product would be zero
9)the horizontal component of the force would be equal to centripetal force= mw^2r=mw^2L/2
7 Mar 2008 09:14:56 IST
Like
1 people liked this
Gravitation
9)
let us consider an element of length dx and a distance x from the centre
dm=M/L dx
dE=G dm / (d^2 + x^2)
total dE = integ 2 dE sin t
since horizontal components would cancel each other
=integ ( 2 G dm / (d^2 + x^2) * d/ root(d^2 + x^2) )
= integ (2GM d dx / L(d^2 + x2)^1.5)
integrating the above from 0 to L/2 would give the answer
9)
let us consider an element of length dx and a distance x from the centre
dm=M/L dx
dE=G dm / (d^2 + x^2)
total dE = integ 2 dE sin t
since horizontal components would cancel each other
=integ ( 2 G dm / (d^2 + x^2) * d/ root(d^2 + x^2) )
= integ (2GM d dx / L(d^2 + x2)^1.5)
integrating the above from 0 to L/2 would give the answer
7 Mar 2008 13:05:52 IST
Like
0 people liked this
rotational motion
objective I
2)
IN THIS CASE BY CONSIDERING INERTIAL FRAME OF REFRENCE i.e no tangential acceleration about that axis of rotation. Thus net force will be towards centre of rotation. so, unit vector B will be perpendicular to unit vector A which is along the axis of rotation. Therefore dot product between A & B will be 0 as cos90 is 0.
objective I
2)
IN THIS CASE BY CONSIDERING INERTIAL FRAME OF REFRENCE i.e no tangential acceleration about that axis of rotation. Thus net force will be towards centre of rotation. so, unit vector B will be perpendicular to unit vector A which is along the axis of rotation. Therefore dot product between A & B will be 0 as cos90 is 0.
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 |
1316 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011